Question

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

int function(void* dat){
    sleep(5);
    printf("Executed Child Process...\n");
    return 5;
}

void dupliFork(int *childpid, int *funcRet, int *func(void*), void* data){
    int id = fork();
    if(id==0) *funcRet = *func(data);
    else *childpid=id;
    printf("Returned dupliFork with id %d\n", id);
}

int main(void){
    int childpid=0, funcRet=0;
    dupliFork(&childpid, &funcRet, function, NULL);
    printf("Waiting for Child Process with id %d...\n", childpid);
    waitpid(childpid, NULL, 0);
    printf("Child Process Terminated with return value %d...\n", funcRet);
}

您好，我无法理解上述代码的工作原理...... 当执行上面的代码时，我得到输出：

Returned dupliFork with id 4941
Waiting for Child Process with id 4941...
Executed Child Process...
Child Process Terminated with return value 0..."

这非常令人沮丧，因为我期待以下输出：

Returned dupliFork with id 4941
Waiting for Child Process with id 4941...
Executed Child Process...
Returned dupliFork with id 0
Child Process Terminated with return value 5...

这很令人沮丧，因为看来叉子之后，原来的duplifork功能完美无缺。但是，它的副本，它调用另一个函数，并需要返回值才能继续执行，从来没有真正表现那样。有人可以向我解释实际上是什么去吧？提前谢谢。

Answer 1

您的函数指针语法错误。

int *func(void*)

应该是：

int (*func)(void*)

和

if(id==0) *funcRet = *func(data);

应该是：

if(id==0) *funcRet = func(data);

进行这些更改，你会得到类似的东西：

paul@horus:~/src/sandbox$ ./fork
Returned dupliFork with id 69707
Waiting for Child Process with id 69707...
Executed Child Process...
Returned dupliFork with id 0
Waiting for Child Process with id 0...
Child Process Terminated with return value 5...
Child Process Terminated with return value 0...
paul@horus:~/src/sandbox$

因为您（也可能错误地）除父母之外还有子进程wait()。

Unix Fork奇怪的行为

1 个答案: