Warren  P Warren P - 1 year ago 98
C++ Question

When should you use constexpr capability in C++11?

It seems to me that having a "function that always returns 5" is breaking or diluting the meaning of "calling a function". There must be a reason, or a need for this capability or it wouldn't be in C++11. Why is it there?

// preprocessor.
#define MEANING_OF_LIFE 42

// constants:
const int MeaningOfLife = 42;

// constexpr-function:
constexpr int MeaningOfLife () { return 42; }

It seems to me that if I wrote a function that return a literal value, and I came up to a code-review, someone would tell me, I should then, declare a constant value instead of writing return 5.

Goz Goz
Answer Source

Suppose it does something a little more complicated.

constexpr int MeaningOfLife ( int a, int b ) { return a * b; }

const int meaningOfLife = MeaningOfLife( 6, 7 );

Now you have something that can be evaluated down to a constant while maintaining good readability and allowing slightly more complex processing than just setting a constant to a number.

It basically provides a good aid to maintainability as it becomes more obvious what you are doing. Take max( a, b ) for example:

template< typename Type > constexpr Type max( Type a, Type b ) { return a < b ? b : a; }

Its a pretty simple choice there but it does mean that if you call max with constant values it is explicitly calculated at compile time and not at runtime.

Another good example would be a DegreesToRadians function. Everyone finds degrees easier to read than radians. While you may know that 180 degrees is in radians it is much clearer written as follows:

const float oneeighty = DegreesToRadians( 180.0f );

Lots of good info here:


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