Question

How would I only return one dimension of an array, while ignoring the other?

Such as:

int map[4][8];

int MapManager::getMapX(int x)
{
    return map[x];
}

Answer 1

Since none of the other answers here return an actual array (pointers are not arrays), I thought I might show how to really return an array. Or the closest thing possible, which is a reference to an array.

typedef int row_type[8];

row_type& MapManager::getMapX(int x) {
    return map[x];
}

What’s the point of this? The pointer works the same!

No, it doesn’t. The pointer loses type information, namely, the size. You can make begin() and end() functions work with arrays, but you can’t with pointers:

// pointer version
int* MapManager::getMapX_ptr(int x) {
    return map[x];
}

row_type& row = getMapX(0);
// row is of type int(&)[8]
// notice the size is not lost in the type!
std::sort(begin(row), end(row));
// compiles fine! end() correctly finds the end of the row

int* ptr = getMapX_ptr(0);
// notice how ptr has no size information at all
std::sort(begin(ptr), end(ptr));
// can't possible be made to work!

You can’t write end for int*.

template <typename T, std::size_t N>
T* end(T(&arr)[N]) {
    return &arr[0] + N;
}

template <typename T>
T* end(T* ptr) {
    // what here?
    // guess?
    // pick a random number?
}