Question

I have a piece of code where I am trying to return the square of the value pointed to by *ptr.

int square(volatile int *ptr)
{
  int a,b;
  a = *ptr;
  b = *ptr;
  return a * b;
}

  main()
  {
    int a=8,t;
    t=square(&a);
    printf("%d",t);
  }

Its working fine for me but author of this code said it might not work because of following reason:
Because it’s possible for the value of *ptr to change unexpectedly, it is possible for a and b to be different. Consequently, this code could return a number that is not a square!. The correct way to do is

long square(volatile int *ptr)
{
  int a;
  a = *ptr;
  return a * a;
}

I really wanted to know why he said like that?

Answer 1

The idea of the volatile keyword is exactly to indicate to the compiler that a variable marked as such can change in unexpected ways during the program execution.

However, that does not make it a source of “random numbers” – it just advises the compiler – what is responsible for actually changing the variable contents should be another process, thread, some hardware interrupt – anything that would write to the process memory but not inlined in the function where the volatile declaration finds itself. In “older times” (compilers with less magic) everything it did was preventing the compiler from caching the variable value in one of the CPU registers. I have no idea on the optimisations/de-optimistions strategies triggered by it by modern compilers – but it will at least do that.

In the absense of any such external factor, a “volatile” variable is just like any other. Actually – it is just like any other variable – as variables not marked as volatile can also be changed by the same external causes (but the compiled C code would not be prepared for that in this case, which might lead to incorrect values being used).