67. constexpr-ii

Question.6

Two developers implement a bitmask calculation:

Developer A -- inline:

inline uint32_t make_mask(int bit) {
   return 1U << bit;
}

Developer B -- constexpr:

constexpr uint32_t make_mask(int bit) {
   return 1U << bit;
}

When called as make_mask(5) where 5 is a literal, what is the difference?

Need Help? Refer to the Quick Guide below

No difference -- both compute at compile time
inline expands the code at the call site (shift still executes at runtime); constexpr computes 1<<5=32 at compile time and stores the literal
constexpr is slower because it adds type checking
inline computes at compile time; constexpr at runtime

Restart quiz!