According to the C++23 standard, what is the output of this program?
#include <iostream>
size_t get_size_1(int* arr)
{
return sizeof arr;
}
size_t get_size_2(int arr[])
{
return sizeof arr;
}
size_t get_size_3(int (&arr)[10])
{
return sizeof arr;
}
int main()
{
int array[10];
// Assume sizeof(int*) != sizeof(int[10])
std::cout << (sizeof(array) == get_size_1(array));
std::cout << (sizeof(array) == get_size_2(array));
std::cout << (sizeof(array) == get_size_3(array));
}§[dcl.fct]¶5 in the standard:
any parameter of type “array of
T” (...) is adjusted to be “pointer toT”".
001
This question compares three ways for a function to take an array as parameter, but two of them are actually the same.
In main, the array object is of array type, therefore the sizeof operator returns the size of the array in terms of bytes. §[expr.sizeof]¶2:
When applied to an array, the result is the total number of bytes in the array. This implies that the size of an array of n elements is n times the size of an element.
In get_size_1, arr is a pointer to int, therefore the sizeof operator returns the size of the pointer. §[expr.sizeof]¶1:
The sizeof operator yields the number of bytes occupied by a non-potentially-overlapping object of the type of its operand.
In get_size_2, arr is in fact also a pointer to int. It looks like an array, but §[dcl.fct]¶5 says:
(...) any parameter of type “array of
T” or of function typeTis adjusted to be “pointer toT”.
So this case is the same as get_size_1, and the size of the pointer is returned again.
In get_size_3, arr is a reference to an array of size 10, therefore the sizeof operator returns the size of the array in terms of bytes, just like in main. §[expr.sizeof]¶2:
When applied to a reference or a reference type, the result is the size of the referenced type.
The referenced type is an array of 10 int, and as we saw above, §[expr.sizeof]¶2:
When applied to an array, the result is the total number of bytes in the array.