为什么与管道信号量的同步会以这种奇怪的方式表现？

Question

我正在尝试解决一些简单的进程同步示例；在这个特定的例子中，我想要三个进程：一个打印“A”，另一个打印“B”，最后一个打印“C”。我想先打印“A”，然后打印“B”或“C”，再打印“A”、“B”或“C”。更清楚地说，我想要一个像“ABACACABACABABACA....”这样的输出模式

下面是我的示例代码，终止不是管理的，因为这只是一个了解事情如何工作的示例。

#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-B-B

相反，如果我在字符后面加上“\n”，我会得到正确的输出。

有人可以解释一下原因吗？

Answer 1

在许多实现中，包括 Linux，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填满或刷新流。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，即使排序按照您的预期是正确的，但是由于存在三个单独的缓冲区，最终输出结果与您预期的不一样。每个进程最终都会将它们的缓冲区刷新到stdout，但每个缓冲区只重复包含相同的字母。

解决方案是在printf 中添加一个换行符（如您所见）或显式地使用fflush(stdout)。另一种选择是使用默认情况下不缓冲的stderr：例如fprintf(stderr, "A-");