if constexpr - why is discarded statement fully checked?

Question

I was messing around with c++20 consteval in GCC 10 and wrote this code

#include 
#include 
#include 

template          


        

Answer 1

This is just how constexpr if works. If we check [stmt.if]/2

If the if statement is of the form if constexpr, the value of the condition shall be a contextually converted constant expression of type bool; this form is called a constexpr if statement. If the value of the converted condition is false, the first substatement is a discarded statement, otherwise the second substatement, if present, is a discarded statement. During the instantiation of an enclosing templated entity ([temp.pre]), if the condition is not value-dependent after its instantiation, the discarded substatement (if any) is not instantiated.[...]

^{emphasis mine}

So we can see that we only do not evaluate the discarded expression if we are in a template and if the condition is value-dependent. main is not a function template so the body of the if statement is still checked by the compiler for correctness.

Cppreference also says this in their section about constexpr if with:

If a constexpr if statement appears inside a templated entity, and if condition is not value-dependent after instantiation, the discarded statement is not instantiated when the enclosing template is instantiated .

template<typename T, typename ... Rest>
void g(T&& p, Rest&& ...rs) {
    // ... handle p
    if constexpr (sizeof...(rs) > 0)
        g(rs...); // never instantiated with an empty argument list.
}

Outside a template, a discarded statement is fully checked. if constexpr is not a substitute for the #if preprocessing directive:

void f() {
    if constexpr(false) {
        int i = 0;
        int *p = i; // Error even though in discarded statement
    }
}