c++ - 使用指针算法计算类型大小的替代方法

Question

以下代码是否 100% 可移植？

int a=10;
size_t size_of_int = (char *)(&a+1)-(char*)(&a); // No problem here?

std::cout<<size_of_int;// or printf("%zu",size_of_int);

P.S:此题仅供学习。所以请不要给出像 Use sizeof() 等

这样的答案

Answer 1

来自 ANSI-ISO-IEC 14882-2003，p.87 (c++03):

"75) Another way to approach pointer arithmetic is first to convert the pointer(s) to character pointer(s): In this scheme the integral value of the expression added to or subtracted from the converted pointer is first multiplied by the size of the object originally pointed to, and the resulting pointer is converted back to the original type. For pointer subtraction, the result of the difference between the character pointers is similarly divided by the size of the object originally pointed to."

这似乎表明指针差异等于对象大小。

如果我们从递增指向标量 a 的指针中删除 UB'ness 并将 a 转换为数组:

int a[1];
size_t size_of_int = (char*)(a+1) - (char*)(a);

std::cout<<size_of_int;// or printf("%zu",size_of_int);

那么这看起来没问题。关于对齐要求的条款与脚注一致，如果对齐要求总是能被对象的大小整除。

更新:很有趣。大多数人可能都知道，GCC 允许将类型的显式对齐指定为扩展。但我不能用它破坏 OP 的“sizeof”方法，因为 GCC 拒绝编译它:

#include <stdio.h>

typedef int a8_int __attribute__((aligned(8)));

int main()
{
 a8_int v[2];

 printf("=>%d\n",((char*)&v[1]-(char*)&v[0]));
}

消息是错误:数组元素的对齐方式大于元素大小。