为什么该程序显示“无法访问的代码”警告？ 以及我如何抑制它？

Question

#include<iostream.h>
#include<conio.h>
#include<process.h>

void function(void);

int main(void)
{
 clrscr();

 int ch;

 while(1)
 {
  cin>>ch;

  if(ch==2)
    exit(0);
  else
    function();
 }//while
 return 0;
}//main

void function(void)
{
 cout<<"Hello";

 return;
}

The above code is working fine, but why I'm getting the "unreachable code" warning? I really don't understand what am I doing wrong. The compiler shows no warning when I comment/remove the return 0; statement in main() . Why is it so? Please tell me what is it that I'm doing wrong and what is the correct way to do it.

Answer 1

The while (1) loop has no option to leave.

Thereby, the exit(0) is not recognized as the dataflow analysis doesn't consider it as an option to jump to code behind of while (1) (and actually it doesn't).

Hence, there is no way to get to the return 0; .

If you replace exit(0) by a break than it changes. The break will cause to leave the while (1) and the return 0; becomes reachable.

Answer 2

why I'm getting the "unreachable code" warning? I really don't understand what am I doing wrong.

The loop has no return condition: while(1) and the loop body doesn't contain a break (or goto ) that would jump out of the loop othwewise. Nevertheless, you have a return 0; statement after the loop. Since the execution never jumps out of the loop, it can never reach the return statement.

The compiler warns you that the line has no effect on the behaviour of the program, since the execution can never reach it. You can get rid of the warning by changing the logic of your program in a way that can potentially jump out of the loop. I suggest following:

if(ch==2)
  break;
else
  function();

The compiler shows no warning when I comment/remove the return 0; statement in main(). Why is it so?

That statement was the "unreachable code" that the warning referred to. If there is no unreachable code, then there is no need to warn about it.

It is safe to remove the line. main is special in the regard that it implicitly returns 0 in the absence of return statement (if the execution ever returns from main which your program never does).