array of pointers as function parameter

Question

I have a basic question on array and pointer in C/C++.

Say I have:

Foo* fooPtrArray[4];

How to pass the fooPtrArray into a function? I have tried:

int getResult(Foo** fooPtrArray){}  //  failed
int getResult(Foo* fooPtrArray[]){} // failed

How can I deal with pointer array?

EDIT: I once thought the error msg is from passing the wrong pointer array, but from all the responses, I realize that it's something else... (pointer assignment)

Error msg:
Description Resource Path Location Type incompatible types in assignment of 
`Foo**' to `Foo*[4]' tryPointers.cpp tryPointers line 21 C/C++ Problem

I don't quite get why it says: Foo* * to Foo*[4]. If as function parameter they are inter-change with each other, why during assignment, it give me compilation error?

I tried to duplicate the error msg with minimum code as follows:

#include <iostream>

using namespace std;

struct Foo
{
int id;
};

void getResult(Foo** fooPtrArray)
{
cout << "I am in getResult" << endl;
Foo* fooPtrArray1[4];
fooPtrArray1 = fooPtrArray;
}

int main()
{
Foo* fooPtrArray[4];
getResult(fooPtrArray);
}

Answer 1

Both

int getResult(Foo** fooPtrArray)

and

int getResult(Foo* fooPtrArray[])

as well as

int getResult(Foo* fooPtrArray[4])

will work perfectly fine (they are all equivalent).

It is not clear from your question what was the problem. What "failed"?

When passing arrays like that it normally makes sense to pass the element count as well, since the trick with allowing the array type to declay to pointer type is normally used specifically to allow passing arrays of different sizes

int getResult(Foo* fooPtrArray[], unsigned n);
...
Foo* array3[3];
Foo* array5[5];
getResult(array3, 3);
getResult(array5, 5);

But if you always going to pass arrays of strictly 4 elements, it might be a better idea to use a differently-typed pointer as a parameter

int getResult(Foo* (*fooPtrArray)[4])

In the latter case the function call will loook as follows

Foo* array[4];
getResult(&array);

(note the & operator applied to the array object).

And, finally, since this question is tagged as C++, in the latter case a reference can also be used instead of a pointer

int getResult(Foo* (&fooPtrArray)[4]);
...
Foo* array[4];
getResult(array);

Answer 2

Both of those function signatures look like they should work. When calling the function, you would do something like this:

int returnValue;
returnValue = getResult(fooPtrArray);

Answer 3

What you have declared with the following line:

Foo* fooPtrArray[4];

is an array of pointers to Foo objects (in other words an array of Foo*).

In C/C++, the name of the array is defined as a pointer to the start of the array. This is because arrays aren't "real" types in these languages but simply a contiguous sequence of values of a specific type in memory.

The name fooPtrArray in this case will be a "pointer to the first pointer" in the array, which is the start of the memory location where the array of pointers is stored.

So either function prototype you have outlined above should work. However, when passing in arrays to a function, you will always need to pass in the size as well so the function knows how many elements are in there. So you should define your method like this:

int getResult(Foo** fooPtrArray, int arraySize);

Inside this function, you can access the individual Foo pointers (and subsequently Foo objects) in the array like this:

for (int i=0; i < arraySize; i++)
{
    Foo* fooPtr = fooPtrArray[i];
    if (fooPtr)
    {
        fooPtr->memberFunction();
        fooPtr->memberVariable;
    }
}

Answer 4

The top one compiles OK:

http://codepad.org/KOcnpmtv

The second one is OK too:

http://codepad.org/7OSqprYI

Answer 5

I don't see how this can compile without an int returned ?!

@Kinopiko your code does not compile in C++.

@Lily what is the error message you get? in your example getResult needs to return an int. It's probably what is failing.