4

考虑以下程序的输出：

int main()
{
    int ret;
    ret=fork();
    ret=fork();
    ret=fork();
    ret=fork();

    if(!ret)
            printf("one\n");
    else
            printf("two\n");
    return 0;
}

我得到的输出为： 

two
one
two
two

http://ideone.com/omgKm

AFAIT，输出应该是8 times one& 8 times two。

one's其余的&在哪里two's？

4

1 回答 1

4

考虑这个替代代码，它可以更好地跟踪发生的事情：

#include <stdio.h>
#include <unistd.h>

int main(void)
{
    int ret1 = fork();
    int ret2 = fork();
    int ret3 = fork();
    int ret4 = fork();

    if (ret4 == 0)
        printf("one: (%d: %d, %d, %d, %d)\n", (int)getpid(), ret1, ret2, ret3, ret4);
    else
        printf("two: (%d: %d, %d, %d, %d)\n", (int)getpid(), ret1, ret2, ret3, ret4);
    return 0;
}

向我们展示这个变体的输出，我们可以看到哪些有效，哪些失败。

看到替代输出后，我在我的 Mac 上得到了这个（Isis JL:我的提示在哪里，rmk是 的替代实现make）：

Isis JL: rmk fb && ./fb
    /usr/bin/gcc -O3 -g -std=c99 -Wall -Wextra fb.c -o fb  
two: (38068: 38069, 38070, 38071, 38072)
one: (38072: 38069, 38070, 38071, 0)
two: (38071: 38069, 38070, 0, 38075)
two: (38070: 38069, 0, 38074, 38077)
two: (38073: 0, 0, 38078, 38079)
two: (38069: 0, 38073, 38076, 38080)
one: (38075: 38069, 38070, 0, 0)
one: (38077: 38069, 0, 38074, 0)
Isis JL: two: (38074: 38069, 0, 0, 38081)
two: (38078: 0, 0, 0, 38082)
one: (38079: 0, 0, 38078, 0)
one: (38081: 38069, 0, 0, 0)
two: (38076: 0, 38073, 0, 38083)
one: (38080: 0, 38073, 38076, 0)
one: (38083: 0, 38073, 0, 0)
one: (38082: 0, 0, 0, 0)

Isis JL:

请注意交错提示 - 最后的空白行是我在输出完成后点击返回的地方。

假设：

初始进程停止后，不会捕获 ideone 上的输出。

试试这个替代方案，它等待孩子们在退出之前死去：

#include <stdio.h>
#include <unistd.h>
#include <sys/wait.h>

int main(void)
{
    int ret1 = fork();
    int ret2 = fork();
    int ret3 = fork();
    int ret4 = fork();

    if (ret4 == 0)
        printf("one: (%d: %d, %d, %d, %d)\n", (int)getpid(), ret1, ret2, ret3, ret4);
    else
        printf("two: (%d: %d, %d, %d, %d)\n", (int)getpid(), ret1, ret2, ret3, ret4);
    while (wait(0) > 0)
        ;
    return 0;
}

再次在 Mac 上输出：

Isis JL: rmk fb && ./fb
    /usr/bin/gcc -O3 -g -std=c99 -Wall -Wextra fb.c -o fb  
two: (38111: 38112, 38113, 38114, 38115)
one: (38115: 38112, 38113, 38114, 0)
two: (38114: 38112, 38113, 0, 38119)
two: (38113: 38112, 0, 38117, 38121)
two: (38117: 38112, 0, 0, 38123)
one: (38119: 38112, 38113, 0, 0)
two: (38118: 0, 38116, 0, 38124)
one: (38121: 38112, 0, 38117, 0)
one: (38125: 0, 0, 38122, 0)
two: (38116: 0, 0, 38122, 38125)
two: (38122: 0, 0, 0, 38126)
two: (38112: 0, 38116, 38118, 38120)
one: (38120: 0, 38116, 38118, 0)
one: (38123: 38112, 0, 0, 0)
one: (38124: 0, 38116, 0, 0)
one: (38126: 0, 0, 0, 0)
Isis JL:

Hypothesis Proven

To the extent you can prove anything... http://ideone.com/zFoLn ...

This shows all 16 lines of output. The problem must have been 'premature termination'.

于 2012-07-17T13:10:26.253 回答