了解指针引用的工作原理

发布于 2024-08-10 03:27:33 字数 654 浏览 2 评论 0原文

我目前正在阅读“COM 和 ATL 3.0 的开发人员研讨会”。第 3 章介绍 GUID、引用和比较。指针很痛苦。我可以使用一些帮助来破译 REFGUID #define （见下文）以及 IsEqualGUID 中的 memcmp 如何针对指针工作。

给定：

  typedef struct_GUID{ unsigned long Data1;  
    unsigned short Data2;  
    unsigned short Data3;  
    unsigned char Data4[8]; } GUID;

我如何解释这个#define？：

 #define REFGUID const GUID * const

&rguid1如何寻址传入变量？

   BOOL IsEqualGUID(REFGUID rguid1, REFGUID rguid2)  
    {  
      return !memcmp(&rguid1, &rguid2, sizeof(GUID));  
    }

原文

I am currently reading "Developer's Workshop to COM and ATL 3.0". Chapter 3 introduces GUIDs, referencing and comparisons. Pointers are painful. I could use some help in deciphering the REFGUID #define (see below) and how memcmp in IsEqualGUID works against the pointers.

Given:

  typedef struct_GUID{ unsigned long Data1;  
    unsigned short Data2;  
    unsigned short Data3;  
    unsigned char Data4[8]; } GUID;

How do I interpret this #define?:

 #define REFGUID const GUID * const

How is the &rguid1 addressing the incoming variable?

   BOOL IsEqualGUID(REFGUID rguid1, REFGUID rguid2)  
    {  
      return !memcmp(&rguid1, &rguid2, sizeof(GUID));  
    }

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别再吹冷风 2024-08-17 03:27:33

REFGUID 是常量 guid 的常量 ptr（即两者都不能改变）。

代码不应该是吗？

 BOOL IsEqualGUID(REFGUID rguid1, REFGUID rguid2)  
 {        
   return !memcmp(rguid1, rguid2, sizeof(GUID));      
 }

正如 memcmp 所采取的那样：

int memcmp(const void *s1, const void *s2, size_t n);

memcmp 应该传递指针（rguidx）而不是指针的地址。

如果看起来代码最初是用 REGUID 定义为 const GUID 或 const GUID 引用（C++）编写的

The REFGUID is constant ptr to a constant guid (ie neither can change).

Shoud the code not be?

 BOOL IsEqualGUID(REFGUID rguid1, REFGUID rguid2)  
 {        
   return !memcmp(rguid1, rguid2, sizeof(GUID));      
 }

as memcmp takes:

int memcmp(const void *s1, const void *s2, size_t n);

The memcmp should be passed the pointers (rguidx) not the address of the pointer.

if looks like the code was originally written with REGUID defined as a const GUID or const GUID reference (C++) perhaps

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静谧幽蓝 2024-08-17 03:27:33

REFGUID 在 C++ 和 C 上下文中的定义不同。如果你看一下它的定义，那就是：

#ifdef __cplusplus
#define REFGUID const GUID &
#else
#define REFGUID const GUID * 
#endif

IsEqualGUID()函数也有不同的实现。

我不喜欢这个主意。我猜这个人发明这个只是为了让它“C++正确”，因为C++发明者相信引用比指针更好。

REFGUID is defined differently in C++ and C context. If you look at its definition, it is:

#ifdef __cplusplus
#define REFGUID const GUID &
#else
#define REFGUID const GUID * 
#endif

IsEqualGUID() function also has different implementations.

I do not like this idea. I guess that the person invented this just to make it "C++ right" because the C++ inventor believes that reference is better than pointer.

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