这是使用std :: array未定义的行为吗？ [英] Is this use of std::array undefined behavior?

问题描述

可能重复：

允许使用`char *`进行别名`T *`。

Possible Duplicate:
Aliasing `T*` with `char*` is allowed. Is it also allowed the other way around?

我使用的是 std :: array of char 用于保存未知原始类型的值，长度不超过10个字节，像这样：

I'm using a std::array of chars to hold a value of unknown primitive type, which is no more than 10 bytes long, like so:

std::array<char, 10> val;
*reinterpret_cast<double*>(val.data()) = 6.3;
//blah blah blah...
double stuff = *reinterpret_cast<double*>(val.data());

我已经读过通过 char * 未定义，因为编译器假设 char * 可以别名任何类型的值。当值放在（我假设是）数组 char 在对象内时，这仍然工作？

I have read that casting back and forth through char * is not undefined, because the compiler assumes a char * may alias a value of any type. Does this still work when the value is placed in (what I assume is) an array of chars inside the object?

注意：我知道我可以在这里使用联合，但这将导致大量的样板代码，我正在做什么，我想避免它，如果必要，因此的问题。 p>

Note: I am aware that I could be using a union here, but that would result in a large amount of boilerplate code for what I am doing, and I would like to avoid it if necessary, hence the question.

推荐答案

是的， std :: array< char，10> 不满足 double 的对齐要求，因此 reinterpret_cast UB。

Yes, std::array< char, 10 > does not satisfy the alignment requirements of double so that reinterpret_cast provokes UB.

请尝试 std： ：aligned_storage 。

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