c++ - C(和 C++)中 char 的对齐是否保证为 1？

Question

alignof(char) 可以不是 1 吗？

来自 unofficial cppreference.com wiki :

The weakest (smallest) alignment is the alignment of the types char, signed char, and unsigned char, and it is usually 1.

“通常”似乎暗示它可能是别的东西。

C 标准中关于 char 对齐的唯一规定是(C11 N1570 6.2.8 第 1 段):

The alignment requirement of a complete type can be queried using an _Alignof expression. The types char, signed char, and unsigned charshall have the weakest alignment requirement.

但是，请考虑对齐的定义(C11 N1570 6.2.8 第 1 段，并且对于 C++11 的定义类似):

An alignment is an implementation-defined integer value representing the number of bytes between successive addresses at which a given object can be allocated.

据此，由于 sizeof(char) ≡ 1，表示相邻char元素之间的距离只能是1字节。

这有意义吗？

Answer 1

是的。尽管标准中没有明确规定此声明，但我想可以从中推断出:

N1570 6.5.3.4 The sizeof and _Alignof operators

4 When sizeof is applied to an operand that has type char, unsigned char, or signed char, (or a qualified version thereof) the result is 1. When applied to an operand that has array type, the result is the total number of bytes in the array.

以char为例。假设我们有一个 char charArr[2];。 sizeof charArr保证为2，sizeof charArr[0] = sizeof charArr[1] = 1。这意味着两个相邻的 char 对象代替了 2 个字节。

因此，可以推断“连续地址之间可以分配一个char的字节数”至少为1。另外，char的对齐方式必须是正整数，所以不能是1以外的任何数字。