为什么标准本身没有将 sizeof(bool) 定义为 1?

Why the sizeof(bool) is not defined to be one, by the Standard itself?(为什么标准本身没有将 sizeof(bool) 定义为 1?)

问题描述

char、signed char 和 unsigned char 的大小由 C++ 标准本身定义为 1 个字节.我想知道为什么它也没有定义 sizeof(bool) ?

Size of char, signed char and unsigned char is defined to be 1 byte, by the C++ Standard itself. I'm wondering why it didn't define the sizeof(bool) also?

C++03 标准 $5.3.3/1 说，

C++03 Standard $5.3.3/1 says,

sizeof(char), sizeof(signed char) 和sizeof(unsigned char) 为 1；这sizeof 应用于任何其他的结果基本类型(3.9.1)是实现定义.[注:在特别是 sizeof(bool) 和sizeof(wchar_t) 是实现定义.69)

sizeof(char), sizeof(signed char) and sizeof(unsigned char) are 1; the result of sizeof applied to any other fundamental type (3.9.1) is implementation-defined. [Note: in particular,sizeof(bool) and sizeof(wchar_t) are implementation-defined.69)

我理解 sizeof(bool) 不能小于一个字节.但是有什么理由为什么它应该大于 1 个字节吗?我并不是说实现将它定义为大于 1，但标准让它由实现定义好像它可能大于 1.

I understand the rationale that sizeof(bool) cannot be less than one byte. But is there any rationale why it should be greater than 1 byte either? I'm not saying that implementations define it to be greater than 1, but the Standard left it to be defined by implementation as if it may be greater than 1.

如果没有理由 sizeof(bool) 大于 1，那么我不明白为什么标准没有将它定义为 1 字节，因为它已经定义了 sizeof(char)，而且都是变体.

If there is no reason sizeof(bool) to be greater than 1, then I don't understand why the Standard didn't define it as just 1 byte, as it has defined sizeof(char), and it's all variants.

推荐答案

许多平台无法有效加载小于 32 位的值.他们必须加载 32 位，并使用移位和掩码操作来提取 8 位.对于单个 bool，你不希望这样，但对于字符串来说没问题.

Many platforms cannot effectively load values smaller than 32 bits. They have to load 32 bits, and use a shift-and-mask operation to extract 8 bits. You wouldn't want this for single bools, but it's OK for strings.

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