Return pointer to array or make void function with parameter

Question

Till now most of the time I was using functions that return pointer to array, but now I started using void function with reference to array, so I am wondering which one of below code is better to use and why?

void doSomething(int** &ary)
{
    ary = new int*[3];
    for (int i = 0; i < 3; ++i)
        ary[i] = new int[3];
    ary[0][0] = 1;
    ary[0][1] = 2;
}

int** ary=NULL;
doSomething(ary);

or this

int** doSomething1()
{
    int **ary = new int*[3];
    for (int i = 0; i < 3; ++i)
        ary[i] = new int[3];
    ary[0][0] = 1;
    ary[0][1] = 2;
    return ary;
}

int **ary1=doSomething1();

It is a matter of opinion. The second version is less awful than the first one. — juanchopanza
– juanchopanza, Commented Nov 22, 2014 at 14:52
Use a type that represents whatever it is you are trying to return. A pointer to a pointer can mean too many things. — juanchopanza
– juanchopanza, Commented Nov 22, 2014 at 16:09
Neither is better, return a std::vector until you have a real reason to do otherwise. Then, switch to C++ 11 move semantics. — Ulrich Eckhardt
– Ulrich Eckhardt, Commented Nov 22, 2014 at 16:39

cmaster - reinstate monica · Accepted Answer · 2014-11-22 14:50:44Z

1

This is a question of opinion, however, here is mine:

I think the version that returns the pointer is better. Why? Simply because it makes the calling site less magic. With the return variant, you call the function like this:

int** my2dArray = doSomething();

It is perfectly clear, that my2dArray is initialized to point to some array that the function supplies. Without a single look at the function definition.

On the other hand, the call

int** my2dArray;
doSomething(my2dArray);

should always ring an alarm bell for the reader: It looks as if you are passing an uninitialized pointer to the function. Even after looking up the function definition, and seeing that the pointer is passed by reference, a reader still can't be certain that the function does not interpret the value that is passed in. It requires a close look at the code to make sure that this call is indeed legitimate.

So, for the sake of debugability, I avoid passing reference arguments. I pass by value, I pass by const reference (which has the same semantic as pass by value), or I pass by explicit pointer. That way no function call can modify a value that is not explicitly visible at the calling site.

answered Nov 22, 2014 at 14:50

cmaster - reinstate monica

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2 Comments

user1403588 Over a year ago

What about memory leak or performance? Maybe any other better option?

cmaster - reinstate monica Over a year ago

To prevent memory leaks, you should use smart pointers (std::unique_ptr<> and std::shared_ptr, or similar). In terms of performance you need to worry that your objects are not copied needlessly, something you can easily achieve by passing by const reference. Pass by reference and pass by explicit pointer are equivalent with respect to performance.

user4281841 · Accepted Answer · 2014-11-22 14:57:14Z

0

There is no big difference between the two examples, however I'd prefer the second example.

In the first example, you take a reference to ary, which means your program has to dereference all pointers each time it accesses an element somewhere in ary. (A reference is equivalent to a pointer after the code has been compiled)

In the second example, you only need to copy the pointer when the function returns.

First example:

reference to pointer to array -> pointer to array -> element in array

Second example:

pointer to array -> element in array

answered Nov 22, 2014 at 14:57

user4281841

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