Question

审议一个类型的<代码>T和一个点至T,T。

是否有办法复制<代码>。 T* ,at a given address x, without using memcpy ( and in a portable way?

Edit:

Just to be clear, I don t want to construct T at address x. I want to copy the value of the pointer to T, T*, at address x, as in

char* p = new char[s];
memcpy(&p[0], t, sizeof(T*)); // t is T*

Answer 1

www.un.org/Depts/DGACM/index_spanish.htm 如果你想在具体地址上构造一个物体,你将重新定位:

#include <new>

T oldT = /* .. */;
T* newT = new (x) T(oldT); // copy-constructed at address x

It s just like new T(), except you specify the buffer. (So take care that it s aligned and lives long enough.)

If you want to copy a value pointed to by a pointer, just dereference the pointer:

T* x = /* .. */;
T copyT = *x; // dereference x, make copy

在这两种情况下,将<条码>改为>。 U或U*等。

Answer 2

页: 1 你们要求的是:

void **address = ...; // address points to a location which can store a void*
T *t = ...;
// note T* and void* are the same size 
// now any of these three should work:
static_cast<T*&>(*address) = t;
*reinterpret_cast<T**>(address) = t;
void *t_addr = t;
*address = t_addr;

Answer 3

我以前从未使用过这个词,但我认为,安排新职位可能有助于解决你的问题。

http://www.parashift.com/c++-faq-lite/dtors.html#faq-11.10

Answer 4

T *& newVal = new (x) T**(oldVal);

Answer 5

void* X;
typeT* T;

X=x;
* (typeT*) X = *T;

This should be portable. Note however that enough space must appropriately allocated before the copy (=) operation takes place.

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