C++ constexpr – Should It Be Used Everywhere?

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 3.14159265 (Pi) in radians it is much clearer written as follows:

const float oneeighty = DegreesToRadians( 180.0f );

Lots of good info here:

http://en.cppreference.com/w/cpp/language/constexpr

There is an easy trick:

template<int n>struct i{};
int foo(int){return 0;}
constexpr int foo(char){return 'a';}

template<class T=int, T x=1,i<foo(x)>* =nullptr>
bool bar(){return true;}
template<class T=int, T x=1,class...Ts>
bool bar(Ts...){return false;}

if int foo(int) is constexpr, a different overload of bar is chosen by default.

With different code running, any behaviour change can occur.

live example (simply change which #define X is commented out).

Design of the example:

The char overload prevents the above code from being ill-formed, no diagnostic required, as all templates must have a valid specialization. foo<char> supplies that. In practice, its existence is not required: ADL could find a foo from far away, overloaded on a some_type*, then pass some_type* as T. Which means no compilation unit could prove the code was ill-formed.

The Ts... makes that bar overload less-preferred. So if the first one matches, there is no ambiguity. Only if the first one fails to match (due to a SFINAE caused by foo(x) not being constexpr) does the second overload get called (or if, say, someone passed arguments to it).