从Foo **到void **的转换无效 - 为什么隐式类型转换允许void *，而不是void **？ [英] Invalid conversion from Foo** to void** - why is implicit type conversion allowed to void* but not to void**?

问题描述

  struct Foo {}; 
 ... 
 void * p =（Foo *）0; // OK 
 void ** pp =（Foo **）0; //无效的转换
  

据我所知，指向任何非指针类型的指针可以是隐式的在C ++中转换为 void * 。为什么同样不允许为 void ** ？

指定类型的指针类型

指针可以隐式转换为 void * ，因为 void * 是通用指针。然而， void ** 不是指针的通用指针。

C FAQ 4.9 解释了为什么在C中没有指向指针类型的通用指针，我认为它也适用于C ++。 p>

struct Foo {};
...
void * p = (Foo*)0; // OK
void ** pp = (Foo**)0; // Invalid conversion

As far as I recall, a pointer to any non-pointer type can be implicitly cast to void* in C++. Why then is the same not allowed for casting a ponter to pointer type to void**?

解决方案

A pointer can be implicitly cast to void * because void * is the generic pointer. However, void ** isn't the generic pointer to pointer.

C FAQ 4.9 explains why there is no generic pointer to pointer type in C, I think it applies to C++ as well.

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