我正在嘗試實作一個代碼ls| grep "pipes"| wc -l。為此，我創建了 3 個子行程并使用了 2 個管道。請找到使用的代碼：
#include <stdio.h>
#include <errno.h>
#include <unistd.h>
#include <sys/wait.h>
#include <stdlib.h>
int main(void)
{
//pid_t p;
int pfds1[2],pfds2[2],s;
pipe(pfds1);
pipe(pfds2);
if(!fork()) //first child ls
{
//printf("ls ppid is:%d\n", getppid());
dup2(pfds1[1],1);
close(pfds1[0]);
close(pfds2[0]);
close(pfds2[1]);
close(pfds1[1]);
if(execlp("ls", "ls", NULL)==-1)
{
perror("Error in exec line 24\n");
exit(1);
}
}
else{
if(!fork())
{
//printf("grep ppid is:%d\n", getppid());
dup2(pfds1[0],0);
dup2(pfds2[1],1);
close(pfds1[0]);
close(pfds2[0]);
close(pfds2[1]);
close(pfds1[1]);
if(execlp("grep","grep","pipes",NULL)==-1)
{
perror("Error in exec line 41\n");
exit(1);
}
}
else{
if(!fork())
{ //printf("wc ppid is:%d\n", getppid());
dup2(pfds2[0],0);
close(pfds1[0]);
close(pfds2[0]);
close(pfds2[1]);
close(pfds1[1]);
if(execlp("wc","wc","-l",NULL)==-1)
{
perror("Error in exec line 56\n");
exit(1);
}
}
else{
close(pfds1[0]);
close(pfds2[0]);
close(pfds1[1]);
close(pfds1[1]);
wait(&s);
wait(&s);
wait(&s);
//printf("parent pid is:%d\n", getpid());
//printf("grandparent pid is:%d\n", getppid());
exit(0);
}
}
}
}
當僅wait(&s)使用 2 個而不是 3 個時，即使存在三個子行程，代碼也能按預期作業。當前代碼卡住并且沒有完成執行。有人可以詳細說明為什么會這樣嗎？
謝謝
uj5u.com熱心網友回復：
首先，您的代碼中有一個錯字，導致最后一個行程中管道的一端保持打開狀態，這使得阻塞發生。見下文。
問題是，當你分叉你的兩個檔案描述符（你正在關閉一個，未使用的，在子行程中，在呼叫 exec 之前）轉換為四個（兩個在子行程中，兩個在父行程中，你不'不關閉父行程中的檔案描述符之一）
如果您不關閉不使用的描述符（在父行程或子行程中），則管道的兩個描述符仍將打開，并且在這種情況下，讀取程序被阻止，等待一些輸入（或EOF）到來。當你創建一個管道時（通過使用pipe(2)系統呼叫，一旦你fork()關閉你不打算使用的管道的檔案描述符，因為你可以阻塞，因為一切都完成了，但你仍然等待輸入的到來（被父行程或子行程，具體取決于您如何組織資訊流）
#include <stdio.h>
#include <errno.h>
#include <unistd.h>
#include <sys/wait.h>
#include <stdlib.h>
int main(void)
{
//pid_t p;
int pfds1[2],pfds2[2], s;
pipe(pfds1);
pipe(pfds2);
if(!fork()) { //first child ls
//printf("ls ppid is:%d\n", getppid());
dup2(pfds1[1],1);
在這里，您已將管道的寫入端安裝為 ls 的標準輸出，但您沒有讀取它。
close(pfds1[0]);
close(pfds2[0]);
close(pfds2[1]);
close(pfds1[1]);
還行吧。所有檔案描述符將在退出時關閉，因此無需顯式關閉它們。
if(execlp("ls", "ls", NULL)==-1) {
perror("Error in exec line 24\n");
exit(1);
}
} else { // parent
if(!fork()) {
//printf("grep ppid is:%d\n", getppid());
dup2(pfds1[0],0);
dup2(pfds2[1],1);
Here you connect the standard input to the reading side of pipe1 and the standard output of pipe2 to the standard input. grep will wait for input on that pipe.
close(pfds1[0]);
close(pfds2[0]);
close(pfds2[1]);
close(pfds1[1]);
as said before, all pipes will be closed when grep finishes... so no need to explicitly close them here.
if(execlp("grep","grep","pipes",NULL)==-1) {
perror("Error in exec line 41\n");
exit(1);
}
} else { // parent
if(!fork()) {
//printf("wc ppid is:%d\n", getppid());
dup2(pfds2[0],0);
you connect also the input edge of pipe2 to standard input, so wc -l will wait for your input on pipe2.
close(pfds1[0]);
close(pfds2[0]);
close(pfds2[1]);
close(pfds1[1]);
as always, no problem if you don't close because this process will close all of them when it finishes.
if(execlp("wc","wc","-l",NULL)==-1) {
perror("Error in exec line 56\n");
exit(1);
}
} else{ // parent
close(pfds1[0]);
close(pfds2[0]);
close(pfds1[1]);
close(pfds1[1]); /* mistake!!!! */
You have a typo above, you close twice the descriptor pdfs1[1] and pdfs2[1] remains open, so the last process remains waiting for more input on the pipe, and never ends.
wait(&s);
wait(&s);
wait(&s);
//printf("parent pid is:%d\n", getpid());
//printf("grandparent pid is:%d\n", getppid());
exit(0);
}
}
}
}
Edit
The pipe requires that all the descriptors belonging to the write side to be closed, in order to notify the readers that end of file has occured and so, unlock the processes read()ing on them. In the fork, the descriptors of the pipe are implicitly dup()ed, and you again dup() them, when you redirect the input or output, by doing an explicit loop. The children processes just need to close the descriptors they are not going to use (you do well, when you do the dup() to redirect, and then close all the pipe descriptors)
In your case you use the pipes just to connect the children, and no communication is made from the parent... but its pipe descriptors also count to consider is EOF will be signalled to a reader. For a reader to be unlocked from read with EOF, you need to close all the dupped writing descriptors. This includes the ones of the parent, and the ones of the other children.
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