投给指针-为什么要先投?(如p = (void*) 42;)

In the GLib documentation, there is a chapter on type conversion macros. In the discussion on converting an int to a *void pointer it says (emphasis mine):

在GLib文档中，有一个关于类型转换宏的章节。在讨论将int转换为空指针时，它说(强调我的):

Naively, you might try this, but it's incorrect:

很天真地，你可以尝试一下，但这是错误的:
gpointer p;
int i;
p = (void*) 42;
i = (int) p;
Again, that example was not correct, don't copy it. The problem is that on some systems you need to do this:

同样，这个例子不正确，不要复制它。问题是在某些系统上你需要这样做:
gpointer p;
int i;
p = (void*) (long) 42;
i = (int) (long) p;

(source: GLib Reference Manual for GLib 2.39.92, chapter Type Conversion Macros ).

(来源:GLib 2.39.92的GLib参考手册，章节类型转换宏)。

Why is that cast to long necessary?

为什么要把它扔到很长时间?

Should any required widening of the int not happen automatically as part of the cast to a pointer?

是否需要将int类型的扩展作为转换的一部分而自动转换为指针?

2 个解决方案

#1

As according to the C99: 6.3.2.3 quote:

根据C99: 6.3.2.3报价:

5 An integer may be converted to any pointer type. Except as previously specified, the result is implementation-defined, might not be correctly aligned, might not point to an entity of the referenced type, and might be a trap representation.56)

一个整数可以转换为任何指针类型。除了前面指定的，结果是实现定义的，可能不正确地对齐，可能不会指向引用类型的实体，也可能是一个陷阱代表。

6 Any pointer type may be converted to an integer type. Except as previously specified, the result is implementation-defined. If the result cannot be represented in the integer type, the behavior is undefined. The result need not be in the range of values of any integer type.

任何指针类型都可以转换为整数类型。除了前面指定的，结果是实现定义的。如果结果不能在整数类型中表示，则行为是未定义的。结果不需要在任何整数类型的值范围内。

According to the documentation at the link you mentioned:

根据你提到的链接文件:

Pointers are always at least 32 bits in size (on all platforms GLib intends to support). Thus you can store at least 32-bit integer values in a pointer value.

指针总是至少32位(在所有平台上，GLib都打算支持)。因此，您可以在一个指针值中存储至少32位整数值。

And further more long is guaranteed to be atleast 32-bits.

更长的时间保证至少32位。

So,the code

所以,代码

gpointer p;
int i;
p = (void*) (long) 42;
i = (int) (long) p;

is safer,more portable and well defined for upto 32-bit integers only, as advertised by GLib.

更安全，更便携，并且仅为32位整数定义，正如GLib所宣传的那样。

#2

I think it is because this conversion is implementation dependendant. It is better to use uintptr_t for this purposes, because it is exactly pointer type in particular implementation.

我认为这是因为这种转换是依赖于实现的。出于这个目的，最好使用uintptr_t，因为它是特定实现中的指针类型。

#1