Question

I am passing a number int into this method but it is not storing the int in the method that calls the intchange method. Any pointers?

int intchange(const char *prompt, int number){
  char buffer[LINESIZE];

  while(1){
    printf("%s", prompt);
    if(!fgets(buffer,LINESIZE, stdin)){
      clearerr(stdin);
      return 0;
    }
   if(sscanf(buffer, "%d", &number) == 1){
     /*exit variable */
   if(number== -1)
      return 0;
   else if (number<= 110 && number>= 0)
      return 1;
       }
     }
  }

Thanks!

Answer 1

In order for the change to be visible to the caller you need to pass the address of the int into intchange(). Otherwise a copy of the int provided by the caller is being changed, not the int the caller provided as C passes by value:

int intchange(const char* prompt, int* number)
{
    /* snip */
    if(sscanf(buffer, "%d", number) == 1){ /* Removed address of operator. */

        if(*number == -1)                  /* Dereference to query int value. */
        /* snip */
    }
}

Invoked:

int i;
if (intchange("14", &i))
{
    printf("i=%d\n", i);
}

See Does C even have “pass by reference”?.