C#NOT(〜)按位运算符返回负值
问题描述：
为什么C#的按位NOT运算符返回(the_number*-1)-1?
Why does C#'s bitwise NOT operator return (the_number*-1)-1?
byte a = 1;
Console.WriteLine(~a); //equals -2
byte b = 9;
Console.WriteLine(~b); //equals -10
// Shouldn't a=0 and b=6?
我将如何在C#中执行此操作?
How would I do this in C#?
9 = 0b1001 -> NOT
= 0b0110 = 6
答
按位操作返回类型为int(带符号)的值.有符号整数使用二进制补码表示负数.从字节到整数时使用符号扩展.
Bitwise operations return a value of type int (signed). Signed integers use two's-complement to represent negative numbers. Sign extension is used when going from byte to int.
byte a = 1; // 0b00000001
int notA = ~a; // 0b11111110 = -128 + 64 + 32 + 16 + 8 + 4 + 2 = -2 (actually 0b11111111 11111111 11111111 11111110)
byte b = 9; // 0b00001001
int notB = ~9; // 0b11110110 = -128 + 64 + 32 + 16 + 4 + 2 = -10 (actually 0b11111111 11111111 11111111 11110110)
投射回字节将为您提供0b11110110
Casting back to byte will give you the "expected" result for 0b11110110
byte notB = unchecked((byte)(~b)); // 0b11110110 = 128 + 64 + 32 + 16 + 4 + 2
Console.WriteLine(notB); // 246