64 位 Windows API:C/C++“DWORD"的大小是多少? [英] 64-bit Windows API: what is the size of a C/C++ "DWORD"?

问题描述

我只安装了 32 位 Windows，所以我无法自己验证.

I only have 32-bit Windows installed, so I cannot verify this myself.

如果我没理解错的话，Microsoft API 中各处使用的 DWORD 是参考了原来的 16 位字，与当前的硬件架构无​​关?

If I understand correctly, the DWORD used in various places in the Microsoft API is in reference to the original 16-bit word, and has nothing to do with the current hardware architecture?

因此，即使我最终编译并链接我的应用程序以在 64 位 Windows 中运行，看起来 32 位的 DWORD 仍将是 32 位吗?还是 DWORD 会变成 128 位宽?

So DWORD which seems to be 32 bits, will remain 32 bits even when I eventually compile and link my app to run in 64-bit Windows? Or will DWORD become 128 bits wide?

推荐答案

唯一能在 32 到 64 之间改变大小的是指针.所以 DWORD 保持 32 位宽.

The only thing that changes size between 32 and 64 are pointers. So DWORD stays 32 bits wide.

有些东西不是立即明显的指针，例如手柄、LPARAM、WPARAM.但是这三个在实际持有指针时会改变宽度.

Some things are not immediately obviously pointers, e.g. HANDLE, LPARAM, WPARAM. But these three change width as they actually hold pointers.

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