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  • How is printf() implemented in c?

    - by Mask
    Disassembling printf doesn't give much info: (gdb) disas printf Dump of assembler code for function printf: 0x00401b38 <printf+0>: jmp *0x405130 0x00401b3e <printf+6>: nop 0x00401b3f <printf+7>: nop End of assembler dump. How is it implemented under the hood? Why disassembling doesn't help? What does * mean before 0x405130?

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  • How to use Java varargs with the GWT Javascript Native Interface? (aka, "GWT has no printf()")

    - by markerikson
    I'm trying to quickly learn GWT as part of a new project. I found out that GWT doesn't implement Java's String.format() function, so there's no printf()-like functionality. I knew that some printf() implementations exist for Javascript, so I figured I could paste one of those in as a GWT Javascript Native Interface function. I ran into problems, and decided I'd better make sure that the varargs values were being passed in correctly. That's where things got ugly. First, some example code: // From Java, call the JSNI function: test("sourceString", "params1", "params2", "params3"); .... public static native void test(Object... params) /*-{ // PROBLEM: this kills GWT! // alert(params.length); // returns "function" alert(typeof(params)); // returns "[Ljava.lang.Object;@b97ff1" alert(params); }-*/; The GWT docs state that "calling a varargs JavaScript method from Java will result in the callee receiving the arguments in an array". I figured that meant I could at least check params.length, but accessing that throws a JavascriptException wrapped in an UmbrellaException, with no real information. When I do "typeof(params)", it returns "function". As if that weren't odd enough, if I check the string value of params, it returns what appears to be a string version of a Java reference. So, I guess I'm asking a few different questions here: 1) How do GWT/JSNI varargs actually work, and do I need to do something special to pass in values? 2) What is actually going on here? 3) Is there any easier way to get printf()-style formatting in a GWT application?

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  • How can I temporarily redirect printf output to a c-string?

    - by Ben S
    I'm writing an assignment which involves adding some functionality to PostgreSQL on a Solaris box. As part of the assignment, we need to print some information on the client side (i.e.: using elog.) PostgreSQL already has lots of helper methods which print out the required information, however, the helper methods are packed with hundreds of printf calls, and the elog method only works with c-style strings. Is there I way that I could temporarily redirect printf calls to a buffer so I could easily send it over elog to the client? If that's not possible, what would be the simplest way to modify the helper methods to end up with a buffer as output?

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  • How to center string output using printf() and variable width? [java]

    - by Khan
    I'm creating an output in Java using printf() to create table headers. One of the columns needs variable width. Basically it should look like this: //two coords Trial Column Heading 1 (20,30)(30,20) //three coords Trial Column Heading 1 (20,40)(50,10)(90,30) I tried to use: int spacing = numCoords * 7; //size of column printf("Trial %*^s", column, "Column Heading"); But I keep getting output errors when I try to use * or ^ in the conversion statement. Does anyone have any idea what the correct formatting string should be?

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  • printf("... %c ...",'\0') and family - what will happen?

    - by SF.
    How will various functions that take printf format string behave upon encountering the %c format given value of \0/NULL? How should they behave? Is it safe? Is it defined? Is it compiler-specific? e.g. sprintf() - will it crop the result string at the NULL? What length will it return? Will printf() output the whole format string or just up to the new NULL? Will va_args + vsprintf/vprintf be affected somehow? If so, how? Do I risk memory leaks or other problems if I e.g. shoot this NULL at a point in std::string.c_str()? What are the best ways to avoid this caveat (sanitize input?)

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  • How can I line up columns with Perl's printf when the value might be a number or a string?

    - by user317203
    I am trying to output a document that looks like this (more at http://pastebin.com/dpBAY8Sb): 10.1.1.1 100 <unknown> <unknown> <unknown> <unknown> <unknown> 72.12.148.186 94 Canada Hamilton ON 43.250000 -79.833300 0.00 72.68.209.149 24 United States Richmond Hill NY 40.700500 -73.834500 611.32 72.192.33.34 4 United States Rocky Hill CT 41.657800 -72.662700 657.48 I cannot find how to format the output I have to have a floating poing and format the distance between columns. My current code looks something like this. if (defined $longitude){ printf FILE ("%-8s %.6f","",$longitude); }else{ $longitude = "<unknown>"; printf FILE ("%-20s ",$longitude); } The extra "" throws off the whole column and it looks like this (more at http://pastebin.com/kcwHyNwb). 10.1.1.1 100 <unknown> <unknown> <unknown> <unknown> <unknown> 72.12.148.186 94 Canada Hamilton ON 43.250000 -79.833300 0.00 72.68.209.149 24 United States Richmond Hill NY 40.700500 -73.834500 571.06

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  • Wordpress related question about printf

    - by Denny Mueller
    Today i work through a theme to get a better understanding of wordpress and templating. I discovered this <?php printf( __('Designed by %s', 'Anyword'), '<a href="http://www.blub.tld" target="_blank">Blub</a>', '' ); ?> I know it shows "Designed by Blub" (Where Blub is linked) But what does the __ mean or why is there a string insert? What for is the Anyword? Can someone exactly explain this line for me? thanks in advance best regards

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  • C - Error with read() of a file, storage in an array, and printing output properly

    - by ns1
    I am new to C, so I am not exactly sure where my error is. However, I do know that the great portion of the issue lies either in how I am storing the doubles in the d_buffer (double) array or the way I am printing it. Specifically, my output keeps printing extremely large numbers (with around 10-12 digits before the decimal point and a trail of zeros after it. Additionally, this is an adaptation of an older program to allow for double inputs, so I only really added the two if statements (in the "read" for loop and the "printf" for loop) and the d_buffer declaration. I would appreciate any input whatsoever as I have spent several hours on this error. #include <stdio.h> #include <fcntl.h> #include <sys/types.h> #include <unistd.h> #include <string.h> struct DataDescription { char fieldname[30]; char fieldtype; int fieldsize; }; /* ----------------------------------------------- eof(fd): returns 1 if file `fd' is out of data ----------------------------------------------- */ int eof(int fd) { char c; if ( read(fd, &c, 1) != 1 ) return(1); else { lseek(fd, -1, SEEK_CUR); return(0); } } void main() { FILE *fp; /* Used to access meta data */ int fd; /* Used to access user data */ /* ---------------------------------------------------------------- Variables to hold the description of the data - max 10 fields ---------------------------------------------------------------- */ struct DataDescription DataDes[10]; /* Holds data descriptions for upto 10 fields */ int n_fields; /* Actual # fields */ /* ------------------------------------------------------ Variables to hold the data - max 10 fields.... ------------------------------------------------------ */ char c_buffer[10][100]; /* For character data */ int i_buffer[10]; /* For integer data */ double d_buffer[10]; int i, j; int found; printf("Program for searching a mini database:\n"); /* ============================= Read in meta information ============================= */ fp = fopen("db-description", "r"); n_fields = 0; while ( fscanf(fp, "%s %c %d", DataDes[n_fields].fieldname, &DataDes[n_fields].fieldtype, &DataDes[n_fields].fieldsize) > 0 ) n_fields++; /* --- Prints meta information --- */ printf("\nThe database consists of these fields:\n"); for (i = 0; i < n_fields; i++) printf("Index %d: Fieldname `%s',\ttype = %c,\tsize = %d\n", i, DataDes[i].fieldname, DataDes[i].fieldtype, DataDes[i].fieldsize); printf("\n\n"); /* --- Open database file --- */ fd = open("db-data", O_RDONLY); /* --- Print content of the database file --- */ printf("\nThe database content is:\n"); while ( ! eof(fd) ) { /* ------------------ Read next record ------------------ */ for (j = 0; j < n_fields; j++) { if ( DataDes[j].fieldtype == 'I' ) read(fd, &i_buffer[j], DataDes[j].fieldsize); if ( DataDes[j].fieldtype == 'F' ) read(fd, &d_buffer[j], DataDes[j].fieldsize); if ( DataDes[j].fieldtype == 'C' ) read(fd, &c_buffer[j], DataDes[j].fieldsize); } double d; /* ------------------ Print it... ------------------ */ for (j = 0; j < n_fields; j++) { if ( DataDes[j].fieldtype == 'I' ) printf("%d ", i_buffer[j]); if ( DataDes[j].fieldtype == 'F' ) d = d_buffer[j]; printf("%lf ", d); if ( DataDes[j].fieldtype == 'C' ) printf("%s ", c_buffer[j]); } printf("\n"); } printf("\n"); printf("\n"); } Post edits output: 16777216 0.000000 107245694331284094976.000000 107245694331284094976.000000 Pi 33554432 107245694331284094976.000000 2954938175610156848888276006519501238173891974277081114627768841840801736306392481516295906896346039950625609765296207682724801406770458881439696544971142710292689518104183685723154223544599940711614138798312668264956190761622328617992192.000000 2954938175610156848888276006519501238173891974277081114627768841840801736306392481516295906896346039950625609765296207682724801406770458881439696544971142710292689518104183685723154223544599940711614138798312668264956190761622328617992192.000000 Secret Key 50331648 2954938175610156848888276006519501238173891974277081114627768841840801736306392481516295906896346039950625609765296207682724801406770458881439696544971142710292689518104183685723154223544599940711614138798312668264956190761622328617992192.000000 -0.000000 -0.000000 The number E Expected Output: 3 rows of data ending with the number "e = 2.18281828" To reproduce the problem, the following two files need to be in the same directory as the lookup-data.c file: - db-data - db-description

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  • How to tell endianness from this output?

    - by Nick Rosencrantz
    I'm running this example program and I'm suppossed to be able to tell from the output what machine type it is. I'm certain it's from inspecting one or two values but how should I perform this inspection? /* pointers.c - Test pointers * Written 2012 by F Lundevall * Copyright abandoned. This file is in the public domain. * * To make this program work on as many systems as possible, * addresses are converted to unsigned long when printed. * The 'l' in formatting-codes %ld and %lx means a long operand. */ #include <stdio.h> #include <stdlib.h> int * ip; /* Declare a pointer to int, a.k.a. int pointer. */ char * cp; /* Pointer to char, a.k.a. char pointer. */ /* Declare fp as a pointer to function, where that function * has one parameter of type int and returns an int. * Use cdecl to get the syntax right, http://cdecl.org/ */ int ( *fp )( int ); int val1 = 111111; int val2 = 222222; int ia[ 17 ]; /* Declare an array of 17 ints, numbered 0 through 16. */ char ca[ 17 ]; /* Declare an array of 17 chars. */ int fun( int parm ) { printf( "Function fun called with parameter %d\n", parm ); return( parm + 1 ); } /* Main function. */ int main() { printf( "Message PT.01 from pointers.c: Hello, pointy World!\n" ); /* Do some assignments. */ ip = &val1; cp = &val2; /* The compiler should warn you about this. */ fp = fun; ia[ 0 ] = 11; /* First element. */ ia[ 1 ] = 17; ia[ 2 ] = 3; ia[ 16 ] = 58; /* Last element. */ ca[ 0 ] = 11; /* First element. */ ca[ 1 ] = 17; ca[ 2 ] = 3; ca[ 16 ] = 58; /* Last element. */ printf( "PT.02: val1: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &val1, val1, val1 ); printf( "PT.03: val2: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &val2, val2, val2 ); printf( "PT.04: ip: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &ip, (long) ip, (long) ip ); printf( "PT.05: Dereference pointer ip and we find: %d \n", *ip ); printf( "PT.06: cp: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &cp, (long) cp, (long) cp ); printf( "PT.07: Dereference pointer cp and we find: %d \n", *cp ); *ip = 1234; printf( "\nPT.08: Executed *ip = 1234; \n" ); printf( "PT.09: val1: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &val1, val1, val1 ); printf( "PT.10: ip: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &ip, (long) ip, (long) ip ); printf( "PT.11: Dereference pointer ip and we find: %d \n", *ip ); printf( "PT.12: val1: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &val1, val1, val1 ); *cp = 1234; /* The compiler should warn you about this. */ printf( "\nPT.13: Executed *cp = 1234; \n" ); printf( "PT.14: val2: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &val2, val2, val2 ); printf( "PT.15: cp: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &cp, (long) cp, (long) cp ); printf( "PT.16: Dereference pointer cp and we find: %d \n", *cp ); printf( "PT.17: val2: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &val2, val2, val2 ); ip = ia; printf( "\nPT.18: Executed ip = ia; \n" ); printf( "PT.19: ia[0]: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &ia[0], ia[0], ia[0] ); printf( "PT.20: ia[1]: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &ia[1], ia[1], ia[1] ); printf( "PT.21: ip: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &ip, (long) ip, (long) ip ); printf( "PT.22: Dereference pointer ip and we find: %d \n", *ip ); ip = ip + 1; /* add 1 to pointer */ printf( "\nPT.23: Executed ip = ip + 1; \n" ); printf( "PT.24: ip: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &ip, (long) ip, (long) ip ); printf( "PT.25: Dereference pointer ip and we find: %d \n", *ip ); cp = ca; printf( "\nPT.26: Executed cp = ca; \n" ); printf( "PT.27: ca[0]: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &ca[0], ca[0], ca[0] ); printf( "PT.28: ca[1]: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &ca[1], ca[1], ca[1] ); printf( "PT.29: cp: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &cp, (long) cp, (long) cp ); printf( "PT.30: Dereference pointer cp and we find: %d \n", *cp ); cp = cp + 1; /* add 1 to pointer */ printf( "\nPT.31: Executed cp = cp + 1; \n" ); printf( "PT.32: cp: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &cp, (long) cp, (long) cp ); printf( "PT.33: Dereference pointer cp and we find: %d \n", *cp ); ip = ca; /* The compiler should warn you about this. */ printf( "\nPT.34: Executed ip = ca; \n" ); printf( "PT.35: ca[0]: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &ca[0], ca[0], ca[0] ); printf( "PT.36: ca[1]: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &ca[1], ca[1], ca[1] ); printf( "PT.37: ip: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &ip, (long) ip, (long) ip ); printf( "PT.38: Dereference pointer ip and we find: %d \n", *ip ); cp = ia; /* The compiler should warn you about this. */ printf( "\nPT.39: Executed cp = ia; \n" ); printf( "PT.40: cp: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &cp, (long) cp, (long) cp ); printf( "PT.41: Dereference pointer cp and we find: %d \n", *cp ); printf( "\nPT.42: fp: stored at %lx (hex); value is %ld (dec), %lx (hex)\n", (long) &fp, (long) fp, (long) fp ); printf( "PT.43: Dereference fp and see what happens.\n" ); val1 = (*fp)(42); printf( "PT.44: Executed val1 = (*fp)(42); \n" ); printf( "PT.45: val1: stored at %lx (hex); value is %d (dec), %x (hex)\n", (long) &val1, val1, val1 ); return( 0 ); } Output Message PT.01 from pointers.c: Hello, pointy World! PT.02: val1: stored at 21e50 (hex); value is 111111 (dec), 1b207 (hex) PT.03: val2: stored at 21e54 (hex); value is 222222 (dec), 3640e (hex) PT.04: ip: stored at 21eb8 (hex); value is 138832 (dec), 21e50 (hex) PT.05: Dereference pointer ip and we find: 111111 PT.06: cp: stored at 21e6c (hex); value is 138836 (dec), 21e54 (hex) PT.07: Dereference pointer cp and we find: 0 PT.08: Executed *ip = 1234; PT.09: val1: stored at 21e50 (hex); value is 1234 (dec), 4d2 (hex) PT.10: ip: stored at 21eb8 (hex); value is 138832 (dec), 21e50 (hex) PT.11: Dereference pointer ip and we find: 1234 PT.12: val1: stored at 21e50 (hex); value is 1234 (dec), 4d2 (hex) PT.13: Executed *cp = 1234; PT.14: val2: stored at 21e54 (hex); value is -771529714 (dec), d203640e (hex) PT.15: cp: stored at 21e6c (hex); value is 138836 (dec), 21e54 (hex) PT.16: Dereference pointer cp and we find: -46 PT.17: val2: stored at 21e54 (hex); value is -771529714 (dec), d203640e (hex) PT.18: Executed ip = ia; PT.19: ia[0]: stored at 21e74 (hex); value is 11 (dec), b (hex) PT.20: ia[1]: stored at 21e78 (hex); value is 17 (dec), 11 (hex) PT.21: ip: stored at 21eb8 (hex); value is 138868 (dec), 21e74 (hex) PT.22: Dereference pointer ip and we find: 11 PT.23: Executed ip = ip + 1; PT.24: ip: stored at 21eb8 (hex); value is 138872 (dec), 21e78 (hex) PT.25: Dereference pointer ip and we find: 17 PT.26: Executed cp = ca; PT.27: ca[0]: stored at 21e58 (hex); value is 11 (dec), b (hex) PT.28: ca[1]: stored at 21e59 (hex); value is 17 (dec), 11 (hex) PT.29: cp: stored at 21e6c (hex); value is 138840 (dec), 21e58 (hex) PT.30: Dereference pointer cp and we find: 11 PT.31: Executed cp = cp + 1; PT.32: cp: stored at 21e6c (hex); value is 138841 (dec), 21e59 (hex) PT.33: Dereference pointer cp and we find: 17 PT.34: Executed ip = ca; PT.35: ca[0]: stored at 21e58 (hex); value is 11 (dec), b (hex) PT.36: ca[1]: stored at 21e59 (hex); value is 17 (dec), 11 (hex) PT.37: ip: stored at 21eb8 (hex); value is 138840 (dec), 21e58 (hex) PT.38: Dereference pointer ip and we find: 185664256 PT.39: Executed cp = ia; PT.40: cp: stored at 21e6c (hex); value is 138868 (dec), 21e74 (hex) PT.41: Dereference pointer cp and we find: 0 PT.42: fp: stored at 21e70 (hex); value is 69288 (dec), 10ea8 (hex) PT.43: Dereference fp and see what happens. Function fun called with parameter 42 PT.44: Executed val1 = (*fp)(42); PT.45: val1: stored at 21e50 (hex); value is 43 (dec), 2b (hex)

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  • strange output in ubuntu terminal when running a lex program

    - by Max
    Hi. I'm running a lexical analyzer using lex, and I've got it mostly correct, but my terminal gives strange output once I take out an ECHO statement I was using to help debug the code. With that statement, my output looks like this: max@Max-Ubuntu:~/Desktop/Compiler Project/project2$ ./a.out <../cmmFiles/expression.cmm VOIDID(){ INTID,ID,ID; BOOLID,ID,ID; ID(ID); ID(ID); ID(ID); ID(ID); ID=-ID-NUM+ID/NUM*(-NUM+ID*IDNUM); ID(ID); ID=ID>ID||ID>=ID; IF(ID)ID(NUM);ELSEID(NUM); ID=ID<ID&&ID<=ID; IF(ID==TRUE)ID(NUM);ELSEID(NUM); ID=ID&&!ID||!ID&&ID; IF(ID!=FALSE)ID(NUM);ELSEID(NUM); } While hard to read, that output is correct. Once I take out the ECHO statement, I instead get this: max@Max-Ubuntu:~/Desktop/Compiler Project/project2$ ./a.out <../cmmFiles/expression.cmm }F(ID!=FALSE)ID(NUM);ELSEID(NUM);; It looks like it's only outputting the final line, except with an extraneous } near the beginning, what looks like half an IF token immediately after, and an extraneous ; at the end. Is this some quirk of my terminal, or does removing that ECHO cause my lexer to screw up that badly? I'm hesitant to keep working until I know for sure what's going on here. Thanks for any answers. Here's my lexer: %{ /* definitions of manifest constants -reserved words- BOOL, ELSE, IF, TRUE, WHILE, DO, FALSE, INT, VOID -Punctuation and operators- LPAREN, RPAREN, LBRACK, RBRACK, LBRACE, RBRACE, SEMI, COMMA, PLUS, MINUS, TIMES, DIV, MOD, AND, OR, NOT, IS, ADDR, EQ, NE, LT, GT, LE, GE -Other tokens- NUMBER, ID, PUNCT, OP */ #include <stdio.h> #include <stdlib.h> //#include "y.tab.h" //int line = 1, numAttr; //char *strAttr; %} /* regular definitions */ delim [ \t] ws {delim}+ start "/*" one [^*] two "*" three [^*/] end "/" comment {start}({one}*{two}+{three})*{one}*{two}+{end} letter [A-Za-z] digit [0-9] id ({letter}|_)({letter}|{digit}|_)* number {digit}+ %% {ws} { /*no action and no return */} {comment} { /*no action and no return */} [\n] {ECHO; /*no action */} // <-- this is the ECHO in question. bool { printf("BOOL");} else { printf("ELSE");} if { printf("IF");} true { printf("TRUE");} while { printf("WHILE");} do { printf("DO");} false { printf("FALSE");} int { printf("INT");} void { printf("VOID");} {id} { printf("ID");} {number} { printf("NUM");} "(" { printf("(");} ")" { printf(")");} "[" { printf("[");} "]" { printf("]");} "{" { printf("{");} "}" { printf("}");} ";" { printf(";");} "," { printf(",");} "+" { printf("+");} "-" { printf("-");} "*" { printf("*");} "/" { printf("/");} "%" { printf("%");} "&" { printf("&");} "&&" { printf("&&");} "||" { printf("||");} "!" { printf("!");} "!=" { printf("!=");} "=" { printf("=");} "==" { printf("==");} "<" { printf("<");} "<=" { printf("<=");} ">" { printf(">");} ">=" { printf(">=");} %% int main() { yylex(); printf("\n"); } int yywrap(void) { return 1; } here's the file it's analyzing: /* this program * illustrates evaluation of * arithmetic and boolean * expressions */ void main( ) { int m,n,p; bool a,b,c; scan(m); print(m); scan(n); print(n); p=-m-3+n/2*(-5+m*n%4); print(p); a=m>n || n>=p; if (a) print(1); else print(0); b=m<n && n<=p; if (b==true) print(1); else print(0); c=a && !b || !a && b; if (c!=false) print(1); else print(0); }

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  • printf + print param and values

    - by yael
    I need to print the following values with printf as the follwoing around like this: printf "[date +%d"/"%b"/"%G"-"%T] [WARN] $PARAM1 $PARAM2 $PARAM3 The required output: [02/Jun/2010-11:08:42] [WARN] val1....val2...val3 the gap between val1 to val2 and from val2 to val3 must be const gap not depend the length of the values

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  • c++: use a #define in printf?

    - by John
    I was wanting to use a constant of some kind for the application ID (so I can use it in printf). I had this: #define _APPID_ "Hello World!" And then the simple printf, calling it into %s (string). It put this out: simple.cpp:32: error: cannot convert ‘_IO_FILE*’ to ‘const char*’ for argument ‘1’ to ‘int printf(const char*, ...)’ What would I use to define the application ID to use in printf? I tried: static const char _APPID_[] = "Hello World"` but it didn't work, same error I think.

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  • Mysterious c debugging problem when trying to utilize printfs

    - by O_O
    Ok, folks. I've never encountered this before and it boggles the mind and is illogical. I have a somewhat complex loop and I want to try and see if everything is working by putting some printf statements. I look the intermediate products using printf and verify that the answer is ok. Then, when I comment out the printf to the intermediate products, the answer is WRONG. Has anyone ever encountered this? This is driving me insane and I don't see how the printfs could change an answer.... X_x If it helps, I am using a c/c++ compiler for a DSP. Thanks for any advice.. Here is a snippet... printf("splitBackground = %d, numWindowPoints = %d\n", splitBackground, numWindowPoints); splitBackground = splitBackground/numWindowPoints; printf("%d ", splitBackground); This is good but when I comment out the first line of code, it turns out to be hugely incorrect. :(

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  • getaddrinfo appears to return different results between Windows and Ubuntu?

    - by MrDuk
    I have the following two sets of code: Windows #undef UNICODE #include <winsock2.h> #include <ws2tcpip.h> #include <stdio.h> // link with Ws2_32.lib #pragma comment (lib, "Ws2_32.lib") int __cdecl main(int argc, char **argv) { //----------------------------------------- // Declare and initialize variables WSADATA wsaData; int iResult; INT iRetval; DWORD dwRetval; argv[1] = "www.google.com"; argv[2] = "80"; int i = 1; struct addrinfo *result = NULL; struct addrinfo *ptr = NULL; struct addrinfo hints; struct sockaddr_in *sockaddr_ipv4; // struct sockaddr_in6 *sockaddr_ipv6; LPSOCKADDR sockaddr_ip; char ipstringbuffer[46]; DWORD ipbufferlength = 46; /* // Validate the parameters if (argc != 3) { printf("usage: %s <hostname> <servicename>\n", argv[0]); printf("getaddrinfo provides protocol-independent translation\n"); printf(" from an ANSI host name to an IP address\n"); printf("%s example usage\n", argv[0]); printf(" %s www.contoso.com 0\n", argv[0]); return 1; } */ // Initialize Winsock iResult = WSAStartup(MAKEWORD(2, 2), &wsaData); if (iResult != 0) { printf("WSAStartup failed: %d\n", iResult); return 1; } //-------------------------------- // Setup the hints address info structure // which is passed to the getaddrinfo() function ZeroMemory( &hints, sizeof(hints) ); hints.ai_family = AF_UNSPEC; hints.ai_socktype = SOCK_STREAM; // hints.ai_protocol = IPPROTO_TCP; printf("Calling getaddrinfo with following parameters:\n"); printf("\tnodename = %s\n", argv[1]); printf("\tservname (or port) = %s\n\n", argv[2]); //-------------------------------- // Call getaddrinfo(). If the call succeeds, // the result variable will hold a linked list // of addrinfo structures containing response // information dwRetval = getaddrinfo(argv[1], argv[2], &hints, &result); if ( dwRetval != 0 ) { printf("getaddrinfo failed with error: %d\n", dwRetval); WSACleanup(); return 1; } printf("getaddrinfo returned success\n"); // Retrieve each address and print out the hex bytes for(ptr=result; ptr != NULL ;ptr=ptr->ai_next) { printf("getaddrinfo response %d\n", i++); printf("\tFlags: 0x%x\n", ptr->ai_flags); printf("\tFamily: "); switch (ptr->ai_family) { case AF_UNSPEC: printf("Unspecified\n"); break; case AF_INET: printf("AF_INET (IPv4)\n"); sockaddr_ipv4 = (struct sockaddr_in *) ptr->ai_addr; printf("\tIPv4 address %s\n", inet_ntoa(sockaddr_ipv4->sin_addr) ); break; case AF_INET6: printf("AF_INET6 (IPv6)\n"); // the InetNtop function is available on Windows Vista and later // sockaddr_ipv6 = (struct sockaddr_in6 *) ptr->ai_addr; // printf("\tIPv6 address %s\n", // InetNtop(AF_INET6, &sockaddr_ipv6->sin6_addr, ipstringbuffer, 46) ); // We use WSAAddressToString since it is supported on Windows XP and later sockaddr_ip = (LPSOCKADDR) ptr->ai_addr; // The buffer length is changed by each call to WSAAddresstoString // So we need to set it for each iteration through the loop for safety ipbufferlength = 46; iRetval = WSAAddressToString(sockaddr_ip, (DWORD) ptr->ai_addrlen, NULL, ipstringbuffer, &ipbufferlength ); if (iRetval) printf("WSAAddressToString failed with %u\n", WSAGetLastError() ); else printf("\tIPv6 address %s\n", ipstringbuffer); break; case AF_NETBIOS: printf("AF_NETBIOS (NetBIOS)\n"); break; default: printf("Other %ld\n", ptr->ai_family); break; } printf("\tSocket type: "); switch (ptr->ai_socktype) { case 0: printf("Unspecified\n"); break; case SOCK_STREAM: printf("SOCK_STREAM (stream)\n"); break; case SOCK_DGRAM: printf("SOCK_DGRAM (datagram) \n"); break; case SOCK_RAW: printf("SOCK_RAW (raw) \n"); break; case SOCK_RDM: printf("SOCK_RDM (reliable message datagram)\n"); break; case SOCK_SEQPACKET: printf("SOCK_SEQPACKET (pseudo-stream packet)\n"); break; default: printf("Other %ld\n", ptr->ai_socktype); break; } printf("\tProtocol: "); switch (ptr->ai_protocol) { case 0: printf("Unspecified\n"); break; case IPPROTO_TCP: printf("IPPROTO_TCP (TCP)\n"); break; case IPPROTO_UDP: printf("IPPROTO_UDP (UDP) \n"); break; default: printf("Other %ld\n", ptr->ai_protocol); break; } printf("\tLength of this sockaddr: %d\n", ptr->ai_addrlen); printf("\tCanonical name: %s\n", ptr->ai_canonname); } freeaddrinfo(result); WSACleanup(); return 0; } Ubuntu /* ** listener.c -- a datagram sockets "server" demo */ #include <stdio.h> #include <stdlib.h> #include <unistd.h> #include <errno.h> #include <string.h> #include <sys/types.h> #include <sys/socket.h> #include <netinet/in.h> #include <arpa/inet.h> #include <netdb.h> #define MYPORT "4950" // the port users will be connecting to #define MAXBUFLEN 100 // get sockaddr, IPv4 or IPv6: void *get_in_addr(struct sockaddr *sa) { if (sa->sa_family == AF_INET) { return &(((struct sockaddr_in*)sa)->sin_addr); } return &(((struct sockaddr_in6*)sa)->sin6_addr); } int main(void) { int sockfd; struct addrinfo hints, *servinfo, *p; int rv; int numbytes; struct sockaddr_storage their_addr; char buf[MAXBUFLEN]; socklen_t addr_len; char s[INET6_ADDRSTRLEN]; memset(&hints, 0, sizeof hints); hints.ai_family = AF_UNSPEC; // set to AF_INET to force IPv4 hints.ai_socktype = SOCK_DGRAM; hints.ai_flags = AI_PASSIVE; // use my IP if ((rv = getaddrinfo(NULL, MYPORT, &hints, &servinfo)) != 0) { fprintf(stderr, "getaddrinfo: %s\n", gai_strerror(rv)); return 1; } // loop through all the results and bind to the first we can for(p = servinfo; p != NULL; p = p->ai_next) { if ((sockfd = socket(p->ai_family, p->ai_socktype, p->ai_protocol)) == -1) { perror("listener: socket"); continue; } if (bind(sockfd, p->ai_addr, p->ai_addrlen) == -1) { close(sockfd); perror("listener: bind"); continue; } break; } if (p == NULL) { fprintf(stderr, "listener: failed to bind socket\n"); return 2; } freeaddrinfo(servinfo); printf("listener: waiting to recvfrom...\n"); addr_len = sizeof their_addr; if ((numbytes = recvfrom(sockfd, buf, MAXBUFLEN-1 , 0, (struct sockaddr *)&their_addr, &addr_len)) == -1) { perror("recvfrom"); exit(1); } printf("listener: got packet from %s\n", inet_ntop(their_addr.ss_family, get_in_addr((struct sockaddr *)&their_addr), s, sizeof s)); printf("listener: packet is %d bytes long\n", numbytes); buf[numbytes] = '\0'; printf("listener: packet contains \"%s\"\n", buf); close(sockfd); return 0; } When I attempt www.google.com, I don't get the ipv6 socket returned on Windows - why is this? Outputs: (ubuntu) caleb@ub1:~/Documents/dev/cs438/mp0/MP0$ ./a.out www.google.com IP addresses for www.google.com: IPv4: 74.125.228.115 IPv4: 74.125.228.116 IPv4: 74.125.228.112 IPv4: 74.125.228.113 IPv4: 74.125.228.114 IPv6: 2607:f8b0:4004:803::1010 Outputs: (win) Calling getaddrinfo with following parameters: nodename = www.google.com servname (or port) = 80 getaddrinfo returned success getaddrinfo response 1 Flags: 0x0 Family: AF_INET (IPv4) IPv4 address 74.125.228.114 Socket type: SOCK_STREAM (stream) Protocol: Unspecified Length of this sockaddr: 16 Canonical name: (null) getaddrinfo response 2 Flags: 0x0 Family: AF_INET (IPv4) IPv4 address 74.125.228.115 Socket type: SOCK_STREAM (stream) Protocol: Unspecified Length of this sockaddr: 16 Canonical name: (null) getaddrinfo response 3 Flags: 0x0 Family: AF_INET (IPv4) IPv4 address 74.125.228.116 Socket type: SOCK_STREAM (stream) Protocol: Unspecified Length of this sockaddr: 16 Canonical name: (null) getaddrinfo response 4 Flags: 0x0 Family: AF_INET (IPv4) IPv4 address 74.125.228.112 Socket type: SOCK_STREAM (stream) Protocol: Unspecified Length of this sockaddr: 16 Canonical name: (null) getaddrinfo response 5 Flags: 0x0 Family: AF_INET (IPv4) IPv4 address 74.125.228.113 Socket type: SOCK_STREAM (stream) Protocol: Unspecified Length of this sockaddr: 16 Canonical name: (null)

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  • Hi, I have a C hashing routine which is behaving strangely?

    - by aks
    Hi, In this hashing routine: 1.) I am able to add strings. 2.) I am able to view my added strings. 3.) When i try to add a duplicate string, it throws me an error saying already present. 4.) But, when i try to delete the same string which is already present in hash table, then the lookup_routine calls hash function to get an index. At this time, it throws a different hash index to the one it was already added. Hence, my delete routine is failing? I am able to understand the reason why for same string, hash fucntion calculates a different index each time (whereas the same logic works in view hash table routine)? Can someone help me? This is the Output, i am getting: $ ./a Press 1 to add an element to the hashtable Press 2 to delete an element from the hashtable Press 3 to search the hashtable Press 4 to view the hashtable Press 5 to exit Please enter your choice: 1 Please enter the string :gaura enters in add_string DEBUG purpose in hash function: str passed = gaura Hashval returned in hash func= 1 hashval = 1 enters in lookup_string str in lookup_string = gaura DEBUG purpose in hash function: str passed = gaura Hashval returned in hash func= 1 hashval = 1 DEBUG ERROR :element not found in lookup string DEBUG Purpose NULL Inserting... DEBUG1 : enters here hashval = 1 String added successfully Press 1 to add an element to the hashtable Press 2 to delete an element from the hashtable Press 3 to search the hashtable Press 4 to view the hashtable Press 5 to exit Please enter your choice: 1 Please enter the string :ayu enters in add_string DEBUG purpose in hash function: str passed = ayu Hashval returned in hash func= 1 hashval = 1 enters in lookup_string str in lookup_string = ayu DEBUG purpose in hash function: str passed = ayu Hashval returned in hash func= 1 hashval = 1 returns NULL in lookup_string DEBUG Purpose NULL Inserting... DEBUG2 : enters here hashval = 1 String added successfully Press 1 to add an element to the hashtable Press 2 to delete an element from the hashtable Press 3 to search the hashtable Press 4 to view the hashtable Press 5 to exit Please enter your choice: 1 Please enter the string :gaurava enters in add_string DEBUG purpose in hash function: str passed = gaurava Hashval returned in hash func= 7 hashval = 7 enters in lookup_string str in lookup_string = gaurava DEBUG purpose in hash function: str passed = gaurava Hashval returned in hash func= 7 hashval = 7 DEBUG ERROR :element not found in lookup string DEBUG Purpose NULL Inserting... DEBUG1 : enters here hashval = 7 String added successfully Press 1 to add an element to the hashtable Press 2 to delete an element from the hashtable Press 3 to search the hashtable Press 4 to view the hashtable Press 5 to exit Please enter your choice: 4 Index : i = 1 String = gaura ayu Index : i = 7 String = gaurava Press 1 to add an element to the hashtable Press 2 to delete an element from the hashtable Press 3 to search the hashtable Press 4 to view the hashtable Press 5 to exit Please enter your choice: 2 Please enter the string you want to delete :gaura String entered = gaura enters in delete_string DEBUG purpose in hash function: str passed = gaura Hashval returned in hash func= 0 hashval = 0 enters in lookup_string str in lookup_string = gaura DEBUG purpose in hash function: str passed = gaura Hashval returned in hash func= 0 hashval = 0 DEBUG ERROR :element not found in lookup string DEBUG Purpose Item not present. So, cannot be deleted Item found in list: Deletion failed Press 1 to add an element to the hashtable Press 2 to delete an element from the hashtable Press 3 to search the hashtable Press 4 to view the hashtable Press 5 to exit Please enter your choice: My routine is pasted below: include include struct list { char *string; struct list *next; }; struct hash_table { int size; /* the size of the table */ struct list *table; / the table elements */ }; struct hash_table * hashtable = NULL; struct hash_table *create_hash_table(int size) { struct hash_table *new_table; int i; if (size<1) return NULL; /* invalid size for table */ /* Attempt to allocate memory for the table structure */ if ((new_table = malloc(sizeof(struct hash_table))) == NULL) { return NULL; } /* Attempt to allocate memory for the table itself */ if ((new_table->table = malloc(sizeof(struct list *) * size)) == NULL) { return NULL; } /* Initialize the elements of the table */ for(i=0; i<size; i++) new_table->table[i] = '\0'; /* Set the table's size */ new_table->size = size; return new_table; } unsigned int hash(struct hash_table *hashtable, char *str) { printf("\n DEBUG purpose in hash function:\n"); printf("\n str passed = %s", str); unsigned int hashval = 0; int i = 0; for(; *str != '\0'; str++) { hashval += str[i]; i++; } hashval = hashval % 10; printf("\n Hashval returned in hash func= %d", hashval); return hashval; } struct list *lookup_string(struct hash_table *hashtable, char *str) { printf("\n enters in lookup_string \n"); printf("\n str in lookup_string = %s",str); struct list * new_list; unsigned int hashval = hash(hashtable, str); printf("\n hashval = %d \n", hashval); if(hashtable->table[hashval] == NULL) { printf("\n DEBUG ERROR :element not found in lookup string \n"); return NULL; } /* Go to the correct list based on the hash value and see if str is * in the list. If it is, return return a pointer to the list element. * If it isn't, the item isn't in the table, so return NULL. */ for(new_list = hashtable->table[hashval]; new_list != NULL;new_list = new_list->next) { if (strcmp(str, new_list->string) == 0) return new_list; } printf("\n returns NULL in lookup_string \n"); return NULL; } int add_string(struct hash_table *hashtable, char *str) { printf("\n enters in add_string \n"); struct list *new_list; struct list *current_list; unsigned int hashval = hash(hashtable, str); printf("\n hashval = %d", hashval); /* Attempt to allocate memory for list */ if ((new_list = malloc(sizeof(struct list))) == NULL) { printf("\n enters here \n"); return 1; } /* Does item already exist? */ current_list = lookup_string(hashtable, str); if (current_list == NULL) { printf("\n DEBUG Purpose \n"); printf("\n NULL \n"); } /* item already exists, don't insert it again. */ if (current_list != NULL) { printf("\n Item already present...\n"); return 2; } /* Insert into list */ printf("\n Inserting...\n"); new_list->string = strdup(str); new_list->next = NULL; //new_list->next = hashtable->table[hashval]; if(hashtable->table[hashval] == NULL) { printf("\n DEBUG1 : enters here \n"); printf("\n hashval = %d", hashval); hashtable->table[hashval] = new_list; } else { printf("\n DEBUG2 : enters here \n"); printf("\n hashval = %d", hashval); struct list * temp_list = hashtable->table[hashval]; while(temp_list->next!=NULL) temp_list = temp_list->next; temp_list->next = new_list; // hashtable->table[hashval] = new_list; } return 0; } int delete_string(struct hash_table *hashtable, char *str) { printf("\n enters in delete_string \n"); struct list *new_list; struct list *current_list; unsigned int hashval = hash(hashtable, str); printf("\n hashval = %d", hashval); /* Does item already exist? */ current_list = lookup_string(hashtable, str); if (current_list == NULL) { printf("\n DEBUG Purpose \n"); printf("\n Item not present. So, cannot be deleted \n"); return 1; } /* item exists, delete it. */ if (current_list != NULL) { struct list * temp = hashtable->table[hashval]; if(strcmp(temp->string,str) == 0) { if(temp->next == NULL) { hashtable->table[hashval] = NULL; free(temp); } else { hashtable->table[hashval] = temp->next; free(temp); } } else { struct list * temp1; while(temp->next != NULL) { temp1 = temp; if(strcmp(temp->string, str) == 0) { break; } else { temp = temp->next; } } if(temp->next == NULL) { temp1->next = NULL; free(temp); } else { temp1->next = temp->next; free(temp); } } } return 0; } void free_table(struct hash_table *hashtable) { int i; struct list *new_list, *temp_list; if (hashtable==NULL) return; /* Free the memory for every item in the table, including the * strings themselves. */ for(i=0; i<hashtable->size; i++) { new_list = hashtable->table[i]; while(new_list!=NULL) { temp_list = new_list; new_list = new_list->next; free(temp_list->string); free(temp_list); } } /* Free the table itself */ free(hashtable->table); free(hashtable); } void view_hashtable(struct hash_table * hashtable) { int i = 0; if(hashtable == NULL) return; for(i =0; i < hashtable->size; i++) { if((hashtable->table[i] == 0) || (strcmp(hashtable->table[i]->string, "*") == 0)) { continue; } printf(" Index : i = %d\t String = %s",i, hashtable->table[i]->string); struct list * temp = hashtable->table[i]->next; while(temp != NULL) { printf("\t %s",temp->string); temp = temp->next; } printf("\n"); } } int main() { hashtable = create_hash_table(10); if(hashtable == NULL) { printf("\n Memory allocation failure during creation of hash table \n"); return 0; } int flag = 1; while(flag) { int choice; printf("\n Press 1 to add an element to the hashtable\n"); printf("\n Press 2 to delete an element from the hashtable\n"); printf("\n Press 3 to search the hashtable\n"); printf("\n Press 4 to view the hashtable\n"); printf("\n Press 5 to exit \n"); printf("\n Please enter your choice: "); scanf("%d",&choice); if(choice == 5) flag = 0; else if(choice == 1) { char str[20]; printf("\n Please enter the string :"); scanf("%s",&str); int i; i = add_string(hashtable,str); if(i == 1) { printf("\n Memory allocation failure:Choice 1 \n"); return 0; } else if(i == 2) { printf("\n String already prsent in hash table : Couldnot add it again\n"); return 0; } else { printf("\n String added successfully \n"); } } else if(choice == 2) { int i; struct list * temp_list; char str[20]; printf("\n Please enter the string you want to delete :"); scanf("%s",&str); printf("\n String entered = %s", str); i = delete_string(hashtable,str); if(i == 0) { printf("\n Item found in list: Deletion success \n"); } else printf("\n Item found in list: Deletion failed \n"); } else if(choice == 3) { struct list * temp_list; char str[20]; printf("\n Please enter the string :"); scanf("%s",&str); temp_list = lookup_string(hashtable,str); if(!temp_list) { printf("\n Item not found in list: Deletion failed \n"); return 0; } printf("\n Item found: Present in Hash Table \n"); } else if(choice == 4) { view_hashtable(hashtable); } else if(choice == 5) { printf("\n Exiting ...."); return 0; } else printf("\n Invalid choice:"); }; free_table(hashtable); return 0; }

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  • ksh + printf stat gap of print

    - by yael
    need to print the follwoing: need smart way by printf to pring this example param1 ............... value1 param2 ............... value2 param3 ............... value1 param4 ............... value2 THX

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  • c++ Mixing printf with wprintf (or cout with wcout)

    - by Bo Jensen
    I know you should not mix printing with printf,cout and wprintf,wcout, but have a hard time finding a good answer why and if it is possible to get round it. The problem is I use a external library that prints with printf and my own uses wcout. If I do a simple example it works fine, but from my full application it simply does not print the printf statements. If this is really a limitation, then there would be many libraries out there which can not work together with wide printing applications. Any insight on this is more than welcome. Update : I boiled it down to : #include <stdio.h> #include <stdlib.h> #include <iostream> #include <readline/readline.h> #include <readline/history.h> int main() { char *buf; std::wcout << std::endl; /* ADDING THIS LINE MAKES PRINTF VANISH!!! */ rl_bind_key('\t',rl_abort);//disable auto-complete while((buf = readline("my-command : "))!=NULL) { if (strcmp(buf,"quit")==0) break; std::wcout<<buf<< std::endl; if (buf[0]!=0) add_history(buf); } free(buf); return 0; } So I guess it might be a flushing problem, but it still looks strange to me, I have to check up on it.

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  • Why does printf report an error on all but three (ASCII-range) Unicode Codepoints, yet is fine with all others?

    - by fred.bear
    The 'printf' I refer to is the standard-issue "program" (not the built-in): /usr/bin/printf I was testing printf out as a viable method of convert a Unicode Codepoint Hex-literal into its Unicoder character representation, I was looking good, and seemed flawless..(btw. the built-in printf can't do this at all (I think)... I then thought to test it at the lower extreme end of the code-spectrum, and it failed with an avalanche of errors.. All in the ASCII range (= 7 bits) The strangest thing was that 3 value printed normally; they are: $ \u0024 @ \u0040 ` \u0060 I'd like to know what is going on here. The ASCII character-set is most definitely part of the Unicode Code-point sequence.... I am puzzled, and still without a good way to bash script this particular converion.. Suggestions are welcome. To be entertained by that same avalanche of errors, paste the following code into a terminal... # Here is one of the error messages # /usr/bin/printf: invalid universal character name \u0041 # ...for them all, run the following script ( for nib1 in {0..9} {A..F}; do for nib0 in {0..9} {A..F}; do [[ $nib1 < A ]] && nl="\n" || nl=" " $(type -P printf) "\u00$nib1$nib0$nl" done done echo )

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  • gdb+osx: no output when using printf/CFShow

    - by yairchu
    I attached to a program with gdb in OSX and I want to use CFShow in the gdb console etc. However, nothing shows up. printf shows nothing as well: (gdb) call (int) printf("Hello\n") $10 = 6 (gdb) call (int) printf("Hello World!\n") $11 = 13 Apple suggests the following tip for when attaching with gdb, to make the output appear in the gdb console: (gdb) call (void) close(1) (gdb) call (void) close(2) (gdb) shell tty /dev/ttyp1 (gdb) call (int) open("/dev/ttyp1", 2, 0) $1 = 1 (gdb) call (int) open("/dev/ttyp1", 2, 0) $2 = 2 In xcode's gdb console tty gives "not a tty", so I tried it in gdb in a terminal. There tty does work but after redirecting stdout there's still no output. Also no output if I direct stdout to a file.. :/ Any salvation?

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  • printf not passing correct Hex Address to stack

    - by kriss
    I have a hickup in using printf . I am on ubuntu 10.04. Basically i have a C program asking for some input and then prints it back. It is OK for printing something after inputing. I tried to insert some Hex Address to Stack by following format:- printf "hello world!\x12\x23\x34" | ./input1 But i don't know what is the problem. If i give only string beyond 12 bytes it overwrites BUT If I give hex address(through printf), it doesn't overwrite on return address. Instead it stores some other thing. Could anyone help??? I can't proceed further becoz of this. Thanks in advance

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  • I do not understand -printf script

    - by jerzdevs
    I have taken over the responsibility of RHLE5 scripting and I've not had any training in this platform or BASH scripting. There's a script that has multiple pieces to it and I will ask only about the second piece but also show you the first, I think it will help with my question below. The first part of the script shows the output of users on a particular server: cut -d : -f 1 /etc/passwd The output will look something like: root bin joe rob other... The second script requires me to fill in each of the accounts listed from the above script and run. From what I can gather, and from my search on the man pages and other web searches, it goes out and finds the group owner of a file or directory and obviously sorts and picks out just unique records but not really sure - so that's my question, what does the below script really do? (The funny thing is, is that if I plug in each name from the output above, I'll sometimes receive a "cannot find username blah, blah, blah" message.) find username -printf %G | sort | uniq

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