c - 函数的隐式声明 ‘wait’

Question

我收到一条警告 > 函数“等待”的隐式声明 < 当我运行该程序时它正常工作，我想了解为什么我收到此警告？

提前致谢

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");
  }
}

Answer 1

您可能缺少 wait(2) 的 header :

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