C ++ offsetof char *算术 [英] C++ offsetof char* arithmetic

问题描述

在此答案中，int8_t*用于(字节)指针算术:

In this answer, int8_t* is used for (byte) pointer arithmetic:

std::size_t offset = offsetof(Thing, b);
Thing* thing = reinterpret_cast<Thing*>(reinterpret_cast<int8_t*>(ptr) - offset);

过去我一直使用char*，但是评论确实令人困惑，没有人回应，所以我发布了这个单独的问题.

I've always used char* in the past but the comments are really confusing, and nobody responded, so I posted this separate question.

char*是否有效，并且是进行这些计算的首选方式?

Is char* valid, and the preferred way of doing these calculations?

推荐答案

您必须使用char*:将reinterpret_cast与int8_t*一起使用的行为指向指向以下内容的指针不是未定义 .强制转换为char*可以视为规则的例外.

You must use char*: the behaviour on using a reinterpret_cast with int8_t* on a pointer to something that is not an int8_t is undefined. Casting to char* can be viewed as an exception to the rule.

Pre C ++ 14，char可以是-127到+127的1's补码类型. int8_t必须为2的补码.即使是C ++ 14及更高版本，我也看不到为什么类型相关:char仍然可以是有符号或无符号类型.

Pre C++14, char can be a 1's complement type with range -127 to +127. int8_t must be 2's complement. Even C++14 and onwards, I can't see why the types are related: char can still be either a signed or an unsigned type.

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