我经常假设枚举的大小与其基础类型的大小相同。但它是标准强制要求的吗？

标准 (C++14, n4296) 表示每个枚举都有一个基础类型 (7.2/5)。该标准还表示对象表示为字节序列，并且对象的大小与其表示有关:

3.9/4 The object representation of an object of type T is the sequence of N unsigned char objects taken up by the object of type T, where N equals sizeof(T).

5.3.3/1 The sizeof operator yields the number of bytes in the object representation of its operand.

但是，我无法找到枚举的基础类型与对象表示之间的任何关系。有没有？如果不是，那么我认为枚举的 sizeof 不必是其基础类型的 sizeof。

所以我的问题是:

1. 枚举的底层类型与其对象表示之间是否存在任何关系？

2. 标准是否真的需要 sizeof(std::underlying_type_t<E>) == sizeof(E)对于任何枚举 E?

最佳答案

取自:What is the underlying type of a c++ enum? , 7.2/5 中声明的旧 C++ 标准:

The underlying type of an enumeration is an integral type that can represent all the enumerator values defined in the enumeration. It is implementation-defined which integral type is used as the underlying type for an enumeration except that the underlying type shall not be larger than int unless the value of an enu- merator cannot fit in an int or unsigned int. If the enumerator-list is empty, the underlying type is as if the enumeration had a single enumerator with value 0. The value of sizeof() applied to an enu- meration type, an object of enumeration type, or an enumerator, is the value of sizeof() applied to the underlying type.

从 draft n4606 我能找到的最接近的是 7.2/7 + 8，它指出:

7) For an enumeration whose underlying type is not fixed, the underlying type is an integral type that can represent all the enumerator values defined in the enumeration. If no integral type can represent all the enumerator values, the enumeration is ill-formed. It is implementation-defined which integral type is used as the underlying type except that the underlying type shall not be larger than int unless the value of an enumerator cannot fit in an int or unsigned int. If the enumerator-list is empty, the underlying type is as if the enumeration had a single enumerator with value 0.

8) For an enumeration whose underlying type is fixed, the values of the enumeration are the values of the underlying type. Otherwise, for an enumeration where emin is the smallest enumerator and emax is the largest, the values of the enumeration are the values in the range bmin to bmax, defined as follows: Let K be 1 for a two’s complement representation and 0 for a ones’ complement or sign-magnitude representation. bmax is the smallest value greater than or equal to max(|emin| − K, |emax|) and equal to 2M − 1, where M is a non-negative integer. bmin is zero if emin is non-negative and −(bmax + K) otherwise. The size of the smallest bit-field large enough to hold all the values of the enumeration type is max(M, 1) if bmin is zero and M + 1 otherwise. It is possible to define an enumeration that has values not defined by any of its enumerators. If the enumerator-list is empty, the values of the enumeration are as if the enumeration had a single enumerator with value 0

一方面看起来足够接近，另一方面，删除了对 sizeof() 运算符的特定需求。 我仍然认为可以肯定地说这两个问题的答案都是肯定的。

关于c++ - 在 C++ 中，枚举的大小是否必须等于其基础类型的大小？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/47444081/