Pointers as arguments to a function that calls scanf - c

Question

I am having some trouble with pointers.
The gist of it is, I am trying to define pointers in one function and call that function in my main to use those pointers.
The exact instructions for my assignment are as follows:
Write two functions, one that reads three numbers from the keyboard
and one that prints some information about these three numbers.
Input Function
Write a function that has three integer pointer parameters, and that
asks the user to enter three whole numbers. The values entered at the
keyboard should be read into the addresses stored in the pointer
parameters.
Note: recall that scanf requires the address of a variable and that
pointers store addresses.
Printing Values
Write a second function called a2question2, with no return value and
no parameters. The function should declare three integer variables
and then use your input function to read values into these variables.
The function should then print the sum, the average, the product, and
the smallest and largest of these numbers.
Here is what I have so far:
int pntloc (int *x, int *y, int *z){
int a = 0;
int b = 0;
int c = 0;
printf("Please enter integer #1: ");
scanf ("%d", & a);
printf ("Please enter integer #2: ");
scanf ("%d", & b);
printf("Please enter integer #3: ");
scanf ("%d", & c);
*x = &a;
*y = &b;
*z = &c;
return *x, *y, *z;
}
// Fourth function
main (){
int x, y, z;
pntloc(x, y, z);
int sum = 0;
int average = 0;
int product = 0;
int smallest = 0;
int largest = 0;
printf ("%d", x);
}
However, after the program asks me for the three integers, it crashes without doing anything.
The first function works fine by its self (tested it by making it the main function without parameters and printed the pointer values) ie:
printf ("%d", *x);
So I guess the values are just not passing from one function to the next. I've tried various ways of writing the first and second function but nothing seems to work.
The best I got was getting the program not to crash but the printed value was nowhere to what I inputted before.
Any ideas how to do this?

Answer 1

Your program is probably crashing because of two errors:
1) You are returning the local address of the variables a, b and c:
*x = &a; // This line says follow the 'x' pointer, and set the value there to
// the address of 'a'
Since a is defined locally (i.e. inside the function), that address is invalid once the function returns.
What you probably meant is:
*x = a; // Follow the 'x' pointer, and set the value there to the value of 'a'
2) You're not passing pointers to pntloc() (your compiler should be warning you about this one)
int x, y, z;
pntloc(x, y, z); // The passes the VALUES of x, y and z
You probably meant:
pntloc(&x, &y, &z); // Pass the ADDRESSES of x, y and z
Some other improvements that aren't causing your crash:
You can massively shorten pntloc() by not using the local variables:
void pntloc (int *x, int *y, int *z){
printf("Please enter integer #1: ");
scanf ("%d", x);
printf ("Please enter integer #2: ");
scanf ("%d", y);
printf("Please enter integer #3: ");
scanf ("%d", z);
}
Note that the & has been removed inside the scanf() call. You asked about it in comments, so here's a bit more explanation: &x says "the address of x", but when you have a pointer, you already have an address. A quick example:
int a; // 'a' is an integer variable
int *b = &a; // 'b' is a pointer to the integer variable 'a'
scanf("%d",&a); // This statement reads an integer into 'a'.
// We pass it the address of 'a', which is written &a
scanf("%d",b); // This statement also reads an integer into 'a'.
// We pass it the address of 'a', which is stored
// in the pointer 'b'.
Since we have pointers passed in to the function:
void pntloc (int *x, int *y, int *z){ // Three pointers to ints
we can pass them straight in to scanf(), and don't need to (and shouldn't) use & when we do.
Note that I also removed the return statement:
return *x, *y, *z;
I don't think this return statement is doing what you think it is. Remember that C only allows one return value from a function.
But why does it compile, you ask? Well, here's what's happening - but feel free to ignore this bit if it is confusing: The comma operator evaluates left to right, discarding the left hand result as it goes. So, your return statement is equivalent to:
*x;
*y;
return *z;
The comma operator is useful when the left hand statement has a side effect, and you want to write everything on one line. In my opinion, it makes code much harder to read, but there are one or two situations where it makes things cleaner. For a beginner, I recommend the following rule: Only use commas inside the round brackets of functions.
Since you weren't using the return value from the function when you called it:
pntloc(&x,&y,&z);
I removed the return entirely, and set the return type to void.

Answer 2

*x = &a;
*y = &b;
*z = &c;
Will cause mayhem and death! a b and c are local variables, so you are setting the contents of x y and z to addresses that will be invalid once you return from the function where a b and c are defined.
Perhaps you mean:
*x = a;
*y = b;
*z = c;

Answer 3

in main(),
pntloc(&x, &y, &z);
and
*x = a;
...
printf ("%d", x);
while you pass the *x as a parameter in pntloc(), you can't change *x 's value after calling the function.
and pntloc() don't need to return them, returning 0 or 1 is enough.