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

问题

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?

回答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

回答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);

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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));

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来源：https://stackoverflow.com/questions/33234979/how-to-write-a-template-function-that-takes-an-array-and-an-int-specifying-array

标签

c++

arrays

function

template-function