Assignment to a prvalue in C++

I was exploring a cppreference page about value categories and found some strange syntax in an example (the second extended content):

#include <iostream>
 
struct S
{
    S() : m{42} {}
    S(int a) : m{a} {}
    int m;
};
 
int main()
{
    S s;
 
    // Expression `S{}` is prvalue
    // May appear on the right-hand side of an assignment expression
    s = S{};
 
    std::cout << s.m << '\n';
 
    // Expression `S{}` is prvalue
    // Can be used on the left-hand side too
    std::cout << (S{} = S{7}).m << '\n';
}

Look at the last line of main(). There is an assignment S{} = S{7} that is very awkward. We can do this even simpler:

#include <iostream>

struct S {
    int m = 1;
};

int main()
{
    std::cout << (S{} = S{}).m << '\n';
}

Compiler shows no warning or errors and everything is completely fine. That could be okay, but I think the documentation says it must be impossible. Quoting the same page:

An rvalue can’t be used as the left-hand operand of the built-in assignment or compound assignment operators.

I guess S{} is a prvalue, and all other conditions are met also. So, how does C++ converts this prvalue into an lvalue?