Warning while passing array by reference to function

**Jdo300** · 06-10-2008

Hello All,

I'm a novice to C programming and I am working on a program that keeps giving me a warning that I can't seem to get rid of. But rather than posting the chunks of code from the original program, I managed to create a simplified example of what I was doing that still causes the same Warning. First, here's my code:

Code:

#include <stdio.h>

int myFunc (char* inputArray);

int main()
{
	char myArray[4];

	myFunc(&myArray);
	return(0);
}

int myFunc (char *inputArray)
{
	inputArray[3] = 'A';
	return(0);
}

When attempting to compile this in Visual Studio (it is saved as a .c file) I get the following two warnings:

1>.\Code.c(9) : warning C4047: 'function' : 'char *' differs in levels of indirection from 'char (*)[4]'
1>.\Code.c(9) : warning C4024: 'myFunc' : different types for formal and actual parameter 1

I know the issue has something to do with passing the array by reference to the function but I'm not sure how to fix the issue. Any help would be greatly appreciated.

God Bless,
Jason O

**QuantumPete** · 06-10-2008

You don't need the ampersand (&), myArray is already a pointer.

QuantumPete

**Elysia** · 06-10-2008

I would like to point out that it's better to say that you're passing the array via pointer, not by reference (that's a C++ term).
Also note that when passing an array, you pass a pointer to its first element, so using the ampersand (&) on it, will make it a pointer to pointer, which isn't what you want.

**vlrk** · 06-10-2008

instead of

Code:

myFunc(&myArray);


myFunc(myArray);

would work

myArray ==> myArray[0]

so first element address will be passed in above case.

**jason_m** · 06-10-2008

Pointers and arrays in C are *almost* the same thing. You can even use array index notation (the []'s) on a pointer! The array name is a varable containing (pointing to) the address of the first array element in memory. So, you don't need to use the & operator:

Code:

myFunc(myArray);

You could also (but i'm not sure why you would want to) do this:

Code:

myFunc(&myArray[0]);

The similarity between pointers and arrays should become obvious inside myFunc(). inputArray is declared as a pointer to a character, yet you are using array notation on it here:

Code:

inputArray[3] = 'A';

If you wanted to do this in "pointer notation", I think the correct code would look like:

Code:

*(inputArray + 3) = 'A';

But, either method works, and the former is probably easier to read.

**Salem** · 06-10-2008

> Also note that when passing an array, you pass a pointer to its first element, so using the ampersand (&) on it, will make it a pointer to pointer, which isn't what you want.
No, it's type would be pointer to array.

As in
int myFunc ( char (*inputArray)[4] );
would be called with
myFunc(&myArray);

Having
int myFunc ( char **inputArray );
cannot be called with either
myFunc(myArray);
or
myFunc(&myArray);

Instead, you would need
char *temp = myArray;
myFunc( &myArray );

**Elysia** · 06-10-2008

Originally Posted by Salem

No, it's type would be pointer to array.

Which I classified as a pointer-to-pointer, which is close to the truth, but not 100%.
Although I haven't really tested it myself, so I'm not 100% sure of the behavior either...

**Salem** · 06-10-2008

pointer to array != pointer to pointer.
I strongly suggest you try it

**Elysia** · 06-10-2008

Indeed... it's a pointer with set bounds.

**QuantumPete** · 06-10-2008

Originally Posted by Elysia

Indeed... it's a pointer with set bounds.

Though nothing stops you from reading and writing data outside those bounds...

**Elysia** · 06-10-2008

Indeed... it's like a suggestion more than set dimensions.

**Jdo300** · 06-10-2008

Hi Everyone,

Thank you very much for the fast replies! To solve my problem, I simply used Jason_m's suggestion of removing the ampersand from in front of the array in the function call and that fixed everything. Again thanks for all the great input!

God Bless,
Jason O

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