为什么与管道信号量同步以这种奇怪的方式表现？

Question

我正在尝试解决一些简单的流程同步示例；在此特定示例中，我需要三个过程：一个打印“ A”，另一个打印“ B”，最后一个打印“ C”。我要先打印“ A”，然后再打印“ B”或“ C”，然后再打印“ A”，“ B”或“ C”。更清楚地说，我想要一个输出模式，例如“ ABACACABACABABACA ....”

下面有我的简单代码，不对终止进行管理，因为这只是了解事情运作方式的一个示例。

#include <stdio.h>
#include <stdlib.h>
#include <fcntl.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>




void semaphore_init (int* sw){
if (pipe(sw) == -1) {
printf ("Error\n");
exit (-1);
     }
}

void semaphore_wait (int* sw){
char buffer;
if (read(sw[0],&buffer,1) != 1) {
printf ("Error\n");
exit (-1);
     }
}


void semaphore_signal (int* sw){
if (write(sw[1],"X",1) != 1) {
printf ("Error\n");
exit (-1);
     }
}

int s1[2];
int s2[2];

void childA();
void childB();
void childC();

int main(){

pid_t pid;

semaphore_init(s1);
semaphore_init(s2);

semaphore_signal(s1);

pid=fork();
if(pid==0) childA();
else{
    pid=fork();
    if (pid==0) childB();
    else {
        pid=fork();
        if(pid==0) childC();
    }
}
wait(NULL);
return 0;
}

void childA(){

while(1){
    semaphore_wait(s1);
    printf("A-");
    semaphore_signal(s2);
   }
exit(0);
}

void childB(){

while(1){
    semaphore_wait(s2);
    printf("B-");
    semaphore_signal(s1);
   }
exit(0);
}

void childC(){

while(1){
    semaphore_wait(s2);
    printf("C-");
    semaphore_signal(s1);
   }
exit(0);
}

我不明白为什么这段代码的输出是这样的：

A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A- A-A-A-A-A-A-A-A-A-A-A-A-A-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C- C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-C-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A- A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-A-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B- B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-B-

如果相反，我在字符后面加上“ \ n”，得到正确的输出。

有人可以请您解释一下原因吗？

我正在尝试解决一些简单的流程同步示例；在此特定示例中，我需要三个过程：一个打印“ A”，另一个打印“ B”，最后一个打印“ C”。我想打印“ A” ...

Answer 1

[在包括Linux在内的许多实现中，stdout是行缓冲的。这意味着任何printf字符都存储到（内存）缓冲区中，直到遇到换行符或行缓冲区已满或流被刷新为止。在示例代码中，这导致每个进程在每次获得信号量时都写入自己的本地缓冲区。因此，即使排序是正确的，但由于有三个单独的缓冲区，最终输出结果也不如您预期。每个进程最终都会将其缓冲区刷新到stdout，但是每个缓冲区仅重复包含相同的字母。