Void * a = & a 如何合法？

考虑下面的 C + + 代码:

void* a = &a;

为什么编译器不抱怨使用未声明的标识符？

还有，编译器认为变量 a是什么？它是指向 void 对象的指针还是指向 void*指针的指针？

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It is equivalent to

void* a;
a = &a;

Therefore, a has been declared. So a gets the address of a written in a. So it is a pointer to a void pointer. (You did not define any objects yet.)

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最佳答案

The scope of declaration of variables in C++ can be pretty surprising:

void* a =               &a;
^~~~~~~~~~~~~~~~~
a declared as `void*` from here on

Therefore, &a is void** but since any pointer type is implicitly convertible to void*...

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In void* a, a is declared as a pointer not to a void type but to "any" type (special case). An address (position in memory) is assigned to a, as to any other variable being declared, of course.

After that, expression &a is evaluated to initialize the variable (also a, but this is not relevant) just declared. The type of &a is "pointer to pointer to any type", which is a special case of "pointer to any type", fully compatible with the type of a. Ergo, no compiler message.

Corollary: do not use void* if you want strong type checking. Anything can be converted to it. Just the opposite in the reverse direction, except for void* itself (it would be an unnecessary exception that a type was incompatible with itself).

Also, AFAIR this really comes from C.


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