Using templates to pass array with N elements in C++

Question

Here we pass a reference to an entire array using templates, and it compiles fine:

template <int N>
void test(int (&in)[N]){
}

int main(){
 int a[2] = {1,2};
 test(a);
 return 0;
}

But doing the same to pass a regular array doesn't work, the template can't figure out what N is on its own, why?

template <int N>
void test(int in[N]){
}

int main(){
 int a[2] = {1,2};
 test(a);
 return 0;
}

error: couldn't deduce template parameter 'N'

Use `std::array a;` instead of `int a[2];`. There is no overhead, they can be copied like normal objects and never accidentally decay to pointers. — François Andrieux, Apr 20 '21 at 15:15
Answering the "why" - https://stackoverflow.com/questions/33291624 — Drew Dormann, Apr 20 '21 at 15:17
Related:Dupe: [Length of array in function argument](https://stackoverflow.com/questions/8269048/) — Remy Lebeau, Apr 20 '21 at 15:47

463035818_is_not_an_ai · Accepted Answer · 2021-04-21T08:11:01.690

It is not "doing the same way". When a function takes an array like this:

void foo( int x[] )

Then that is just a slightly obfuscated way to write this:

void foo( int* x)

Arrays decay to pointers when passed to functions. You cannot pass arrays by value.

You can pass arrays by reference and thats what your first code does. Maybe the template is confusing you. Consider that if you know the dimension you can write:

void test(int (&in)[2]){
}

Then in inside test will be a reference to an array with two elements, but if you pass a pointer then its just a pointer and information on the array size is lost.

score 2 · Answer 2 · answered Apr 20 '21 at 15:15

2

C-arrays cannot be passed as value and decay to pointer, making N non deducible.

answered Apr 20 '21 at 15:15

Jarod42

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Using templates to pass array with N elements in C++

2 Answers2