无符号字符的按位求反 [英] Bitwise negation of unsigned char

问题描述

这是一个与c99标准相关的问题，涉及整数提升和未签名字符的按位求反.

This is a question relating the c99 standard and concerns integer promotions and bitwise negations of unsigned char.

在6.5.3.3节中指出:

In section 6.5.3.3 it states that:

对操作数执行整数提升，结果 具有提升的类型.如果提升的类型是无符号类型，则 表达式〜E等于其中可表示的最大值 输入减号E.

The integer promotions are performed on the operand, and the result has the promoted type. If the promoted type is an unsigned type, the expression ~E is equivalent to the maximum value representable in that type minus E.

我说的那句话我理解正确吗?

Am I understanding that correctly when I say that, that means that:

unsigned int ui = ~ (unsigned char) ~0; // ui is now 0xFF00.

我的困惑源于与一位同事的讨论，他在我们的编译器中看到以下行为.

My confusion stems from a discussion with a collegue where he is seeing the following behaviour in our compiler.

unsigned char uc = 0;
unsigned char ucInverted = ~0;

if( ~uc == ~0 )              // True, obviously
if( ~ucInverted == ~(~0) )   // False as it evaluates to: 0xFF00 == 0x0000
if( ~uc == ucInverted )      // False as it evaluates to: 0xFFFF == 0x00FF
if( uc == ~ucInverted )      // False as it evaluates to: 0x0000 == 0xFF00

此行为已通过gcc确认.

This behaviour is confirmed with gcc.

是否有可能获得适当的预期字符比较，以使这些情况中的每一个都为真?

Is it possible to get proper expected char comparisons where each of these cases will evaluate to true?

推荐答案

~uc的类型为int(值0xFFFF). ucInverted的类型为unsigned char(值0xFF)，然后提升为int(值0x00FF).因此它们是不相等的.

~uc is of type int (value 0xFFFF). ucInverted is of type unsigned char (value 0xFF), which is then promoted to int (value 0x00FF). Thus they are not equal.

我想你可以if ((unsigned char)~uc == ucInverted).这两个操作数仍将进行提升，但是在提升之前它们将具有相同的值.

I guess you could do if ((unsigned char)~uc == ucInverted). Both operands will still undergo promotion, but they will have identical values before the promotion.

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