Error narrow conversion when used with auto - c++

Question

Getting error in use of auto for the implicit conversion.
Capturing the return of v.size() using int variable is ok but auto complains.
Compiler error does inform it is because of narrowing conversion .But I would like to understand in terms memory how this happens and why auto is not allowing it do that whereas normal conversion is ok
#include <iostream>
#include <vector>
int main()
{
auto v = std::vector<int>{ 1, 2, 3 };
auto c = 'h';
auto n2 = int{c};
std::cout<<n2<<" "<<c;
auto size = int{v.size()};
std::cout<<size;
int size_1 = v.size();
std::cout<<size_1;
}
Error because of below line
auto size = int{v.size()};
main.cpp: In function 'int main()': main.cpp:11:27: error: narrowing
conversion of 'v.std::vector::size()' from
'std::vector::size_type {aka long unsigned int}' to 'int' inside
{ } [-Wnarrowing]
When that line is commented it works perfectly
#include <iostream>
#include <vector>
int main()
{
auto v = std::vector<int>{ 1, 2, 3 };
auto c = 'h';
auto n2 = int{c};
std::cout<<n2<<" "<<c;
//auto size = int{v.size()};
//std::cout<<size;
int size_1 = v.size();
std::cout<<size_1;
}
Output
104 h3
DEMO LINK

Answer 1

This has nothing to do with auto. You will get the same warning if you use
int size = int{v.size()};
The issue here is int{v.size()} is using braced initialization to initialize a temporary int. A narrowing conversion in braced initialization is what causes the diagnostic to be displayed.
Do note that
int size_1 = v.size();
Is also a narrowing conversion, it is just not one where the standard mandates that a warning/error be emitted.