什么是按位运算符和整数提升？ [英] What is going on with bitwise operators and integer promotion?

问题描述

我有一个简单的程序。 请注意，我使用了大小为1字节的未签署固定宽度整数。

I have a simple program. Notice that I use an unsigned fixed-width integer 1 byte in size.

#include <cstdint>
#include <iostream>
#include <limits>

int main()
{
    uint8_t x = 12;
    std::cout << (x << 1) << '\n';
    std::cout << ~x;

    std::cin.clear();
    std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
    std::cin.get();

    return 0;
}

我的输出如下。

24
-13

我测试了更大的数字，运算符< 总是给出正数，而运算符〜我负数。然后我使用 sizeof（）并找到...

I tested larger numbers and operator << always gives me positive numbers, while operator ~ always gives me negative numbers. I then used sizeof() and found...

当我使用左移位向运算符（< ）时，我收到一个无符号的4字节整数。

When I use the left shift bitwise operator(<<), I receive an unsigned 4 byte integer.

当我使用bitwise not运算符（〜）时，我收到一个有符号的4字节整数。

When I use the bitwise not operator(~), I receive a signed 4 byte integer.

似乎bitwise不是运算符（〜）算术运算符。但是，左移位运算符（< <）似乎提升为无符号整数。

It seems that the bitwise not operator(~) does a signed integral promotion like the arithmetic operators do. However, the left shift operator(<<) seems to promote to an unsigned integral.

有义务知道编译器何时改变我背后的东西。如果我在我的分析是正确的，所有的位运算符提升到一个4字节整数？为什么有些签名和一些未签名？我很困惑！

I feel obligated to know when the compiler is changing something behind my back. If I'm correct in my analysis, do all the bitwise operators promote to a 4 byte integer? And why are some signed and some unsigned? I'm so confused!

编辑：我的假设总是得到积极或总是得到负值是错误的。但是，从错误的角度来看，我明白了下面这些伟大的答案真正发生了什么。

My assumption of always getting positive or always getting negative values was wrong. But from being wrong, I understand what was really happening thanks to the great answers below.

推荐答案

[expr.unary.op]

[expr.unary.op]

〜的操作数应具有整数或无范围的枚举类型;
结果是其操作数的一个补码。 整合促销活动为
。

The operand of ~ shall have integral or unscoped enumeration type; the result is the one’s complement of　its operand. Integral promotions are performed.

[expr.shift]
$ b

[expr.shift]

移位运算符< 和>> 组从左到右。 [...]操作数应为整数或无范围的枚举类型，并且执行整数促销。

The shift operators << and >> group left-to-right. [...] The operands shall be of integral or unscoped enumeration type and integral promotions are performed.

uint8_t （幕后通常是 unsigned_char ）的整体推广是什么？

What's the integral promotion of uint8_t (which is usually going to be unsigned_char behind the scenes)?

[conv.prom]

[conv.prom]

除 bool ， char16_t ， char32_t 或
wchar_t ，其整数转换等级（4.13）小于
int 的等级可以转换为类型 int 如果 int 可以表示所有
源类型的值;否则，源prvalue可以
转换为 unsigned int 的prvalue。

A prvalue of an integer type other than bool, char16_t, char32_t, or wchar_t whose integer conversion rank (4.13) is less than the rank of int can be converted to a prvalue of type int if int can represent all the values of the source type; otherwise, the source prvalue can be converted to a prvalue of type unsigned int.

所以 int ，因为 uint8_t 的所有值都可以用 int 。

So int, because all of the values of a uint8_t can be represented by int.

什么是 int（12）< 1 ？ int（24）。

〜int $ c>？ int（-13）。

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