당신이 C.
를 사용하여 파이프 라인을 실행 도움이 될 것입니다이 작업 예에서 참조하시기 바랍니다
#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);
}
코드에서 파이프 라인의 예는
who | awk '{print $1}' | sort | uniq -c | sort -n
당신은 어떤 파이프 라인을 사용할 수있다 .
시험 : 말할 것도 MCVE없이
[email protected] ~/C/>
who | awk '{print $1}' | sort | uniq -c | sort -n
1 dac
[email protected] ~/C/> gcc main.c
[email protected] ~/C/> ./a.out
1 dac
없다. –
나는 숙제를 위해 똑같은 것을 만들었고, 이것은 나를 도왔다 http://stackoverflow.com/questions/948221/does-this-multiple-pipes-code-in-c-makes-sense –
@EugeneSh. MCVE 란 무엇입니까? –