我正在嘗試編寫生成兩個子進程的代碼,這兩個子進程通過管道相互發送消息,然後終止。但是,當我運行下面的代碼時,只有child2打印它的問候語,但是child1仍然打印它從child2獲得的消息,其中child1沒有。兩個子進程與管道之間進行通信
有沒有人知道我的方法有什麼問題?
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <sys/types.h>
#include <unistd.h>
int main(int argc, char** argv) {
char chld_1_send[20] = "hello from child 1";
char chld_1_recv[20];
char chld_2_send[20] = "hi from child 2";
char chld_2_recv[20];
int chld_1_outgoing[2];
int chld_2_outgoing[2];
pipe(chld_1_outgoing);
pipe(chld_2_outgoing);
int chld_1_status, chld_2_status;
pid_t chld_1, chld_2;
chld_1 = fork();
if(chld_1 == 0) {
chld_2 = fork();
}
if(chld_1 == 0 && chld_2 == 0) {
printf("parent [pid:%d] waiting on both children to finish\n", getpid());
while(wait(&chld_1_status) != chld_1 && wait(&chld_2_status) != chld_2) {}
printf("done waiting\n");
}
else if(chld_1 != 0 && chld_2 == 0) {
printf("this is child 1 [pid:%d] with parent [pid:%d]\n", getpid(), getppid());
write(chld_1_outgoing[1], chld_1_send, strlen(chld_1_send));
while(read(chld_1_outgoing[0], &chld_1_recv, sizeof(chld_2_recv)) < 0) {}
printf("child 2 said '%s'\n", chld_1_recv);
exit(0);
}
else if(chld_2 != 0 && chld_1 == 0) {
printf("this is child 2 [pid:%d] with parent [pid:%d]\n", getpid(), getppid());
write(chld_2_outgoing[1], chld_2_send, strlen(chld_2_send));
while(read(chld_2_outgoing[0], &chld_2_recv, sizeof(chld_2_recv)) < 0) {}
printf("child 1 said '%s'\n", chld_2_recv);
exit(0);
}
printf("both children have terminated successfully\n");
return 0;
}
然而,運行此命令打印出該終端和進入一個無限循環:
$ this is child 2 [pid:15713] with parent [pid:1]
child 1 said 'hi from child 2'
parent [pid:15714] waiting on both children to finish
http://developers.sun.com/solaris/articles/named_pipes.html – Anders
是否有沒有命名管道做到這一點?我的意思是一個管道仍然能夠與另一個溝通,而不是另一個孩子出於某種原因。 – David
我改變了waitpid調用等待。但是,現在該程序進入無限循環。 – David