infinite while loop inside a switch statement.

Does any one has any idea if the while statement in the following code is ever executed.

Code:

---

switch XYZ
{
  while(1)
  {
    case X:
          -------
          -------
          return p;
   
      case Y:
          -------
          -------
          return q;
    }
}

---

If that is representative of your real code, then anything that isn't X or Y when you enter the switch, will just loop around within the while-loop, since there is no way out of that loop other than in the case X and case Y options.

--
Mats

ok, I understood.
Now my problem is I am getting a warning when compiling this code.
"loop is not reachable from preceding code".
Is there a way to fix this warning? Any ideas??

What are you ACTUALLY trying to achieve? Normally, you do not put a switch, then a while, and then case-labels inside that, so it would help to understand what you want to achieve...

--
Mats

This compiles, runs and produces expected output. I don't see what would be the purpose of that, though.

Code:

---

#include <stdio.h>

int foo(int n)
{
    switch (n) {
        while (1) {
            case 1:
                return 2;
            case 2:
                return 4;
            case 3:; //a case that doesn't return
        }
    }
    return 7;
}

int main()
{
    printf("&#37;d %d %d %d", foo(1), foo(2), foo(3), foo(4));
    return 0;
}

---

Yes, it is VALID to do that. There is an example of a piece of code that puts case-labels in an if-statement in the Samuel P. Harbison "C - A reference manual" [at least, there was one in my older copy].

--
Mats

yes but u have to declare the return values if u dunt want one of them to break out of the loop
but i prefer forever loops for infinite loops