C and "call by reference"

kpachopoulos · 10-19-2004, 02:01 PM

I will explain my question through an example:
void swap(int *a, int *b)
void swap(int &a, int &b)
Question 1: are both of these techniques (the first technique using pointers and the second using addresses) called "call by reference"?
Question 2: both of these have the same result when they are used; the first one uses addresses as arguments and the second one pointers.
Which is their difference? Which one is faster? Which one uses more memory space? When should i use the first technique and when should i use the second one?

itsme86 · 10-19-2004, 02:08 PM

C has no such thing as "pass by reference". You can only pass values to functions in C. void swap(int &a, int &b); is not valid C code. In C you would do something like this:

Code:

void swap(int *a, int *b)
{
  int temp;

  temp = *a;
  *a = *b;
  *b = temp;
}

int main(void)
{
  int a = 5, b = 7;

  printf("a = %d, b = %d\n", a, b);
  swap(&a, &b);
  printf("a = %d, b = %d\n", a, b);

  return 0;
}

swap(&a, &b); may look like you're passing by reference, but you're really passing a value; the address of the variable.

kpachopoulos · 10-19-2004, 02:10 PM

thanks

SoulSkorpion · 10-20-2004, 12:10 AM

That's not the whole story.

C++ allows passing by reference, using the syntax you've described. Note: C++, NOT C. Don't confuse the two. Enough people do as it is...

itsme86 demonstrated the C way. C++ allows you to do it like this:

Code:

void swap(int& a, int& b)
{
  int temp;

  temp = a;
  a = b;
  b = temp;
}

int main() //no void in the declaration, for it is Bad and Vile.
{
  int a = 5, b = 7;

  printf("a = %d, b = %d\n", a, b);
  swap(a, b);
  printf("a = %d, b = %d\n", a, b);

  //no return code necessary
}

kpachopoulos · 10-20-2004, 04:34 AM

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ta0kira · 10-20-2004, 05:44 AM

Passing using * and & do the same thing internally; they pass a pointer. The use of & is a semantical convenience allowed by the compiler so that you don't have to do *a = or a-> to use the argument; it makes the pointer appear like an ordinary object. Another reason to use a reference is so you don't have to check for a bad pointer in the function: even though the reference operation actually passes a pointer, it must be a valid object (unless you dereference a bad pointer as the argument.) This means that NULL cannot be passed as would be allowed by a pointer argument. The bad thing is temp objects can't be used; if your function asks for int&, you cannot give it a char argument; even though it converts if by value, the references cannot be converted. Therefore; do not use for POD unless you are changing the value.

Pass by reference merely means that the argument is not copied. A * argument is actually a pointer, which is a value, therefore that is a pointer argument being passed by value. You can, however pass a pointer by reference with *&, thereby allowing you to change the pointer value, to say a new[] array. You can have multiple pointer types e.g. int**, but you can only use one reference per type, and cannot take a pointer to a reference; int*& OK int**& OK int&* NOT OK int&& NOT OK.
ta0kira

SoulSkorpion · 10-20-2004, 06:34 AM

Quote:

Originally posted by nocturna_gr
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Whoops. Sorry, mentioned the wrong person. I've edited my original post.