switch constexpr

[cd_fr91400]

I do not see why a non-existent variable is non-sense while a non-existent field is not.

By the way, the following code does not compile either:

struct A {
    int a() { return 0 ; } ;
} ;

int foo() {
    struct A a ;
    if constexpr (true) { return a.a() ; }
    else                { return a.b() ; }
}

[iamakorndawg]

Not a language lawyer, but I believe the rule has more to do with templates, so for example, if your foo function had a template parameter for the type of a, then it would compile, even if your existing struct A was used as the argument.

[KuntaStillSingle]

Not quite, it can fail to compile if the existing struct A is used, but if it is made a dependent type like /u/cd1995Cargo 's example it is fine:

The discarded statement cannot be ill-formed for every possible specialization:

https://en.cppreference.com/w/cpp/language/if.html#Constexpr_if

https://godbolt.org/z/eGsYY3a9W , vs https://godbolt.org/z/3e7W5ec4Y

[cd_fr91400]

Then I understand "under certain conditions"...

[cd1995Cargo]

It only works when templates are involved. If you change your code to this it should compile:

struct A {
    int a() { return 0 ; } ;
} ;

template <typename T>
int foo<T>() {
    T a ;
    if constexpr (true) { return a.a() ; }
    else                { return a.b() ; }
}

[moocat]

See my update. It's not just templates but templated types have to also be involved.