Passing an array to a function declared in two different ways-CodePudding

So, I have declared an array using the <array> header in C , and now I want to pass this array to a function that doesn't return any value but displays each element in a new line on the terminal. Following is the code for that.

// to pass an array to a function
#include <iostream>
#include <array>
using namespace std;

void myFunction(int arr[5]); // function prototype

int main () {
    array <int, 5> arr{1, 2, 3, 4, 5};

    myFunction(arr);
}

// function definition
void myFunction(int arr[5]) {
    for (size_t counter{0}; counter < 5;   counter) {
        cout << arr[counter] << endl;
    }
}

The above code isn't working.

arr no suitable conversion function from "std::array<int, 5ULL>" to "int *" existsC/C  (413)

...

demo.cpp: In function 'int main()':
demo.cpp:11:16: error: cannot convert 'std::array<int, 5>' to 'int*'
   11 |     myFunction(arr);
      |                ^~~
      |                |
      |                std::array<int, 5>
demo.cpp:6:21: note:   initializing argument 1 of 'void myFunction(int*)'
    6 | void myFunction(int arr[5]); // function prototype
      |

These two are the error messages getting shown to me. Now on the other hand, if I declare the array differently, without using the standard library header, it works fine.

arr[5] {1, 2, 3, 4, 5}; // instead of 
                        // array <int, 5> arr{1, 2, 3, 4, 5};

It produces the following result

I just want to know what the array declarations have to do with any of this? If there is a way of doing this using the declaration with <array> header, what is it?

CodePudding user response：

The problem is that your function expects an int* but you're passing a array <int, 5> but since there is no implicit conversion from array <int, 5> to int*, we get the mentioned error.

You can also use std::array::data as myFunction(arr.data()); to make this work.

The reason your second case works when you create array like int arr[5] {1, 2, 3, 4, 5}; and then pass it like myFunction(arr) is because a built in array decays to a pointer to its first element(int* in your example) due to type decay. So in this modified case, the type of the argument and the parameter matches and this works.

CodePudding user response：

Your formal argument must have the same type as your factual.

If you try to call with std::array<int, 5> then declare your function like void myFunction(std::array<int, 5> arr).

Need to know that such declaration means copy your array. It may have no need. Then you may use a reference. If need just to read without change and move then const lvalue reference is enough. Ready prototype:

void myFunction (std::array<int,5> const &arr);

CodePudding user response：

If you can compile with C 20, I would strongly suggest you change your function to:

void myFunction(std::span<const int> arr) {
    for (const int elm : arr) {
        std::cout << elm << std::endl;
    }
}

By using span as the parameter, your function can be called with any type of continous int of any size, eg.

std::vector<int> vec{1,2,3};
std::array arr{1,2,3,4,5};
int oldArray[4] = {1,2,3,4};

myFunc(vec);
myFunc(arr);
myFunc(oldArray);