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

Question

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.

This is the #1 suckiest thing about constexpr.

Joseph Garvin
– Joseph Garvin

2012-10-14 18:12:22 +00:00
Commented Oct 14, 2012 at 18:12 — Joseph Garvin
– Joseph Garvin, Commented Oct 14, 2012 at 18:12

Veloso · Accepted Answer · 2012-08-23 13:37:22Z

6

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);
}

answered Aug 23, 2012 at 13:37

Veloso

1902 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

John K · Accepted Answer · 2015-02-25 21:39:25Z

3

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

edited Feb 25, 2015 at 21:39

answered Aug 14, 2012 at 18:01

John K

5063 silver badges16 bronze badges

3 Comments

Jan Hudec Over a year ago

You should do it the other way around. f2(float temp) { return temp * temp * temp; } and f(float a, float b) { return f2(a + b); }. That way you avoid the 3 calls.

John K Over a year ago

This doesn't matter with constexpr because it indicates to the compiler that the results of the function can be memorized and/or evaluated at compile time. The compiler will call temp once and reuse the value 3 times. Grand total of 1 add, 3 multiplies your way and my way. But, as I said in the intro, I simplified the problem because the question isn't about algebra.

celtschk Over a year ago

@JohnK: It doesn't matter for the compiled code, but it does matter for the humans writing and reading that code. For the one writing it: Less repetition = less opportunity to make an error. For the one reading it: It is easier to figure out that you've got three times the same value.

Collectives™ on Stack Overflow

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

2 Answers 2

Comments

3 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related