Passing const pointers to const by reference in functions

Question

I am playing a bit around with arrays and passing pointers by reference to functions. Consider as an example the following code:

#include<iostream>

void test_fn(const int* const &i){
    std::cout<<*i<<std::endl;
}

int main(){
    int arr_1[5] {1, 3, 6, 4, 5};
    int *int_ptr1 {nullptr};
    int_ptr1=arr_1;
    test_fn(int_ptr1);

    return 0;
}

This code runs properly.

However, if I amend the definition of the test_fn to read:

void test_fn(const int* &i){
    std::cout<<*i<<std::endl;
}

I get the following error:

cannot bind non-const lvalue reference of type 'const int*&' to an rvalue of type 'const int*'

My question is: Why is this an issue in the second example but not in the first, where the only difference is that the pointer is being passed as a const pointer?

Why a C-style array and not a std::array??

Jesper Juhl
– Jesper Juhl

2020-01-29 18:07:56 +00:00
Commented Jan 29, 2020 at 18:07 — Jesper Juhl
– Jesper Juhl, Commented Jan 29, 2020 at 18:07

Remy Lebeau · Accepted Answer · 2020-01-29 18:01:38Z

4

In the 2nd example, the function takes a reference to a non-const const int* pointer, but you are giving it an int* pointer instead. Because the pointer is non-const, the reference cannot be bound to a temporary const int* variable created by the compiler to implicitly convert from int* to const int*. The reference requires a non-const const int* variable at the call site, so you would have to explicitly declare int_ptr1 as const int* instead of int*:

const int *int_ptr1 = arr_1;

In the 1st example, the function takes a reference to a const const int* pointer. Since the pointer is const, the compiler can implicitly convert from int* to const int* and store the result in a temporary variable, and then bind the reference to that temporary.

edited Jan 29, 2020 at 18:01

answered Jan 29, 2020 at 17:51

Remy Lebeau

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1 Comment

François Andrieux Over a year ago

I'm not sure which object you are referring to when you say "it cannot be bound to a temporary variable created by the compiler". Is this a hypothetical variable? Edit : Nevermind, the result of the implicit conversion is the rvalue.

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