为什么我可以投int和BOOL为void *，而不是浮动？ [英] Why can I cast int and BOOL to void*, but not float?

问题描述

无效* 是C和衍生语言非常有用的特性。例如，它可以使用无效* 来的Objective-C的对象指针存储在C ++类。

void* is a useful feature of C and derivative languages. For example, it's possible to use void* to store objective-C object pointers in a C++ class.

我工作的一个类型转换框架和近期由于时间的限制有点懒 - 所以我用无效* ...这就是这个问题怎么来了：

I was working on a type conversion framework recently and due to time constraints was a little lazy - so I used void*... That's how this question came up:

我为什么INT强制转换为void *，但不能上浮到void *？

Why can I typecast int to void*, but not float to void* ?

推荐答案

BOOL不是C ++类型。它可能的typedef或某处定义的，并且在这些情况下，这将是一样中间体例如Windows，有这个在WINDEF.H：

BOOL is not a C++ type. It's probably typedef or defined somewhere, and in these cases, it would be the same as int. Windows, for example, has this in Windef.h:

    typedef int                 BOOL;

所以你的问题减少了，为什么你可以强制转换INT为void *，但不能上浮到void *？

so your question reduces to, why can you typecast int to void*, but not float to void*?

INT为void *是确定的，但一般不建议（和一些编译器会警告一下吧），因为它们本质上是在重新presentation相同。指针基本上是指向内存地址的整数。

int to void* is ok but generally not recommended (and some compilers will warn about it) because they are inherently the same in representation. A pointer is basically an integer that points to an address in memory.

浮动到void *也不行，因为浮点值的相互pretation和实际位重新presenting它是不同的。例如，如果你这样做：

float to void* is not ok because the interpretation of the float value and the actual bits representing it are different. For example, if you do:

   float x = 1.0;

它的作用是它集32位内存为00 00 80 3F（浮点值1.0 IEEE单precision实际重新presentation）。当你施放一个浮点数为void *，国米pretation是模糊的。你的意思是指向位置1在内存中的指针？或者你的意思是指向位置3f800000在内存中（假设小端）？指针

what it does is it sets the 32 bit memory to 00 00 80 3f (the actual representation of the float value 1.0 in IEEE single precision). When you cast a float to a void*, the interpretation is ambiguous. Do you mean the pointer that points to location 1 in memory? or do you mean the pointer that points to location 3f800000 (assuming little endian) in memory?

当然，如果你确信这你想要的两种情况下，总有一种方式来获得解决此问题。例如：

Of course, if you are sure which of the two cases you want, there is always a way to get around the problem. For example:

  void* u = (void*)((int)x);        // first case
  void* u = (void*)(((unsigned short*)(&x))[0] | (((unsigned int)((unsigned short*)(&x))[1]) << 16)); // second case

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