我正在尝试用C语言编写自己的函数。首先,一旦我已经分叉并且在孩子身上,我检查以检测管道输入的位置,如下所示。我循环遍历我自己创建的StringArray(sa),然后将令牌复制到char cmd1 [64](先前已初始化)和令牌之后复制到char cmd2 [64]。 int管道被赋予下一步的值。
if(strcmp(sa[i], "|") == 0)
{
printf("got to the pipe\n");
sa[i]=NULL;
strcpy(cmd1, sa[i-1]);
strcpy(cmd2, sa[i+1]);
piping=2;
}
然后,程序达到了这个声明:
if (piping)
{
if (pipe(fd) != 0){
perror("Failed to create pipe");
}
if ((cpPid = fork()) == -1){
perror("Failed to fork");
}
if (cpPid == 0){
dup2(fd[1], 1);
close(fd[0]);
close(fd[1]);
execvp(cmd1, sa);
error("failed to exec command 1");
}
else
{
dup2(fd[0], 0);
close(fd[0]);
close(fd[1]);
execvp(cmd2, sa);
error("failed to exec command 2");
}
}
我的程序完全崩溃并且只发出未知错误10275024.有人可以帮我弄清楚出了什么问题吗?
答案 0 :(得分:0)
请查看这个可以帮助您使用C执行管道的工作示例。
#include <assert.h>
#include <stdio.h>
#include <sys/wait.h>
#include <unistd.h>
#include <stdlib.h>
#include <memory.h>
#include <errno.h>
typedef int Pipe[2];
/* exec_nth_command() and exec_pipe_command() are mutually recursive */
static void exec_pipe_command(int ncmds, char ***cmds, Pipe output);
static void err_vsyswarn(char const *fmt, va_list args) {
int errnum = errno;
vfprintf(stderr, fmt, args);
if (errnum != 0)
fprintf(stderr, " (%d: %s)", errnum, strerror(errnum));
putc('\n', stderr);
}
static void err_syswarn(char const *fmt, ...) {
va_list args;
err_vsyswarn(fmt, args);;
}
static void err_sysexit(char const *fmt, ...) {
va_list args;
err_vsyswarn(fmt, args);
exit(1);
}
/* With the standard output plumbing sorted, execute Nth command */
static void exec_nth_command(int ncmds, char ***cmds) {
assert(ncmds >= 1);
if (ncmds > 1) {
pid_t pid;
Pipe input;
if (pipe(input) != 0)
err_sysexit("Failed to create pipe");
if ((pid = fork()) < 0)
err_sysexit("Failed to fork");
if (pid == 0) {
/* Child */
exec_pipe_command(ncmds - 1, cmds, input);
}
/* Fix standard input to read end of pipe */
dup2(input[0], 0);
close(input[0]);
close(input[1]);
}
execvp(cmds[ncmds - 1][0], cmds[ncmds - 1]);
err_sysexit("Failed to exec %s", cmds[ncmds - 1][0]);
/*NOTREACHED*/
}
/* exec_nth_command() and exec_pipe_command() are mutually recursive */
/* Given pipe, plumb it to standard output, then execute Nth command */
static void exec_pipe_command(int ncmds, char ***cmds, Pipe output) {
assert(ncmds >= 1);
/* Fix stdout to write end of pipe */
dup2(output[1], 1);
close(output[0]);
close(output[1]);
exec_nth_command(ncmds, cmds);
}
/* Execute the N commands in the pipeline */
static void exec_pipeline(int ncmds, char ***cmds) {
assert(ncmds >= 1);
pid_t pid;
if ((pid = fork()) < 0)
err_syswarn("Failed to fork");
if (pid != 0)
return;
exec_nth_command(ncmds, cmds);
}
/* who | awk '{print $1}' | sort | uniq -c | sort -n */
static char *cmd0[] = {"who", 0};
static char *cmd1[] = {"awk", "{print $1}", 0};
static char *cmd2[] = {"sort", 0};
static char *cmd3[] = {"uniq", "-c", 0};
static char *cmd4[] = {"sort", "-n", 0};
static char **cmds[] = {cmd0, cmd1, cmd2, cmd3, cmd4};
static int ncmds = sizeof(cmds) / sizeof(cmds[0]);
int main(int argc, char **argv) {
exec_pipeline(ncmds, cmds);
return (0);
}
代码中的管道示例是
who | awk '{print $1}' | sort | uniq -c | sort -n
您可以使用任何管道。
测试:
dac@dac-Latitude-E7450 ~/C/>
who | awk '{print $1}' | sort | uniq -c | sort -n
1 dac
dac@dac-Latitude-E7450 ~/C/> gcc main.c
dac@dac-Latitude-E7450 ~/C/> ./a.out
1 dac