为什么是布尔在C ++中int32_t，并在中int8_t常见的类型？ [英] Why is the common type of bool and int8_t an int32_t in C++?

问题描述

我很好奇一些内置的布尔键入C ++的行为。据我了解， 的std :: common_type 决定使用隐式可兑换常见的类型。我期望与布尔和其他类型的前pression将导致布尔转换为该类型。例如，我可以看到布尔+浮动 - > 浮动和布尔+双 - > 双击。然而，布尔+中int8_t - > int32_t 和布尔+ int16_t - > int32_t 。为什么会出现这种情况？

I'm curious about some of the behavior of the builtin bool type in C++. As I understand it, std::common_type determines the common type using implicit convertibility. I would expect that an expression with bool and another type would cause bool to convert to that type. For instance, I can see that bool + float -> float and bool + double -> double. However, bool + int8_t -> int32_t and bool + int16_t -> int32_t. Why is this the case?

推荐答案

简短的回答：积分的推广

在数值运算，小整型（包括布尔，字符， unsigned char型，符号字符，短，无符号短等）被晋升为 INT 如果所有可能的值适合 INT ，否则会被提拔到 unsigned int类型。

In numerical arithmetic, small integral types (including bool, char, unsigned char, signed char, short, unsigned short, etc) are promoted to int if all the possible values fit in int, otherwise they are promoted to unsigned int.

在大多数机器的今天， int32_t 相同 INT 。在的情况下，布尔+中int8_t 或布尔+ int16_t ，两者都提升到 INT 。

On most machines today, int32_t is the same as int. In the case of bool + int8_t or bool + int16_t, both are promoted to int.

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