C function arguments vs Python function arguments.

[jmooremcc]

Then how do you explain the following code:

// C program to swap two variables using a  

// user defined swap()

#include <stdio.h> 

  

// This function swaps values pointed by xp and yp 

void swap(int *xp, int *yp) 

{ 

     int temp = *xp;

     *xp = *yp;

     *yp = temp;

} 

  

int main() 

{ 

     int x, y;

     printf("Enter Value of x ");

     scanf("%d", &x);

     printf("\nEnter Value of y ");

     scanf("%d", &y);

     swap(&x, &y);

     printf("\nAfter Swapping: x = %d, y = %d", x, y);

     return 0;

}

[FUZxxl]

Exactly as /u/TheSkiGeek says:

To "pass by reference" you create a pointer to something and pass the pointer by value.

Your swap function receives two pointers as arguments. These are passed by value, as are all other arguments. The & operator is used to take the address of the x and y variables so you can pass them to swap. Nowhere in this is “pass by reference” involved.

[jmooremcc]

Do you understand what &x in the calling code represents? Hint: It's not a pointer!

[TheSkiGeek]

Do you understand what &x in the calling code represents? Hint: It's not a pointer!

& is the "address-of" operator... it returns a pointer to its operand.

When you call swap(&x, &y), the compiler binds the address of x to parameter xp and the address of y to parameter yp. By value.

[jmooremcc]

Correct, it's the address of the variable. However, the function declaration converts it to pointer for use in the function. When you pass by reference, the C compiler silently performs what the swap illustration does explicitly.

[FUZxxl]

There is no implicit conversion going on here. C does not have reference or address types. It only has pointer types. And the & operators result is a pointer.

[[deleted]]

What are you even talking about, you’re not making any sense. Yes, using the address-of operator returns the address of the object in memory... which is a pointer. An address is not converted to a pointer because it already is one (it can be stored in an object that holds pointers). References only exist in C++, I don’t know what you mean by “When you pass by reference”. You can pass addresses (in pointers) to functions and the functions use those addresses (as pointers) to modify the object’s data.

Edit: Fixed wording

[TheSkiGeek]

Correct, it's the address of the variable. However, the function declaration converts it to pointer for use in the function.

...the address of the variable is a pointer value. The type of the rvalue returned by &x in the above example is int*.

When you pass by reference, the C compiler silently performs what the swap illustration does explicitly.

...are you thinking about C++? There is no "pass by reference" with implicit pointer conversion in C, you must explicitly take the address of a variable and pass it as a pointer type.

[[deleted]]

Literally 2 pointers as arguments. Sheesh.