How to write a template function that takes an array and an int specifying array size

Question

For a university exercise, I have been asked to write a template function "print();", which takes two arguments, 1: an array of a generic type, and 2: an int specifying the size of the array. The function should then print out every item in the array to the console. I am having some trouble with the function arguments. The code I currently have is:

   template <typename Type>
   Type print (Type a, Type b)
    {
        Type items;
        Type array;
        a = array;
        b = items;

        for (int i = 0; i < items; i++) {
        std::cout << std::endl << "The element of the index " << i << " is " << array << std::endl;
        std::cout << std::endl;
    }

and in main():

    print(Array[], 10);

Obviously putting Array as an argument isn't returning a value, so I am not sure what else to do. Any ideas?

Eolithic

Is it fine to pass a pointer (after array-to-pointer decay) or do you want to take the array by reference? – Nudge

(in the first case you could then also pass non-static arrays) – Nudge

Apologies, I am not sure, I have only written in this question exactly what the text book has asked me. – Eolithic

Answer 1

The correct way to write it is

Live On Coliru

#include <iostream>

template <typename T, size_t size> void print(const T (&array)[size])
{
    for(size_t i = 0; i < size; ++i)
        std::cout << array[i] << " ";
}

int main() {
    int arr[] = { 1,2,3,4,99};

    print(arr);
}

Prints

1 2 3 4 99

Mho

Thank you, what exactly would I put in as an argument when I call the function in main? If you could give me a quick example that would be great, I'm new to c++! – Eolithic

Needs more range-for. Boost.Range indexed if indices are needed ;) – Carlotacarlotta

@Eolithic added a live sample – Mho

Why can't we pass the array by value? – Samara

@Samara that's how the language is defined - it decays to a pointer. That's actually just a C-compatibility relic. So, if you want consistency, you totally can: use std::array although in a scenario that simple I'd consider initializer_list – Mho

@sehe: print(arr) passes the arr but how the size gets calculated? – Vulcanize

@RasmiRanjanNayak It's deduced into size_t size by the compiler during overload resolution – Mho

Answer 2

If you want to pass the array by reference, you could

template <typename T, size_t SIZE>
void print(const T(&array)[SIZE])
{
    for (size_t i = 0; i < SIZE; i++)
        std::cout << array[i] << " ";
}

and then, e.g.

int x[] = {1, 2, 3};
print(x);

LIVE

Otherwise, you can pass it by pointer, note that the array will decay to pointer, and you have to guarantee the correctness of SIZE being passed.

template <typename T>
void print(const T array[], size_t SIZE)
{
    for(size_t i = 0; i < SIZE; i++)
        std::cout << array[i] << " ";
}

and then, e.g.

int x[] = {1, 2, 3};
print(x, sizeof(x) / sizeof(int));

LIVE

Gironde

Note that passing (pointer+size) is a bit of an anti-pattern in C++. GSL defines array_view for precisely this reason – Mho