关于小尾数和大尾数的按位非运算符（C中的〜） [英] Bitwise Not Operator (~ in C) with regards to little endian and big endian

问题描述

这与家庭作业有关，但与家庭作业无关。

This is in relation to a homework assignment but this is not the homework assignment.

我很难理解按位不（在C语言中是〜）如何存在差异在大型字节序计算机与小型字节序计算机上编译时，受影响的 signed int 和 unsigned int 。

I'm having difficultly understanding if there is a difference on how the bitwise not (~ in C) would affected signed int and unsigned int when compiled on a big endian machine vs. a little endian machine.

确实是个字节是向后吗？如果不是，按位运算符（和其他运算符）是否会导致不同的结果会根据机器类型产生int 吗？

Are the bytes really "backwards" and if so does the bitwise not (and other operators) cause different resulting ints be produced depending on the machine type?

虽然我们这样做了，但对于每个按位运算符，答案都是一样的吗？ C还是严重依赖？

While we are at it, is the answer the same for each of the bitwise operators in C or does it heavily depend?

我要指的运算符是：

~  /* bitwise Not */
&  /* bitwise And */
|  /* bitwise Or */
^  /* bitwise Exclusive-Or */

谢谢

更新：到目前为止，在阅读我的回答时，我不得不问问按位非运算符是否会影响符号上的符号 signed int 。恐怕我在这方面有些困惑，因为我忘记了所有的静止状态。亚当似乎在说所有值都被视为未签名。是重新应用符号位还是曾经签名的值变为无符号？

Update: In reading my responses thus far, I feel compelled to ask if the bitwise not operator affects the sign bit on a signed int. I'm afraid I've been a bit confused on this part as I forgot about all that stillyness. Adam seems to be stating that all values are treated as unsigned. Is the sign-bit reapplied or does the once signed value become unsigned?

推荐答案

按位运算符和逻辑运算符在大端和小端的机器。同样，它们对带符号和无符号数据也进行相同的运算：结果就好像所有内容都是无符号的一样。因此，如果 x 和 y 是有符号整数，则〜x 等于（int）（〜（（unsigned int）x）和 x& y 等于（int）（（（（unsigned int）x）&（（unsigned int）y），对于所有其他运算符，依此类推。

The bitwise operators and logical operators all operate identically in big- and little-endian machines. Likewise, they also operator identically on signed and unsigned data: the result is as if everything were unsigned. So, if x and y are signed integers, then ~x equals (int)(~((unsigned int)x) and x & y equals (int)(((unsigned int)x) & ((unsigned int)y), and so on for all of the other operators.

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