Thread: Pointers and Functions

Question regarding pointers being used in functions..

Below is some sample code I have written. I understand the principles of points and how they are passed to functions (and used in general).

However I am unsure exactly as to why when you call a function you omit the * before the argument?

My thoughts..
Below I have declared that the argument for function that int a is a pointer (*), therefore when I call the function, I only need to use the variable name and not the specifier * as the function is expecting a pointer??

Code:

#include <stdio.h>


void function(int *a);


int main () {
    int value = 6;
    int *pValue = &value;
    function(pValue);
    
    return 0;
}


void function(int *a) {
    printf("Pointer memory address is %p\n", a);
    printf("The pointer value is %d\n", *a);
}

Hi all..

One thing I would suggest from my limited understanding is...not to think of the (*) as being a pointer. But look at how its presented....the * character is also a dereferencing character for pointers, so look at the format as a whole.

(int * pValue) to me is a variable(pValue) with type ( int*) ie Pointer, just like double, int, char.

And once a variable is declared, to give it a value you don't have to use its type anymore, you decladed it already, jst use its name.
So in your code, would be (pValue = &value) in this case the variable value is a memory address.

As your function is asking for a pointer ( as shown by the type ( int*) you pass pValue to it, as it contains the memory address of the (int) of interest.

Just thought I would try explain also....as much for myself as for Steve.

John.

Originally Posted by laserlight

It might be a little more instructive to try this:

Code:

#include <stdio.h>

void function(int *a);

int main(void) {
    int value = 6;
    int *pValue = &value;
    int **ppValue = &pValue;
    function(&value);
    function(pValue);
    function(*ppValue);

    return 0;
}

void function(int *a) {
    printf("Pointer memory address is %p\n", (void*)a);
    printf("The pointer value is %d\n", *a);
}

Observe that the function named function is called three times with different arguments.

The function(&value) call passes &value as an argument because value is an int, so &value is a pointer to int. The formal parameter named a is a pointer to int, so this matches exactly.

The function(pValue) call passes pValue as an argument because pValue is a pointer to int. The formal parameter named a is a pointer to int, so this matches exactly.

The function(*ppValue) call passes *ppValue as an argument because ppValue is a pointer to a pointer to int, so *ppValue is a pointer to int. The formal parameter named a is a pointer to int, so this matches exactly.

Also, notice that I changed this:

Code:

printf("Pointer memory address is %p\n", a);

to this:

Code:

printf("Pointer memory address is %p\n", (void*)a);

The reason is that the %p format specifier expects a corresponding pointer to void argument, so I cast a to void* to match.

EDIT:
By the way, the text for the printf statements are not correct. I would have written:

Code:

printf("The pointer value is %p\n", (void*)a);
printf("The value of what the pointer points to is %d\n", *a);

The reason is that the value of a pointer is an address, i.e., the address of what the pointer points to. The address of the pointer itself is something else, in this case it would have been &a.

Thanks for this.. Something I picked up on was the = function(&value) ..
Obviously pointers are important, however why are they needed if instead we could just write the code &variable instead ?

Thanks !