Array of Pointers as Function Parameter

array of pointers as function parameter

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 decay 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 are 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);

Passing an array containing pointers to a function properly

Try this:

void printArray(int **iXArray, int iSize) ...

In your example you provide an array of (int*) so reference it as one, you must tell the compiler to expect an array of pointers.

By default passing an array is by reference. If you change the array's content, it changes at the callee's side aswell. The pointer itself is passed by value, so changing the value of the iXArray parameter (iXArray = (int**)123;) will not change the array2 pointer at the callee's side.

If you want to pass the array by value, will need to wrap it in a value type:

typedef struct {
  int A[123];
} Array;

Array incrementArray(Array array, int count) {
    for (int i=0; i<count; i++) {
        array.A[i]++;
    }
    return array;
}

C pass int array pointer as parameter into a function

In your new code,

int func(int *B){
    *B[0] = 5;
}

B is a pointer to int, thus B[0] is an int, and you can't dereference an int. Just remove the *,

int func(int *B){
    B[0] = 5;
}

and it works.

In the initialisation

int B[10] = {NULL};

you are initialising anint with a void* (NULL). Since there is a valid conversion from void* to int, that works, but it is not quite kosher, because the conversion is implementation defined, and usually indicates a mistake by the programmer, hence the compiler warns about it.

int B[10] = {0};

is the proper way to 0-initialise an int[10].

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