Thread: Simple question, trying to understand this code...

> returning a value from 'main' *is* optional
Not in C it isn't.

That little magic cookie only works in C++.

Originally Posted by Salem

Not in C it isn't.

That little magic cookie only works in C++.

It works in C, according to the 1999 edition of the C standard. Other than that it does not.

Originally Posted by matthayzon89

can someone be kind of enough to add comments to the code above so I understand whats goign on within the code, I tried to compile the code and learn from it like that, but I still dont really understand it:/

So far, what are you able to understand of it?

Im really lost because there is a b in the code and also a *b in the code.... Also, b is declared in the very beginning as 11 and then it is redefined...

IS there any way someone can tell me why the first line of output is what it is ( basically explain the logic in one example)

Code:

#include <stdio.h>

int f(int *a, int *b); //this declares and integer a and b in function f. Correct? the '*' next to the variable represents that it is a pointer, right? where exactly is it pointing to?

int main() {

  int a=9, b=11;  

  if (f(&b,&a))
    printf("a = %d,b = %d\n",a,b);
  else
    printf("Zero is the answer.\n"); //so under what conditions is this going to print out?

  b = f(&a, &b);

  printf("a = %d,b = %d\n",a,b);
  return 0; //what is the significance of having a return type?
}

int f(int* a, int* b) {
  int c;  //why does this program print out more than once? there isnt a loop anywhere... 
  c = 2*(*a)%(*b);
  (*a) = c + (3*(*b)%(*a));
  (*b) = c - (2*(*b)%(*a));
  printf("a = %d,b = %d\n",*a,*b);
  return c; // why does this say return c?
}

1. No, it declares a pointer to an integer, to be named a, and another to be named b.
2. Look at the printf. It says what condition will execute it: "zero is the answer".
3. main needs to return a value to your operating system.
4. f is called more than once.
5. The function returns a value which you are testing in #2 of this list, and using again in #4 of this list.


Quzah.

Code:

int f(int *a, int *b);

Prototype a function f which returns an int, has two arguments a and b which are pointers to integers.

The reason you are prototyping the function is so that the lines after that know about it, because you haven't actually defined the function. You're just saying "Hey, if you see f, this is what it looks like...". Then later on, at the bottom, you're saying, "Ok, here is what f does."


Quzah.

If you're going from a book or are in a class, you need to go back and re-read the section on pointers.

Code:

int a;
int *pa;

The first declares an integer. It holds a value, such as 5.
The second declares a pointer to an integer. It holds a value, but the value it holds is a memory address.

If you want to use a variable that you've declared outside a function, inside the function you're calling, and you want any changes made by that function to affect the original variable, then you need to pass its memory address, instead of the value it holds:

Code:

int x = 5;
...
foo( x ); /* pass the value stored in x to foo */
foo( 5 ); /* the same as the above, with the exception of not getting the 5 from the variable x */

Compared to:

Code:

int x = 5; /* store 5 in an int named x */
int *px = &x; /* store the memory address of x in px */
...
bar( px ); /* pass the value in px to bar */
bar( &x ); /* pass the address of x to bar */

Those two lines have the same effect.

Everything you pass to a function in C is just a value. What that value represents, determines what you can do with it. Just like the difference between a float and an int.


Quzah.