Why am I using the deleted function 'void std::ref(const _Tp&&) [with _Tp = int]'

Question

#include<iostream>
#include<utility>
#include<tuple>
#include<functional>
using namespace std;
int main()
{
    int i = 0;
    auto p = make_pair(ref(i), ref(i++));
    p.first++;
    p.second++;
    cout << "i = " << i << endl;
}

For example if I use ref() like this, the compiler will say

use of deleted function 'void std::ref(const _Tp&&) [with _Tp = int]'

however if my code is following

#include<iostream>
#include<utility>
#include<tuple>
#include<functional>
using namespace std;
int main()
{
    int i = 0;
    auto p = make_pair(ref(i), ref(++i));
    p.first++;
    p.second++;
    cout << "i = " << i << endl;
}

I will successfully get the output i = 3, so I can't understand why I get so different answers.

Answer 1

std::ref takes a variable and gives you something that acts like a reference to that variable.

i++ is not a variable; it is a temporary. That's because of how post-increment works; the original value is incremented but the expression evaluates to the old value, and a temporary is required to hold that old value so you can read it.

std::ref doesn't allow you to use a temporary, in order to avoid mistakes like this one. It would be a dangling reference, otherwise.

++i, on the other hand, just gives you back the original variable, so you can take a reference to that just fine.

However, you can't put i and ++i right next to each other like that; the two expressions are indeterminately sequenced (or something) with respect to each other. Avoid this kind of code at all costs.