How can I have a temporary variable in a constexpr function?

问题

This is a simplified version what I would like to do.

constexpr float f(float a, float b){
    constexpr float temp = a+b;
    return temp*temp*temp;
}

In my version, a+b is something much more complicated, so I don't want to cut and paste it three times. Using 3*(a+b) is also not a working solution for the real function. I'm trying to keep the question related to syntax, and not algebra. I can get it to work by moving a+b to it's own constexpr function, but I'd prefer to not pollute the namespace with otherwise useless functions.

回答1:

As you've discovered, you can't declare variables, even constexpr ones, inside the body of a constexpr function.

It's still possible to factor out a common expression, by passing it in as an argument to a second constexpr function. For the example you've given here:

constexpr float pow3(float c) {
    return c*c*c;
}

constexpr float f(float a, float b) {
    return pow3(a+b);
}

回答2:

This is not permitted in C++11, but is now permitted in C++14.

See https://en.wikipedia.org/wiki/C%2B%2B14#Relaxed_constexpr_restrictions

来源：https://stackoverflow.com/questions/11917582/how-can-i-have-a-temporary-variable-in-a-constexpr-function