在c ++中运行shell脚本

Question

Ive been writing the folowing code:

#include <iostream>  
#include <stdlib.h> 
using namespace std; 
  int main() {  
  cout << "The script will be executed"; 
  system("./simple.sh");  
} 

But when I run it the shell script is executed first.
What can I do to execute "cout << "The script will be executed"" first?

Answer 1

Flushing the output stream buffer should be enough. You can do this with

cout << "The script will be executed";
cout.flush();

Alternatively, if you intended to also print a newline character then you can use std::endl which implicitly flushes the buffer:

cout << "The script will be executed" << endl;

Answer 2

You're not flushing the output stream.

Try:

cout << "The script will be executed" << endl; // or cout.flush()
system("./simple.sh");

The script is executed second, the delay between the call to cout and printing to the console is probably throwing you off.

Answer 3

you can use cerr << "The script will be executed";
when you're trying to print something using cout it keeps that in buffer. cerr is printing immediately but @Jon answered better.

Answer 4

尝试在调用system（）syscall之前刷新stdout流。