c++ - 复制位模式: float 到uint32_t

Question

将float值的位模式复制到uint32_t中，反之亦然(不强制转换)，我们可以使用std::copy或memcpy逐位复制位。另一种方法是使用reinterpret_cast如下(？):

float f = 0.5f;
uint32_t i = *reinterpret_cast<uint32_t*>(&f);

要么

uint32_t i;
reinterpret_cast<float&>(i) = 10;

但是，有一个claim表示上面使用的两个reinterpret_cast会调用未定义的行为。真的吗？怎么样？

Answer 1

是的，这是未定义的行为，因为它违反了严格的别名规则:

[basic.lval]/10: If a program attempts to access the stored value of an object through a glvalue of other than one of the following types the behavior is undefined — the dynamic type of the object,

— a cv-qualified version of the dynamic type of the object,

— a type similar (as defined in 4.4) to the dynamic type of the object,

— a type that is the signed or unsigned type corresponding to the dynamic type of the object,

— a type that is the signed or unsigned type corresponding to a cv-qualified version of the dynamic type of the object,

— an aggregate or union type that includes one of the aforementioned types among its elements or non- static data members (including, recursively, an element or non-static data member of a subaggregate or contained union),

— a type that is a (possibly cv-qualifded) base class type of the dynamic type of the object,

— a char or unsigned char type.

由于在尝试访问uint32_t类型的对象时，float并非以上所述，因此行为未定义。