'Why doesn't an if constexpr make this core constant expression error disappear?

In reference to this question. The core constant expression that is used to initialize the constexpr variable y is ill-formed. So much is a given.

But if I try to turn the if into an if constexpr:

template <typename T>
void foo() {
    constexpr int x = -1;
    if constexpr (x >= 0){
        constexpr int y = 1 << x;
    }
}

int main(){
    foo<int>();
}

The error persists. With GCC 7.2 still giving:

error: right operand of shift expression '(1 << -1)' is negative [-fpermissive]

But I thought that the semantic check should be left unpreformed on a discarded branch.

Making an indirection via a constexpr lambda does help, however:

template <typename T>
void foo(){
    constexpr int x = -1;
    constexpr auto p = []() constexpr { return x; };
    if constexpr (x >= 0){
        constexpr int y = 1<<p();
    }
}

The constexpr specifier on y seems to alter how the discarded branch is checked. Is this the intended behavior?

@max66 was kind enough to check other implementations. He reports that the error is reproducible with both GCC (7.2.0 / Head 8.0.0) and Clang (5.0.0 / Head 6.0.0).

c++language-lawyerc++17constexprif-constexpr


Solution 1:[1]

The standard doesn't say much about the discarded statement of an if constexpr. There are essentially two statements in [stmt.if] about these:

  1. In an enclosing template discarded statements are not instantiated.
  2. Names referenced from a discarded statement are not required ODR to be defined.

Neither of these applies to your use: the compilers are correct to complain about the constexpr if initialisation. Note that you'll need to make the condition dependent on a template parameter when you want to take advantage of the instantiation to fail: if the value isn't dependent on a template parameter the failure happens when the template is defined. For example, this code still fails:

template <typename T>
void f() {
    constexpr int x = -1;
    if constexpr (x >= 0){
        constexpr int y = 1<<x;
    }
}

However, if you make x dependent on the type T it is OK, even when f is instantiated with int:

template <typename T>
void f() {
    constexpr T x = -1;
    if constexpr (x >= 0){
        constexpr int y = 1<<x;
    }
}
int main() {
    f<int>();
}

Solution 2:[2]

Note that for the statement discarded by Constexpr If:

the discarded statement can't be ill-formed for every possible specialization:

To fix the issue you can make the statement depending on the template parameter, e.g.

template<typename T, int X> struct dependent_value { constexpr static int V = X; };

template <typename T>
void foo() {
    constexpr int x = -1;
    if constexpr (x >= 0){
        constexpr int y = 1 << dependent_value<T, x>::V;
    }
}

LIVE

Solution 3:[3]

I'm not sure why you expect the branch to not be checked. The only time an if branch is "not checked" is when it is part of a template and not instantiated, as per [stmt.if]p2:

During the instantiation of an enclosing templated entity (Clause 17), if the condition is not value-dependent after its instantiation, the discarded substatement (if any) is not instantiated.

Your code doesn't seem to be in a situation where this applies.

Sources

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Source: Stack Overflow

Solution Source
Solution 1 Potatoswatter
Solution 2 songyuanyao
Solution 3 Kerrek SB

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