C ++ 11用户定义的文字比普通的类型转换慢吗？

Question

1) is any of these faster than the other at runtime? which and why?
2) does this occur at compile time or runtime?

unsigned short operator"" _ushort( unsigned long long arg ){ return arg; }

unsigned short my_var = 0x1234;          // using type and literal
auto my_var = unsigned short(0x1234);    // using auto and casting literal to type
auto my_var = 0x1234_ushort;             // using auto and user defined literal to cast

edit: does using constexpr help it?

constexpr unsigned short operator"" _ushort( unsigned long long arg ){ return arg; }

Answer 1

所有这些都在编译时初始化，因此对它们中的任何一个都没有运行时影响。

Answer 2

Let us see from the generated assemblies... using this tool: https://gcc.godbolt.org

The assembly produced by clang is: (modified your code to compile)

For this input,

inline unsigned char operator "" _kx ( unsigned long long arg ){ return arg; }

unsigned char my_var = 0x14;          // using type and literal
auto my_var2 = (unsigned char) 0x1234;    // using auto and casting literal to type
auto my_var3 = 0x1234_kx;             // using auto and user defined literal to cast

The assembly generated is

my_var:
        .byte   20                      # 0x14

my_var2:
        .byte   52                      # 0x34

my_var3:
        .byte   52                      # 0x34

So, there is no performance hit... rather a gain in flexibility.... Though the operator function seems to still be created in certain compiler versions under certain flags.... The values are initialized at compile-time

https://godbolt.org/g/YuJyQd