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  • Configuring Multi-Tap on Synaptics Touchpad

    - by nunos
    I am having a hard time configuring my notebook's touchpad. The touchpad already works. It successfully responds to one-finger tap, two-finger tap and two-finger vertical scrolling. What I want to accomplish: change two-finger tap action from right-mouse click to middle-mouse click add three-finger tap functionality to yield right-mouse click action (i have checked that the three-finger tap is supported by my laptop's touchpad since it works on Windows) I read on a forum to use this as a guide. I have successfully accomplished point 1 with synclient TapButton2=2. However, I have to do it everytime I log in. I have tried to put that command on /etc/rc.local but the computer always boots and logins with the default configuration. Regarding point 2, I have tried synclient TapButton3=3 but it doesn't do anything when I three-finger tap the touchpad. I am running Ubuntu 11.10 on an Asus N82JV. /etc/X11/xorg.conf: nuno@mozart:~$ cat /etc/X11/xorg.conf Section "InputClass" Identifier "touchpad catchall" Driver "synaptics" MatchIsTouchpad "on" MatchDevicePath "/dev/input/event*" Option "TapButton1" "1" Option "TapButton2" "2" Option "TapButton3" "3" EndSection /usr/share/X11/xorg.conf.d/50-synaptics.conf: nuno@mozart:~$ cat /usr/share/X11/xorg.conf.d/50-synaptics.conf # Example xorg.conf.d snippet that assigns the touchpad driver # to all touchpads. See xorg.conf.d(5) for more information on # InputClass. # DO NOT EDIT THIS FILE, your distribution will likely overwrite # it when updating. Copy (and rename) this file into # /etc/X11/xorg.conf.d first. # Additional options may be added in the form of # Option "OptionName" "value" # Section "InputClass" Identifier "touchpad catchall" Driver "synaptics" MatchIsTouchpad "on" MatchDevicePath "/dev/input/event*" Option "TapButton1" "1" Option "TapButton2" "2" Option "TapButton3" "3" EndSection xinput list: nuno@mozart:~$ xinput list ? Virtual core pointer id=2 [master pointer (3)] ? ? Virtual core XTEST pointer id=4 [slave pointer (2)] ? ? Microsoft Microsoft® Nano Transceiver v2.0 id=12 [slave pointer (2)] ? ? Microsoft Microsoft® Nano Transceiver v2.0 id=13 [slave pointer (2)] ? ? ETPS/2 Elantech Touchpad id=16 [slave pointer (2)] ? Virtual core keyboard id=3 [master keyboard (2)] ? Virtual core XTEST keyboard id=5 [slave keyboard (3)] ? Power Button id=6 [slave keyboard (3)] ? Video Bus id=7 [slave keyboard (3)] ? Video Bus id=8 [slave keyboard (3)] ? Sleep Button id=9 [slave keyboard (3)] ? USB2.0 2.0M UVC WebCam id=10 [slave keyboard (3)] ? Microsoft Microsoft® Nano Transceiver v2.0 id=11 [slave keyboard (3)] ? Asus Laptop extra buttons id=14 [slave keyboard (3)] ? AT Translated Set 2 keyboard id=15 [slave keyboard (3)]

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  • Configuring Multi-Tap on Synpaptics Touchpad

    - by nunos
    I am having a hard time configuring my notebook's touchpad. The touchpad already works. It succesfully responds to one-finger tap, two-finger tap and two-finger vertical scrolling. What I want to accomplish: change two-finger tap action from right-mouse click to middle-mouse click add three-finger tap functionality to yield right-mouse click action I read on a forum to use this as a guide. I have succesfully accomplished point 1 with synclient TapButton2=2. However, I have to do it everytime I log in. I have tried to put that command on /etc/rc.local but the computer always boots and logins with the default configuration. Regarding point 2, I have tried synclient TapButton3=3 but it doesn't do anything when I three-finger tap the touchpad. I am running Ubuntu 11.10 on an Asus N82JV. /etc/X11/xorg.conf: nuno@mozart:~$ cat /etc/X11/xorg.conf Section "InputClass" Identifier "touchpad catchall" Driver "synaptics" MatchIsTouchpad "on" MatchDevicePath "/dev/input/event*" Option "TapButton1" "1" Option "TapButton2" "2" Option "TapButton3" "3" EndSection /usr/share/X11/xorg.conf.d/50-synaptics.conf: nuno@mozart:~$ cat /usr/share/X11/xorg.conf.d/50-synaptics.conf # Example xorg.conf.d snippet that assigns the touchpad driver # to all touchpads. See xorg.conf.d(5) for more information on # InputClass. # DO NOT EDIT THIS FILE, your distribution will likely overwrite # it when updating. Copy (and rename) this file into # /etc/X11/xorg.conf.d first. # Additional options may be added in the form of # Option "OptionName" "value" # Section "InputClass" Identifier "touchpad catchall" Driver "synaptics" MatchIsTouchpad "on" MatchDevicePath "/dev/input/event*" Option "TapButton1" "1" Option "TapButton2" "2" Option "TapButton3" "3" EndSection xinput list: nuno@mozart:~$ xinput list ? Virtual core pointer id=2 [master pointer (3)] ? ? Virtual core XTEST pointer id=4 [slave pointer (2)] ? ? Microsoft Microsoft® Nano Transceiver v2.0 id=12 [slave pointer (2)] ? ? Microsoft Microsoft® Nano Transceiver v2.0 id=13 [slave pointer (2)] ? ? ETPS/2 Elantech Touchpad id=16 [slave pointer (2)] ? Virtual core keyboard id=3 [master keyboard (2)] ? Virtual core XTEST keyboard id=5 [slave keyboard (3)] ? Power Button id=6 [slave keyboard (3)] ? Video Bus id=7 [slave keyboard (3)] ? Video Bus id=8 [slave keyboard (3)] ? Sleep Button id=9 [slave keyboard (3)] ? USB2.0 2.0M UVC WebCam id=10 [slave keyboard (3)] ? Microsoft Microsoft® Nano Transceiver v2.0 id=11 [slave keyboard (3)] ? Asus Laptop extra buttons id=14 [slave keyboard (3)] ? AT Translated Set 2 keyboard id=15 [slave keyboard (3)]

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  • How to pin Eclipse Indigo to Unity?

    - by nunos
    How can I pin eclipse indigo 3.7 I have 'installed' at /opt/ to the taskbar? I have tried launching eclipse and right-clicking to choose the 'keep in launcher' option. But when I click that icon, after closing eclipse, it doesn't start eclipse. I have already looked at How do I add Eclipse Indigo to the launcher? but no answer worked for me. (Mod note: The answers from this thread are now merged into this one)

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  • Configuring Touchpad Multi-Tap on Ubuntu 11.10

    - by nunos
    I am having a hard time configuring my notebook's touchpad. I can do everything I can in Windows with the exception of three-finger tap, which doesn't work, and the action of two-finger tap which is giving me the equivalent to a right-mouse click, when I wanted a middle-mouse click. I read on a forum to use this as a guide. The problem is that I can't even find the configuration file /etc/X11/xorg.conf.d/10-synaptics.conf. I tried running pacman -S xf86-input-synaptics but I don't have the pacman program installed. When I try to install it by sudo apt-get install I get a pacman game instead! I know the guide is for archlinux, so maybe that's why it doesn't work with me. I am running Ubuntu 11.10 on an Asus N82JV. Any help on this is appreciated. Here's the output of xinput list: nuno@mozart:~$ xinput list ? Virtual core pointer id=2 [master pointer (3)] ? ? Virtual core XTEST pointer id=4 [slave pointer (2)] ? ? Microsoft Microsoft® Nano Transceiver v2.0 id=12 [slave pointer (2)] ? ? Microsoft Microsoft® Nano Transceiver v2.0 id=13 [slave pointer (2)] ? ? ETPS/2 Elantech Touchpad id=16 [slave pointer (2)] ? Virtual core keyboard id=3 [master keyboard (2)] ? Virtual core XTEST keyboard id=5 [slave keyboard (3)] ? Power Button id=6 [slave keyboard (3)] ? Video Bus id=7 [slave keyboard (3)] ? Video Bus id=8 [slave keyboard (3)] ? Sleep Button id=9 [slave keyboard (3)] ? USB2.0 2.0M UVC WebCam id=10 [slave keyboard (3)] ? Microsoft Microsoft® Nano Transceiver v2.0 id=11 [slave keyboard (3)] ? Asus Laptop extra buttons id=14 [slave keyboard (3)] ? AT Translated Set 2 keyboard id=15 [slave keyboard (3)]

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  • Partitioning Software for Windows 7

    - by nunos
    I used to use Patition Magic Pro from Norton to do disk partitioning. Now that I have Windows 7, I have been warned, by the OS itself and later confirmed on the internet, that Partition Magic Pro has compatibility issues with Windows 7. So, I am asking you which software partitioning software do you recommend. It would be great if it was freeware but I know it is hard to find a good parititioning app that's free, so, if you know a good one that's paid, there's no problem and please mention it in your reply. Thanks in advance. P.S. I am aware of "create and format disk partitions feature" of Windows 7, but that's not enough for me. I am using Windows 7 Professional 64 bits.

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  • Help debugging c fifos code - stack smashing detected - open call not functioning - removing pipes

    - by nunos
    I have three bugs/questions regarding the source code pasted below: stack smashing deteced: In order to compile and not have that error I have addedd the gcc compile flag -fno-stack-protector. However, this should be just a temporary solution, since I would like to find where the cause for this is and correct it. However, I haven't been able to do so. Any clues? For some reason, the last open function call doesn't work and the programs just stops there, without an error, even though the fifo already exists. I want to delete the pipes from the filesystem after before terminating the processes. I have added close and unlink statements at the end, but the fifos are not removed. What am I doing wrong? Thanks very much in advance. P.S.: I am pasting here the whole source file for additional clarity. Just ignore the comments, since they are in my own native language. server.c: #include <stdio.h> #include <stdlib.h> #include <string.h> #include <unistd.h> #include <sys/types.h> #include <sys/stat.h> #include <fcntl.h> #include <errno.h> #define MAX_INPUT_LENGTH 100 #define FIFO_NAME_MAX_LEN 20 #define FIFO_DIR "/tmp/" #define FIFO_NAME_CMD_CLI_TO_SRV "lrc_cmd_cli_to_srv" typedef enum { false, true } bool; bool background = false; char* logfile = NULL; void read_from_fifo(int fd, char** var) { int n_bytes; read(fd, &n_bytes, sizeof(int)); *var = (char *) malloc (n_bytes); read(fd, *var, n_bytes); printf("read %d bytes '%s'\n", n_bytes, *var); } void write_to_fifo(int fd, char* data) { int n_bytes = (strlen(data)+1) * sizeof(char); write(fd, &n_bytes, sizeof(int)); //primeiro envia o numero de bytes que a proxima instrucao write ira enviar write(fd, data, n_bytes); printf("writing %d bytes '%s'\n", n_bytes, data); } int main(int argc, char* argv[]) { //CRIA FIFO CMD_CLI_TO_SRV, se ainda nao existir char* fifo_name_cmd_cli_to_srv; fifo_name_cmd_cli_to_srv = (char*) malloc ( (strlen(FIFO_NAME_CMD_CLI_TO_SRV) + strlen(FIFO_DIR) + 1) * sizeof(char) ); strcpy(fifo_name_cmd_cli_to_srv, FIFO_DIR); strcat(fifo_name_cmd_cli_to_srv, FIFO_NAME_CMD_CLI_TO_SRV); int n = mkfifo(fifo_name_cmd_cli_to_srv, 0660); //TODO ver permissoes if (n < 0 && errno != EEXIST) //se houver erro, e nao for por causa de ja haver um com o mesmo nome, termina o programa { fprintf(stderr, "erro ao criar o fifo\n"); fprintf(stderr, "errno: %d\n", errno); exit(4); } //se por acaso já existir, nao cria o fifo e continua o programa normalmente //le informacao enviada pelo cliente, nesta ordem: //1. pid (em formato char*) do processo cliente //2. comando /CONNECT //3. nome de fifo INFO_SRV_TO_CLIXXX //4. nome de fifo MSG_SRV_TO_CLIXXX char* command; char* fifo_name_info_srv_to_cli; char* fifo_name_msg_srv_to_cli; char* client_pid_string; int client_pid; int fd_cmd_cli_to_srv, fd_info_srv_to_cli; fd_cmd_cli_to_srv = open(fifo_name_cmd_cli_to_srv, O_RDONLY); read_from_fifo(fd_cmd_cli_to_srv, &client_pid_string); client_pid = atoi(client_pid_string); read_from_fifo(fd_cmd_cli_to_srv, &command); //recebe commando /CONNECT read_from_fifo(fd_cmd_cli_to_srv, &fifo_name_info_srv_to_cli); //recebe nome de fifo INFO_SRV_TO_CLIXXX read_from_fifo(fd_cmd_cli_to_srv, &fifo_name_msg_srv_to_cli); //recebe nome de fifo MSG_TO_SRV_TO_CLIXXX //CIRA FIFO MSG_CLIXXX_TO_SRV char fifo_name_msg_cli_to_srv[FIFO_NAME_MAX_LEN]; strcpy(fifo_name_msg_cli_to_srv, FIFO_DIR); strcat(fifo_name_msg_cli_to_srv, "lrc_msg_cli"); strcat(fifo_name_msg_cli_to_srv, client_pid_string); strcat(fifo_name_msg_cli_to_srv, "_to_srv"); n = mkfifo(fifo_name_msg_cli_to_srv, 0660); if (n < 0) { fprintf(stderr, "error creating %s\n", fifo_name_msg_cli_to_srv); fprintf(stderr, "errno: %d\n", errno); exit(5); } //envia ao cliente a resposta ao commando /CONNECT fd_info_srv_to_cli = open(fifo_name_info_srv_to_cli, O_WRONLY); write_to_fifo(fd_info_srv_to_cli, fifo_name_msg_cli_to_srv); free(logfile); free(fifo_name_cmd_cli_to_srv); close(fd_cmd_cli_to_srv); unlink(fifo_name_cmd_cli_to_srv); unlink(fifo_name_msg_cli_to_srv); unlink(fifo_name_msg_srv_to_cli); unlink(fifo_name_info_srv_to_cli); printf("fim\n"); return 0; } client.c: #include <stdio.h> #include <stdlib.h> #include <string.h> #include <unistd.h> #include <sys/types.h> #include <sys/stat.h> #include <fcntl.h> #include <errno.h> #define MAX_INPUT_LENGTH 100 #define PID_BUFFER_LEN 10 #define FIFO_NAME_CMD_CLI_TO_SRV "lrc_cmd_cli_to_srv" #define FIFO_NAME_INFO_SRV_TO_CLI "lrc_info_srv_to_cli" #define FIFO_NAME_MSG_SRV_TO_CLI "lrc_msg_srv_to_cli" #define COMMAND_MAX_LEN 100 #define FIFO_DIR "/tmp/" typedef enum { false, true } bool; char* nickname; char* name; char* email; void write_to_fifo(int fd, char* data) { int n_bytes = (strlen(data)+1) * sizeof(char); write(fd, &n_bytes, sizeof(int)); //primeiro envia o numero de bytes que a proxima instrucao write ira enviar write(fd, data, n_bytes); printf("writing %d bytes '%s'\n", n_bytes, data); } void read_from_fifo(int fd, char** var) { int n_bytes; read(fd, &n_bytes, sizeof(int)); *var = (char *) malloc (n_bytes); printf("read '%s'\n", *var); read(fd, *var, n_bytes); } int main(int argc, char* argv[]) { pid_t pid = getpid(); //CRIA FIFO INFO_SRV_TO_CLIXXX char pid_string[PID_BUFFER_LEN]; sprintf(pid_string, "%d", pid); char* fifo_name_info_srv_to_cli; fifo_name_info_srv_to_cli = (char *) malloc ( (strlen(FIFO_DIR) + strlen(FIFO_NAME_INFO_SRV_TO_CLI) + strlen(pid_string) + 1 ) * sizeof(char) ); strcpy(fifo_name_info_srv_to_cli, FIFO_DIR); strcat(fifo_name_info_srv_to_cli, FIFO_NAME_INFO_SRV_TO_CLI); strcat(fifo_name_info_srv_to_cli, pid_string); int n = mkfifo(fifo_name_info_srv_to_cli, 0660); if (n < 0) { fprintf(stderr, "error creating %s\n", fifo_name_info_srv_to_cli); fprintf(stderr, "errno: %d\n", errno); exit(6); } int fd_cmd_cli_to_srv, fd_info_srv_to_cli; fd_cmd_cli_to_srv = open("/tmp/lrc_cmd_cli_to_srv", O_WRONLY); char command[COMMAND_MAX_LEN]; printf("> "); scanf("%s", command); while (strcmp(command, "/CONNECT")) { printf("O primeiro comando deverá ser \"/CONNECT\"\n"); printf("> "); scanf("%s", command); } //CRIA FIFO MSG_SRV_TO_CLIXXX char* fifo_name_msg_srv_to_cli; fifo_name_msg_srv_to_cli = (char *) malloc ( (strlen(FIFO_DIR) + strlen(FIFO_NAME_MSG_SRV_TO_CLI) + strlen(pid_string) + 1) * sizeof(char) ); strcpy(fifo_name_msg_srv_to_cli, FIFO_DIR); strcat(fifo_name_msg_srv_to_cli, FIFO_NAME_MSG_SRV_TO_CLI); strcat(fifo_name_msg_srv_to_cli, pid_string); n = mkfifo(fifo_name_msg_srv_to_cli, 0660); if (n < 0) { fprintf(stderr, "error creating %s\n", fifo_name_info_srv_to_cli); fprintf(stderr, "errno: %d\n", errno); exit(7); } // ENVIA COMANDO /CONNECT write_to_fifo(fd_cmd_cli_to_srv, pid_string); //envia pid do processo cliente write_to_fifo(fd_cmd_cli_to_srv, command); //envia commando /CONNECT write_to_fifo(fd_cmd_cli_to_srv, fifo_name_info_srv_to_cli); //envia nome de fifo INFO_SRV_TO_CLIXXX write_to_fifo(fd_cmd_cli_to_srv, fifo_name_msg_srv_to_cli); //envia nome de fifo MSG_TO_SRV_TO_CLIXXX // recebe do servidor a resposta ao comanddo /CONNECT printf("msg1\n"); printf("vamos tentar abrir %s\n", fifo_name_info_srv_to_cli); fd_info_srv_to_cli = open(fifo_name_info_srv_to_cli, O_RDONLY); printf("%s aberto", fifo_name_info_srv_to_cli); if (fd_info_srv_to_cli < 0) { fprintf(stderr, "erro ao criar %s\n", fifo_name_info_srv_to_cli); fprintf(stderr, "errno: %d\n", errno); } printf("msg2\n"); char* fifo_name_msg_cli_to_srv; printf("msg3\n"); read_from_fifo(fd_info_srv_to_cli, &fifo_name_msg_cli_to_srv); printf("msg4\n"); free(nickname); free(name); free(email); free(fifo_name_info_srv_to_cli); free(fifo_name_msg_srv_to_cli); unlink(fifo_name_msg_srv_to_cli); unlink(fifo_name_info_srv_to_cli); printf("fim\n"); return 0; } makefile: CC = gcc CFLAGS = -Wall -lpthread -fno-stack-protector all: client server client: client.c $(CC) $(CFLAGS) client.c -o client server: server.c $(CC) $(CFLAGS) server.c -o server clean: rm -f client server *~

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  • Help with malloc and free: Glibc detected: free(): invalid pointer

    - by nunos
    I need help with debugging this piece of code. I know the problem is in malloc and free but can't find exactly where, why and how to fix it. Please don't answer: "Use gdb" and that's it. I would use gdb to debug it, but I still don't know much about it and am still learning it, and would like to have, in the meanwhile, another solution. Thanks. #include <stdio.h> #include <stdlib.h> #include <ctype.h> #include <unistd.h> #include <string.h> #include <sys/wait.h> #include <sys/types.h> #define MAX_COMMAND_LENGTH 256 #define MAX_ARGS_NUMBER 128 #define MAX_HISTORY_NUMBER 100 #define PROMPT ">>> " int num_elems; typedef enum {false, true} bool; typedef struct { char **arg; char *infile; char *outfile; int background; } Command_Info; int parse_cmd(char *cmd_line, Command_Info *cmd_info) { char *arg; char *args[MAX_ARGS_NUMBER]; int i = 0; arg = strtok(cmd_line, " "); while (arg != NULL) { args[i] = arg; arg = strtok(NULL, " "); i++; } num_elems = i;precisa em free_mem if (num_elems == 0) return 0; cmd_info->arg = (char **) ( malloc(num_elems * sizeof(char *)) ); cmd_info->infile = NULL; cmd_info->outfile = NULL; cmd_info->background = 0; bool b_infile = false; bool b_outfile = false; int iarg = 0; for (i = 0; i < num_elems; i++) { if ( !strcmp(args[i], "<") ) { if ( b_infile || i == num_elems-1 || !strcmp(args[i+1], "<") || !strcmp(args[i+1], ">") || !strcmp(args[i+1], "&") ) return -1; i++; cmd_info->infile = malloc(strlen(args[i]) * sizeof(char)); strcpy(cmd_info->infile, args[i]); b_infile = true; } else if (!strcmp(args[i], ">")) { if ( b_outfile || i == num_elems-1 || !strcmp(args[i+1], ">") || !strcmp(args[i+1], "<") || !strcmp(args[i+1], "&") ) return -1; i++; cmd_info->outfile = malloc(strlen(args[i]) * sizeof(char)); strcpy(cmd_info->outfile, args[i]); b_outfile = true; } else if (!strcmp(args[i], "&")) { if ( i == 0 || i != num_elems-1 || cmd_info->background ) return -1; cmd_info->background = true; } else { cmd_info->arg[iarg] = malloc(strlen(args[i]) * sizeof(char)); strcpy(cmd_info->arg[iarg], args[i]); iarg++; } } cmd_info->arg[iarg] = NULL; return 0; } void print_cmd(Command_Info *cmd_info) { int i; for (i = 0; cmd_info->arg[i] != NULL; i++) printf("arg[%d]=\"%s\"\n", i, cmd_info->arg[i]); printf("arg[%d]=\"%s\"\n", i, cmd_info->arg[i]); printf("infile=\"%s\"\n", cmd_info->infile); printf("outfile=\"%s\"\n", cmd_info->outfile); printf("background=\"%d\"\n", cmd_info->background); } void get_cmd(char* str) { fgets(str, MAX_COMMAND_LENGTH, stdin); str[strlen(str)-1] = '\0'; } pid_t exec_simple(Command_Info *cmd_info) { pid_t pid = fork(); if (pid < 0) { perror("Fork Error"); return -1; } if (pid == 0) { if ( (execvp(cmd_info->arg[0], cmd_info->arg)) == -1) { perror(cmd_info->arg[0]); exit(1); } } return pid; } void type_prompt(void) { printf("%s", PROMPT); } void syntax_error(void) { printf("msh syntax error\n"); } void free_mem(Command_Info *cmd_info) { int i; for (i = 0; cmd_info->arg[i] != NULL; i++) free(cmd_info->arg[i]); free(cmd_info->arg); free(cmd_info->infile); free(cmd_info->outfile); } int main(int argc, char* argv[]) { char cmd_line[MAX_COMMAND_LENGTH]; Command_Info cmd_info; //char* history[MAX_HISTORY_NUMBER]; while (true) { type_prompt(); get_cmd(cmd_line); if ( parse_cmd(cmd_line, &cmd_info) == -1) { syntax_error(); continue; } if (!strcmp(cmd_line, "")) continue; if (!strcmp(cmd_info.arg[0], "exit")) exit(0); pid_t pid = exec_simple(&cmd_info); waitpid(pid, NULL, 0); free_mem(&cmd_info); } return 0; }

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  • Draw arrow on line algorithm

    - by nunos
    Does anyone have an algorithm for drawing an arrow in the middle of a given line. I have searched for google but haven't found any good implementation. P.S. I really don't mind the language, but it would be great if it was Java, since it is the language I am using for this. Thanks in advance.

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  • How to skip parentheses on Netbeans with enter?

    - by nunos
    So I have been programming in C++ with Eclipse and have the habit of hitting enter to skip parentheses (anyone who has ever used eclipse probably knows what I am talking about). I have recently started learning Java and decided to use NetBeans, mostly due to the much more simple interface. However, I would like to know if there a way to skip the (), [], < and "" on enter just like what happens in Eclipse in NetBeans. Thanks.

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  • Pipe implementation

    - by nunos
    I am trying to implement a linux shell that supports piping. I have already done simple commands, commands running in background, redirections, but piping is still missing. I have already read about it and seen some snippets of code, but still haven't been able to sort out a working solution. What I have so far: int fd[2]; pid_t pid = fork(); if (pid == -1) return -1; if (pid == 0) { dup2(0, fd[0]); execlp("sort", "sort", NULL); } I am a novice programmer, as you can probably tell, and when I am programming something I don't know much about, this being obviously the case, I like to start with something really easy and concrete and then build from there. So, before being able to implement three and more different commands in pipeline, I would like to be able to compute "ls names.txt | sort" or something similiar, in which names.txt is a file of names alfabetically unordered. Thanks.

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  • Bidirectional FIFO

    - by nunos
    I would like to implement a bidirectional fifo. The code below is functioning but it is not using bidirectional fifo. I have searched all over the internet, but haven't found any good example... How can I do that? Thanks, WRITER.c: #include <stdio.h> #include <unistd.h> #include <string.h> #include <sys/types.h> #include <sys/wait.h> #include <fcntl.h> #define MAXLINE 4096 #define READ 0 #define WRITE 1 int main (int argc, char** argv) { int a, b, fd; do { fd=open("/tmp/myfifo",O_WRONLY); if (fd==-1) sleep(1); } while (fd==-1); while (1) { scanf("%d", &a); scanf("%d", &b); write(fd,&a,sizeof(int)); write(fd,&b,sizeof(int)); if (a == 0 && b == 0) { break; } } close(fd); return 0; } READER.c: #include <stdio.h> #include <unistd.h> #include <string.h> #include <sys/types.h> #include <sys/wait.h> #include <fcntl.h> #include <sys/stat.h> #define MAXLINE 4096 #define READ 0 #define WRITE 1 int main(void) { int n1, n2; int fd; mkfifo("/tmp/myfifo",0660); fd=open("/tmp/myfifo",O_RDONLY); while(read(fd, &n1, sizeof(int) )) { read(fd, &n2, sizeof(int)); if (n1 == 0 && n2 == 0) { break; } printf("soma: %d\n",n1+n2); printf("diferenca: %d\n", n1-n2); printf("divisao: %f\n", n1/(double)n2); printf("multiplicacao: %d\n", n1*n2); } close(fd); return 0; }

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  • Listing directories in Linux from C

    - by nunos
    I am trying to simulate linux command ls using linux api from c. Looking at the code it does make sense, but when I run it I get "stat error: No such file or directory". I have checked that opendir is working ok. I think the problem is in stat, which is returning -1 even though I think it should return 0. What am I missing? Thanks for your help. #include <stdio.h> #include <stdlib.h> #include <string.h> #include <dirent.h> #include <sys/stat.h> #include <errno.h> int main(int argc, char *argv[]) { DIR *dirp; struct dirent *direntp; struct stat stat_buf; char *str; if (argc != 2) { fprintf( stderr, "Usage: %s dir_name\n", argv[0]); exit(1); } if ((dirp = opendir( argv[1])) == NULL) { perror(argv[1]); exit(2); } while ((direntp = readdir( dirp)) != NULL) { if (stat(direntp->d_name, &stat_buf)==-1) { perror("stat ERROR"); exit(3); } if (S_ISREG(stat_buf.st_mode)) str = "regular"; else if (S_ISDIR(stat_buf.st_mode)) str = "directory"; else str = "other"; printf("%-25s - %s\n", direntp->d_name, str); } closedir(dirp); exit(0); }

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  • FIFOs implementation

    - by nunos
    Consider the following code: writer.c mkfifo("/tmp/myfifo", 0660); int fd = open("/tmp/myfifo", O_WRONLY); char *foo, *bar; ... write(fd, foo, strlen(foo)*sizeof(char)); write(fd, bar, strlen(bar)*sizeof(char)); reader.c int fd = open("/tmp/myfifo", O_RDONLY); char buf[100]; read(fd, buf, ??); My question is: Since it's not know before hand how many bytes will foo and bar have, how can I know how many bytes to read from reader.c? Because if I, for example, read 10 bytes in reader and foo and bar are together less than 10 bytes, I will have them both in the same variable and that I do not want. Ideally I would have one read function for every variable, but again I don't know before hand how many bytes will the data have. I thought about adding another write instruction in writer.c between the write for foo and bar with a separator and then I would have no problem decoding it from reader.c. Is this the way to go about it? Thanks.

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  • why is my 3n+1 problem solution wrong?

    - by nunos
    I have recently started reading "Programming Challenges" book by S. Skiena and believe or not I am kind of stuck in the very first problem. Here's a link to the problem: 3n+1 problem Here's my code: #include <iostream> #include <vector> #include <algorithm> using namespace std; unsigned long calc(unsigned long n); int main() { int i, j, a, b, m; vector<int> temp; while (true) { cin >> i >> j; if (cin.fail()) break; if (i < j) { a = i; b = j; } else { a = j; b = i; } temp.clear(); for (unsigned int k = a; k != b; k++) { temp.push_back(calc(k)); } m = *max_element(temp.begin(), temp.end()); cout << i << ' ' << j << ' ' << m << endl; } } unsigned long calc(unsigned long n) { unsigned long ret = 1; while (n != 1) { if (n % 2 == 0) n = n/2; else n = 3*n + 1; ret++; } return ret; } I know the code is inefficient and I should not be using vectors to store the data. Anyway, I still would like to know what it's wrong with this, since, for me, it makes perfect sense, even though I am getting WA (wrong answer at programming challenges judge and RE (Runtime Error) at UVa judge). Thanks.

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  • Show last command with up arrow on a linux c shell

    - by nunos
    I have implemented a simple linux shell in c. Now, I am adding some features and one I immediately thought about was to be able to show the last commands with the up arrow. Question 1: However, I have no idea how to accomplish this. Do you? Question 2: Any comment on how to store the "history" commands are also appreciated. I suppose something like a queue which allows access to all elements would be a good idea. Am I wrong? Do I have to implement it or is there already some good implementation out there I should know about? Thanks.

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  • Warning: comparison with string literals results in unspecified behaviour

    - by nunos
    So I starting the project of writing a simplified sheel for linux in c. I am not at all proficient with c nor with Linux that's exactly the reason I decided it would be a good idea. Starting with the parser, I have already encountered some problems. The code should be straightforward that's why I didn't include any comments. I am getting a warning with gcc: "comparison with string literals results in unspecified behaviour" at the lines commented with "WARNING HERE" (see code below). I have no idea why this causes an warning, but the real problem is that even though I am comparing an "<" to an "<" is doesn't get inside the if... I am looking for an answer for the problem explained, however if there's something that you see in the code that should be improved please say so. Just take in mind I am not that proficient and that this is still a work in progress (or better yet, a work in start). Thanks in advance. #include <stdio.h> #include <unistd.h> #include <string.h> typedef enum {false, true} bool; typedef struct { char **arg; char *infile; char *outfile; int background; } Command_Info; int parse_cmd(char *cmd_line, Command_Info *cmd_info) { char *arg; char *args[100]; int i = 0; arg = strtok(cmd_line, " \n"); while (arg != NULL) { args[i] = arg; arg = strtok(NULL, " \n"); i++; } int num_elems = i; cmd_info->infile = NULL; cmd_info->outfile = NULL; cmd_info->background = 0; int iarg = 0; for (i = 0; i < num_elems; i++) { if (args[i] == "&") //WARNING HERE return -1; else if (args[i] == "<") //WARNING HERE if (args[i+1] != NULL) cmd_info->infile = args[i+1]; else return -1; else if (args[i] == ">") //WARNING HERE if (args[i+1] != NULL) cmd_info->outfile = args[i+1]; else return -1; else cmd_info->arg[iarg++] = args[i]; } cmd_info->arg[iarg] = NULL; return 0; } void print_cmd(Command_Info *cmd_info) { int i; for (i = 0; cmd_info->arg[i] != NULL; i++) printf("arg[%d]=\"%s\"\n", i, cmd_info->arg[i]); printf("arg[%d]=\"%s\"\n", i, cmd_info->arg[i]); printf("infile=\"%s\"\n", cmd_info->infile); printf("outfile=\"%s\"\n", cmd_info->outfile); printf("background=\"%d\"\n", cmd_info->background); } int main(int argc, char* argv[]) { char cmd_line[100]; Command_Info cmd_info; printf(">>> "); fgets(cmd_line, 100, stdin); parse_cmd(cmd_line, &cmd_info); print_cmd(&cmd_info); return 0; }

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  • Help with simple linux shell implementation

    - by nunos
    I am implementing a simple version of a linux shell in c. I have succesfully written the parser, but I am having some trouble forking out the child process. However, I think the problem is due to arrays, pointers and such, because just started C with this project and am not still very knowledgable with them. I am getting a segmentation fault and don't know where from. Any help is greatly appreciated. #include <stdio.h> #include <stdlib.h> #include <unistd.h> #include <string.h> #include <sys/wait.h> #include <sys/types.h> #define MAX_COMMAND_LENGTH 250 #define MAX_ARG_LENGTH 250 typedef enum {false, true} bool; typedef struct { char **arg; char *infile; char *outfile; int background; } Command_Info; int parse_cmd(char *cmd_line, Command_Info *cmd_info) { char *arg; char *args[MAX_ARG_LENGTH]; int i = 0; arg = strtok(cmd_line, " "); while (arg != NULL) { args[i] = arg; arg = strtok(NULL, " "); i++; } int num_elems = i; if (num_elems == 0) return -1; cmd_info->infile = NULL; cmd_info->outfile = NULL; cmd_info->background = 0; int iarg = 0; for (i = 0; i < num_elems-1; i++) { if (!strcmp(args[i], "<")) { if (args[i+1] != NULL) cmd_info->infile = args[++i]; else return -1; } else if (!strcmp(args[i], ">")) { if (args[i+1] != NULL) cmd_info->outfile = args[++i]; else return -1; } else cmd_info->arg[iarg++] = args[i]; } if (!strcmp(args[i], "&")) cmd_info->background = true; else cmd_info->arg[iarg++] = args[i]; cmd_info->arg[iarg] = NULL; return 0; } void print_cmd(Command_Info *cmd_info) { int i; for (i = 0; cmd_info->arg[i] != NULL; i++) printf("arg[%d]=\"%s\"\n", i, cmd_info->arg[i]); printf("arg[%d]=\"%s\"\n", i, cmd_info->arg[i]); printf("infile=\"%s\"\n", cmd_info->infile); printf("outfile=\"%s\"\n", cmd_info->outfile); printf("background=\"%d\"\n", cmd_info->background); } void get_cmd(char* str) { fgets(str, MAX_COMMAND_LENGTH, stdin); str[strlen(str)-1] = '\0'; //apaga o '\n' do fim } pid_t exec_simple(Command_Info *cmd_info) { pid_t pid = fork(); if (pid < 0) { perror("Fork Error"); return -1; } if (pid == 0) { execvp(cmd_info->arg[0], cmd_info->arg); perror(cmd_info->arg[0]); exit(1); } return pid; } int main(int argc, char* argv[]) { while (true) { char cmd_line[MAX_COMMAND_LENGTH]; Command_Info cmd_info; printf(">>> "); get_cmd(cmd_line); if ( (parse_cmd(cmd_line, &cmd_info) == -1) ) return -1; parse_cmd(cmd_line, &cmd_info); if (!strcmp(cmd_info.arg[0], "exit")) exit(0); pid_t pid = exec_simple(&cmd_info); waitpid(pid, NULL, 0); } return 0; } Thanks.

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  • Simple popup java form with at least two fields

    - by nunos
    This is, I suppose, a very simple question for any Java programmer. My question is as simple as this: When the user clicks a button, I want to show a popup form that should have at least two JTextFields and two JLabels, so using JOptionPane.showInputDialog is not a possibility. Thanks.

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  • C readline function

    - by nunos
    In an assignment for college it was suggested to use the c readline function in an exercise. I have searched for its reference but still haven't found it. Does it really exist? In which header? Can you please post the link to the reference? Thanks.

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  • Declaring a function inside a function?

    - by nunos
    I have came across the following code, and being a c beginner, I came here for your help. This function is from a c implmentation of a queue. Bool queuePut(Queue *q, char c) { void beep(); if (queueFull(q)) { beep(); return false; } //do stuff return true; } So, I am getting a strange error with gcc on the void beep(). Can someone please explain me what is this, declaring a function inside a function. Or is it the void beep() simply out of place? I was given this code and there's always the possibility that it isn't correct. Thanks.

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