void 大小未知时的指针运算 [英] Pointer arithmetic when void has unknown size

问题描述

在 Visual Studio C++ 版本 9(可能还有其他版本)中，以下代码:

In Visual Studio C++ version 9 (and probably other versions too), the following code:

int a = sizeof(void);
void const *b = static_cast<void const *>("hello world");
b += 6;

生成这些 错误:

error C2070: 'void': illegal sizeof operand
error C2036: 'const void *' : unknown size

此代码在 GCC 下工作，它处理 sizeof(void) as 1.

This code works under GCC, which treats sizeof(void) as 1.

有什么办法可以绕过这个限制，因为为了指针算术的目的显式地强制转换为 char * 增加了混乱(void * 被很好地识别并用作指向原始内存的无类型指针).

Is there some way around this limitation, as casting explicitly to char * for purposes of pointer arithmetic adds to the confusion (void * is well recognised and used as a typeless pointer to raw memory).

• 请注意，我很清楚标准的存在.
• 我想要做原始指针算术.
• 我把sizeof(void)表明我很清楚这个事实它不是 1 是问题的原因.
• 代码示例只是为了演示什么需要生成错误.
• 我知道这不是使用 void 的正常"方式，但这是 C，并且会发生这些事情.
• 是的，人们需要在低级别执行此操作.我不是在追求为什么，我在追求如何.如果您想知道原因，请查看一些内核源代码或您友好的 glibc.

• Please note, I'm well aware of the existence of the standard.
• I want to do raw pointer arithmetic.
• I take sizeof(void) to show that I'm well aware the fact that it isn't 1 is the cause of the problem.
• The code example is simply to demonstrate what is required to generate the errors.
• I know this isn't a "normal" way to use void, but this is C, and these things happen.
• Yes, people need to do this at low-level. I'm not after why, I'm after how. If you want to know why, take a look at some kernel source, or your friendly glibc.

这个问题似乎引起了很大的混乱.问题不在于为什么sizeof(void) == 1 不是标准的，而是当它不是标准时该怎么办.

It seems this question has generated a great deal of confusion. The question is not about why having sizeof(void) == 1 is not standard, but what to do when it isn't.

在要进行单字节指针运算的情况下，结果证明强制转换为 char * 是正确答案，而不是因为 *(void *) 没有大小，但因为标准实际上保证 *(char *) 总是 1.因此，char * 的使用总是正确的，并且与 void * 与 GCC 扩展一致，用于原始指针算法.

In the instance that single-byte pointer arithmetic is to be done, it turns out that casting to char * is the correct answer, not because *(void *) has no size, but because the standard actually guarantees that *(char *) is always 1. Therefore the use of char * is always correct, and congruent with void * with the GCC extension for the purposes of raw pointer arithmetic.

为了进一步强调这一点，void * 是指向无类型内存的指针的正确选择，char * 是转换为 的正确类型，用于原始指针算术.

To further reinforce the point, void * is the correct choice for pointers to typeless memory, and char * is the correct type to cast to for raw pointer arithmetic.

推荐答案

它会出现 正确答案 是使用 char * 进行指针运算，因为 sizeof(char) 始终定义为 1，并且在任何平台上都具有最好的可寻址粒度.

It would appear the correct answer is to use char * for pointer arithmetic, because sizeof(char) is always defined to be 1, and to be of the finest addressable granularity on any platform.

所以简而言之，没有办法绕过限制，char *实际上是正确的方法.

So in short, there is no way around the limitation, char * is in fact the proper way to do it.

Matthias Wandel 拥有 正确答案，但理由不同.

Matthias Wandel had the right answer but with a different justification.

这篇关于void 大小未知时的指针运算的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！