windows 64 位机器上的 DWORD 和 DWORD_PTR
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DWORD and DWORD_PTR on 64 bit machine
提问by Nikhil
There are few *_PTRtypes added to the Windows API in order to support Win64's 64bit addressing.
*_PTR为支持 Win64 的 64 位寻址，向 Windows API 添加的类型很少。
SetItemData(int nIndex,DWORD_PTR dwItemData)
This API works for both 64 and 32 bit machines when I pass second parameter as DWORD.
当我将第二个参数作为DWORD.
I want to know, if this particular API will fail on 64 bit machine, if I pass the second parameter as DWORD. How can I test the fail scenario?
我想知道，如果这个特定的 API 会在 64 位机器上失败，如果我将第二个参数作为DWORD. 如何测试失败场景？
Thanks, Nikhil
谢谢，尼基尔
采纳答案by Frédéric Hamidi
The function will not fail if you pass a DWORD, because it fits into a DWORD_PTR. A pointer, however, is guaranteed to fit into a DWORD_PTRbut notinto a DWORDon 64-bit platforms.
如果您传递 a DWORD，该函数不会失败，因为它适合 a DWORD_PTR。一个指针，然而，是保证装配到DWORD_PTR但不为DWORD在64位平台。
Thus, this code is correct:
因此，这段代码是正确的：
int *before_ptr = new int;
yourListBox.SetItemData(index, (DWORD_PTR) before_ptr);
int *after_ptr = (int *) yourListBox.GetItemData(index);
ASSERT(before_ptr == after_ptr); // Succeeds.
delete after_ptr; // Works.
But this code is wrong and will silently truncate the pointer to its lower 32 bits:
但是这段代码是错误的，它会默默地将指针截断到它的低 32 位：
int *before_ptr = new int;
yourListBox.SetItemData(index, (DWORD) before_ptr);
int *after_ptr = (int *) yourListBox.GetItemData(index);
ASSERT(before_ptr == after_ptr); // Fails.
delete after_ptr; // Undefined behavior, might corrupt the heap.
