整数文字如何存储？

Question

Does a small integer literal in C# (eg 12 ) uses 4 bytes of stack like an integer variable‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌‌? Does it need 4 bytes?

Answer 1

Does a short integer literal in C# ...

You cannot declare a literal of type short so the question is iffy. In practice, the C# compiler will treat it like an int (or long if large enough) and readily convert it to byte or short where appropriate without a cast. And complain when such a conversion causes an overflow.

It will end up getting encoded in a processor instruction like MOV or PUSH, depending on how you use the literal. A practical example, use the Debug > Windows > Disasssembly window to see it:

    static void Main(string[] args) {
        Console.WriteLine(12);
    }

Generates:

  005B2DB0 B9 0C 00 00 00       mov         ecx,0Ch  
  005B2DB5 E8 7E 38 BE 72       call        73196638  
  005B2DBA C3                   ret  

Note the MOV instruction and the instruction bytes it generates. B9 is the "move 32-bit immediate" instruction, the next 4 bytes are the value in little-endian order. Otherwise selected because the C# compiler used the WriteLine(Int32) overload, it does not have overloads for byte or short . Use the same technique to see what happens with your specific code.

Answer 2

You can't declare a short numeric literal. You can only declare numeric literals of types int , float ( f suffix), double ( d suffix), uint ( u suffix), long ( l suffix), ulong ( ul suffix) and decimal ( m suffix).

All numeric literals without suffixes eg 12 are inferred to be of the first of these types in which their value can be represented: int , uint , long , ulong . so if 12 were put on the stack, it would require 4 bytes on a 32-bit system.