for()循环中的fork()

Question

我正在尝试做一个我必须使用的家庭作业,fork()但我不知道为什么在通过我的for循环运行它们之后我无法阻止我的叉子:

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

int main(int argc, char *argv[]){
    int limit = argc/2;
    if(argc%2 == 0){

            perror("the number of arguments given must pe even!");
            exit(1);
    }

    int i;
    for(i=0; i<=limit; i++){
        if(fork()==-1){
            perror("childes couldn't be created\n");
            exit(1);
        }
        if(fork()==0){
            printf("fork: %d \n",i);
            exit(1);
        }
        wait(0);
    }


    printf("exiting...\n");
    return 0;
}

输出:

warzaru@ubuntu:~/OS/UprocH$ ./exe a b c d
fork: 0 
fork: 0 
fork: 1 
fork: 1 
fork: 1 
fork: 2 
fork: 2 
fork: 1 
fork: 2 
exiting...
exiting...
fork: 2 
exiting...
exiting...
fork: 2 
exiting...
exiting...
exiting...
warzaru@ubuntu:~/OS/UprocH$ fork: 2 
fork: 2 
fork: 2 
exiting...

Answer 1

Daniel Fischer强迫我提供答案.

更改:

if(fork()==-1){} // first fork
if(fork()==0){}  // second fork

至:

pid = fork();
if(pid == -1)
{
    ... //failed
}
else if(pid == 0)
{
    ... //success
}

或者使用switch语句:

switch(fork())
{
    case -1:
        ... // failed
        break;
    case 0:
        ... // success
        break;
}