c++ dynamic size of the array

I have got a small proble开发者_开发技巧m with 1D array in c++. I have got a function line this:

void func(int (&array)[???])
{
    // some math here;

    "for" loop {
        array[i] = something;
    }
}

I call the functions somewhere in the code, and before I made math I'm not able to know dimension of the array. The array goes to the function as a reference!, because I need it in the main() function. How I can allocate array like this?, so array with ?? dimension goes to the function as reference then I have to put the dimension and write to it some values.

Since you're using C++, why not use a std::vector<> instead?

Other have mentioned that you should use std::vector in C++ and they are right.

But you can make your code work by making func a function template.

template <typename T, size_t N>
void func(T (&array)[N])
{
    // some math here;

    "for" loop {
        array[i] = something;
    }
}

Use a pointer, not a reference:

void func(int *a, int N);

Or, easier, use a vector:

void func(std::vector<int> &a);

Vectors can be allocated by simply saying

std::vector<int> a(10);

The number of elements can be retrieved using a.size().

If the array you pass to func is a stack array, and not a pointer, you can retain its size by using a function template:

template <class T, size_t N>
void func(T(&array)[N])
{
    size_t array_length = N; // or just use N directly
}

int main() 
{
    int array[4];
    func(array);
}

That said, as others have already pointed out, std::vector is probably the best solution here.

As well as vector which has been suggested you could possibly use valarray which is also part of STL and is intended specificially to handle mathematical collections.

What you have to realize, is that arrays are pointers. A definition like int array[5] will allocate space for 5 integers on the stack and array will be the address of the first value. Thus, to access the first value in the array, you can write

array[0] or *array (which is the same as *(array + 0))

In the same way to retrieve the address of the third element, you can write

&array[2] or array + 2

Since arrays are pointers, you don't have to worry about the runtime size of your array if you would like to pass it to a function, simply pass it as a pointer:

void func(int *array)
{
    int size;
    //compute size of the array
    for (int i = 0; i < size; ++i)
    {
        //do whatever you want with array[i]
    }
}