Question

I have:

 #include <stdio.h>

 int sum ( int x, int y );

 main ()
 {
 int theSum = sum (10, 11);
 printf ( "Sum of %i and %i is: %i\n", x, y, theSum );
 }

 int sum ( int x, int y )
 {
 return x + y;
 }

However, when I compile and run it says x and y are undeclared? Any help greatly appreciated. Thanks

Answer 1

In line three all you have done is declare a function sum which takes two parameters, both integers, called x and y. You haven’t declared any variables. Those parameters can only be referred to inside the function itself. Below is a simplification which will help you at this stage, but you should try to read a basic programming book. “The C Programming Language” by Kernighan and Ritchie is a fine place to start.

Variables are chunks of memory that you refer to by name. They can take on any value (of their type) during the life of your program – hence the name ‘variable’. They must be declared before you use them; you do this by telling the compiler their type and their name. int a means ‘reserve me a block of memory big enough to hold any integer, and let me refer to it later with the name a‘. You can assign values to it: a = 10 and you can make use of it: a + 20.

You need to understand the difference between parameters and variables to get what’s going on here. A function’s parameters are basically variables which exist only during the life of that function. Here’s your sum again:

int sum(int x, int y) {
    return x + y;
}

Notice how the top line looks just like a variable declaration int x. That’s because it is. x and y are variables you can use in the function.

You call sum by passing in values. The compiler, in effect, replaces x and y in your function with the values you pass in. In your case, you’re passing literals: 10 and 11. When the program reaches the call to sum, the parameters x and y take on the values 10 and 11, so the return becomes return 10 + 11; which is of course 21.

Just remember that the parameters x and y only exist in that function. You may only refer to them within your function. Why? Because each pair of curly braces { and } define a scope, and anything declared within that scope can only be used within that scope. That includes variables and parameters.

So, here is a more complete example. I have changed the letters so you can see the different ways you use variables and parameters:

#include <stdio.h>

int sum ( int x, int y );

main ()
{
   /* We declare our variables */
   int a;
   int b;

   /* We assign values to them */
   a = 10;
   b = 11;

   /* We pass them as parameters to your sum function */
   int theSum = sum (a, b);
   /* And we use them as parameters again, in a call to the printf function */
   printf ( "Sum of %i and %i is: %i\n", a, b, theSum );
}

int sum ( int x, int y )
{
   return x + y;
}