How Can I Guarantee a Compile-Time Error for an Unreachable `constexpr if-else` Clause?

Ensuring Compile Time Error for Unexhausted Constexpr if-else

When working with constexpr if-else statements, it's crucial to guarantee that all conditions are covered to prevent undefined behavior at runtime. However, some scenarios may arise where the else clause should never be reached. In such cases, how can you raise a compile time error to alert developers of incorrect flow?

Challenging the Traditional Approach

The initial inclination might be to rely on static_assert(false) within the else block. However, this approach is not permitted within constexpr statements. Instead, an alternative solution is necessary to flag the unreachable else clause.

Leveraging a Dependent False Type

To overcome this limitation, we can utilize a dependent false type. By introducing a helper template like constexpr std::false_type always_false{};, we create a type that always evaluates to false.

Integrating the Dependent False Type

Within the constexpr if-else statement, we can now leverage always_false as follows:

if constexpr(condition1){
    ...
} else if constexpr (condition2) {
   ....
} else if constexpr (condition3) {
  ....
} else {       
    static_assert(always_false<T>);
}

This approach relies on the fact that if no valid specialization can be generated for the template always_false, the compiler will raise an error during template instantiation, signaling that the else clause is unreachable.

Conclusion

By adopting this technique, developers can confidently handle situations where all constexpr if-else conditions must be accounted for, avoiding any potential errors during compilation.

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