max path - Page 111 - Developer IT

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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