使用std :: memcpy在内存中移动对象

Question

是否允许将类实例对象从一个位置移动到另一个位置（例如，通过使用std::memcpy或std::memove ？假设源位置和目标位置具有相同的对齐方式。然后将目标“object”转换为源对象的类型并调用它.C ++ 11标准的哪一部分禁止这个？

Answer 1

You should consider the template is_trivially_copyable provided by C++11 that is useful to understand if it safe to memcpy a given type. If the value is false then the result is undefined.

Answer 2

As long as they are is_trivially_copyable , it is safe.

§ 3.9.2

For any object (other than a base-class subobject) of trivially copyable type T , whether or not the object holds a valid value of type T, the underlying bytes (1.7) making up the object can be copied into an array of char or unsigned char.40 If the content of the array of char or unsigned char is copied back into the object, the object shall subsequently hold its original value.

#define N sizeof(T)
char buf[N];
T obj; // obj initialized to its original value
std::memcpy(buf, &obj, N); // between these two calls to std::memcpy,
                           // obj might be modified
std::memcpy(&obj, buf, N); // at this point, each subobject of obj of scalar type
                           // holds its original value

§ 3.9.3

For any trivially copyable type T , if two pointers to T point to distinct T objects obj1 and obj2, where neither obj1 nor obj2 is a base-class subobject, if the underlying bytes (1.7) making up obj1 are copied into obj2,41 obj2 shall subsequently hold the same value as obj1. [ Example:

T* t1p;
T* t2p;
    // provided that t2p points to an initialized object ...
std::memcpy(t1p, t2p, sizeof(T));
    // at this point, every subobject of trivially copyable type in *t1p contains
    // the same value as the corresponding subobject in *t2p