How to return an array from a function?

by

int* test();

but it would be “more C++” to use vectors:

std::vector< int > test();

EDIT
I’ll clarify some point. Since you mentioned C++, I’ll go with new[] and delete[] operators, but it’s the same with malloc/free.

In the first case, you’ll write something like:

int* test() {
    return new int[size_needed];
}

but it’s not a nice idea because your function’s client doesn’t really know the size of the array you are returning, although the client can safely deallocate it with a call to delete[].

int* theArray = test();
for (size_t i; i < ???; ++i) { // I don't know what is the array size!
    // ...
}
delete[] theArray; // ok.

A better signature would be this one:

int* test(size_t& arraySize) {
    array_size = 10;
    return new int[array_size];
}

And your client code would now be:

size_t theSize = 0;
int* theArray = test(theSize);
for (size_t i; i < theSize; ++i) { // now I can safely iterate the array
    // ...
}
delete[] theArray; // still ok.

Since this is C++, std::vector<T> is a widely-used solution:

std::vector<int> test() {
    std::vector<int> vector(10);
    return vector;
}

Now you don’t have to call delete[], since it will be handled by the object, and you can safely iterate it with:

std::vector<int> v = test();
std::vector<int>::iterator it = v.begin();
for (; it != v.end(); ++it) {
   // do your things
}

which is easier and safer.

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