C语言 函数 'wait' 的隐式声明

Implicit declaration of function ‘wait’

提问by AyeJay

I am getting a warning > Implicit declaration of function ‘wait' < and when I run the program it works correctly, I would like to understand why I am getting this warning?

我收到警告 > 函数“wait”的隐式声明 < 并且当我运行程序时它可以正常工作，我想了解为什么会收到此警告？

Thanks in advance

提前致谢

Edit: I forgot to add the library included

编辑：我忘了添加包含的库

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


void create (char* program, char** arg_list)
{
  /* put your code here */
  pid_t childPid;
  int status;

  if((childPid = fork()) < 0){
    printf("Failed to fork() --- exiting...\n");
    exit(1);
  }
  else if (childPid == 0){ // --- inside the child process
    if(execvp(program, arg_list) < 0){ // Failed to run the command
      printf("*** Failed to exec %s\n", program);
      exit(1);
    }
  }
  else{ // --- parent process
    while(wait(&status) != childPid)
      printf("...\n");
  }
}

回答by usr

You are probably missing the headers for wait(2):

您可能缺少以下标题wait(2)：

  #include <sys/types.h>
  #include <sys/wait.h>

回答by Barmar

You need to put:

你需要把：

#include <sys/types.h>
#include <sys/wait.h>

at the top of the program to get the declaration of the function.

在程序的顶部获取函数的声明。

This is shown in the man page

这显示在手册页中