1
我想实现这样的场景：并行处理方案实现停留在等待输入
其中ls -l和cat /etc/group进程中运行并行，并tr /a-z/ /A-Z/需要从他们组合的输入。然后将它们的大写版本转移到cat -n和grep Z（它们也应该并行运行）。 cat -n和grep Z应输出到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
应打印。
我写这个程序来实现我的目标：
#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。我究竟做错了什么？你可以帮我吗？
'small.c：97：29：error：'n'may be uninitialized ...'你编译的代码作为例子吗？发布之前您需要测试它。 – jdarthenay
我用gcc成功编译过。但是对于你，我把你的编译器错误的那一行移开了，你能再试一次吗？ –
什么是您使用的编译命令？我推荐至少'-Wall'，你会被警告你的varriable没有初始化（这会导致未定义的行为）。 – jdarthenay