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我想實現這樣的場景：並行處理方案實現停留在等待輸入
其中ls -l和cat /etc/group進程中運行並行，並tr /a-z/ /A-Z/需要從他們組合的輸入。然後將它們的大寫版本轉移到cat -n和grep Z（它們也應該並行運行）。 cat -n和grep Z應輸出到stdout。
所以在stdoutput，
(ls -l ; cat /etc/group) | tr /a-z/ /A-Z/ | cat -n
(ls -l ; cat /etc/group) | tr /a-z/ /A-Z/ | grep A
應打印。
我寫這個程序來實現我的目標：
#include<stdio.h>
#include<unistd.h>
#include<wait.h>
int main()
{
int dummy, i;
size_t size = 0;
int pipe1[2];
int pipe2[2];
int pipe3[2];
char* lsargs0[] = {"/bin/ls", "-l", NULL};
char* lsargs1[] = {"cat", "/etc/group", NULL};
char* lsargs2[] = {"tr", "/a-z/", "/A-Z/", NULL};
char* lsargs3[] = {"cat", "-n", NULL};
char* lsargs4[] = {"grep", "A", NULL};
char** am[] = {lsargs0, lsargs1, lsargs2, lsargs3, lsargs4};
char buf[4096];
pipe(pipe1);
pipe(pipe2);
pipe(pipe3);
for (i = 0 ; i < 5 ; i++)
{
pid_t pid = fork();
if (pid == 0)
{
if (i == 0 || i == 1) // ls -l & cat
{
// 0 -> stdin, 1 -> pipe write end
close(pipe2[0]);
close(pipe2[1]);
close(pipe3[0]);
close(pipe3[1]);
close(pipe1[0]);
dup2(pipe1[1], 1);
close(pipe1[1]);
execvp(am[i][0], am[i]);
}
else if (i == 2) //TR AZ AZ
{
int extraPipe[2];
pipe(extraPipe);
int forkMaster = fork();
if (forkMaster != 0)
{
printf("TR az AZ fork parent\n");
//parent
close(pipe2[0]);
close(pipe2[1]);
close(pipe3[0]);
close(pipe3[1]);
close(pipe1[1]);
dup2(pipe1[0], 0);
close(pipe1[0]);
close(extraPipe[0]);
dup2(extraPipe[1], 1);
close(extraPipe[1]);
//waitpid(forkMaster, &dummy, WNOHANG);
//wait(&dummy);
execvp(am[i][0], am[i]);
}
else
{
printf("TR az AZ child\n");
//child
read(extraPipe[0], buf, 4095);
//printf("%s\n", buf);
write(pipe2[1], buf, size);
write(pipe3[1], buf, size);
close(pipe1[1]);
close(pipe1[0]);
close(extraPipe[0]);
close(extraPipe[1]);
close(pipe2[0]);
close(pipe2[1]);
close(pipe3[0]);
close(pipe3[1]);
break;
}
}
else if (i == 3) // cat -n
{
close(pipe1[0]);
close(pipe1[1]);
close(pipe3[0]);
close(pipe3[1]);
close(pipe2[1]);
dup2(pipe2[0], 0);
close(pipe2[0]);
execvp(am[i][0], am[i]);
}
else if (i == 4) //grep Z
{
close(pipe1[0]);
close(pipe1[1]);
close(pipe2[0]);
close(pipe2[1]);
close(pipe3[1]);
dup2(pipe3[0], 0);
close(pipe3[0]);
execvp(am[i][0], am[i]);
}
break;
}
//DON'T FORGET TO CLOSE PIPES ON PARENT
else if (i == 4)
{
close(pipe1[0]);
close(pipe1[1]);
close(pipe2[0]);
close(pipe2[1]);
close(pipe3[0]);
close(pipe3[1]);
}
}
for (i = 0 ; i < 5 ; i++)
{
wait(&dummy);
}
return 0;
}
但在read(extraPipe[0], buf, 4096);行程序stucks。我究竟做錯了什麼？你可以幫我嗎？
'small.c：97：29：error：'n'may be uninitialized ...'你編譯的代碼作爲例子嗎？發佈之前您需要測試它。 – jdarthenay
我用gcc成功編譯過。但是對於你，我把你的編譯器錯誤的那一行移開了，你能再試一次嗎？ –
什麼是您使用的編譯命令？我推薦至少'-Wall'，你會被警告你的varriable沒有初始化（這會導致未定義的行爲）。 – jdarthenay