第章22:指向指针的指针:

关于指针和内存分配的指针的一个侧面:尽管由malloc返回的void *类型是通用指针"，适合于分配给任何类型的指针或从任何类型的指针进行分配，假设类型 void **不是通用指针" .

One side point about pointers to pointers and memory allocation: although the void * type, as returned by malloc, is a "generic pointer," suitable for assigning to or from pointers of any type, the hypothetical type void ** is not a "generic pointer to pointer".

因此，只有void *是通用指针. void **不是通用指针.因此，传递给您的函数的参数必须为void **类型.

So, only void * is a generic pointer. void ** is not generic pointer. Therefore the argument passed to your function must be of type void **.

另请参见 C-FAQ :

C中没有通用的指针对指针类型. void *仅作为通用指针，因为当其他指针类型分配给void *或从void *分配时，会自动应用转换(如果需要).如果试图间接指向void *以外的指针类型的void **值，则无法执行这些转换.当您使用void **指针值时(例如，当您使用*运算符访问void **所指向的void *值时)，编译器无法知道 value曾经从其他某种指针类型转换而来.它必须假设它不过是void *；它只不过是void *.它无法执行任何隐式转换.

There is no generic pointer-to-pointer type in C. void * acts as a generic pointer only because conversions (if necessary) are applied automatically when other pointer types are assigned to and from void *'s; these conversions cannot be performed if an attempt is made to indirect upon a void ** value which points at a pointer type other than void *. When you make use of a void ** pointer value (for instance, when you use the * operator to access the void * value to which the void ** points), the compiler has no way of knowing whether that void * value was once converted from some other pointer type. It must assume that it is nothing more than a void *; it cannot perform any implicit conversions.

换句话说，您使用的任何void **值必须是某个地方实际void *值的地址；像(void **)&dp这样的强制类型转换虽然可能会关闭编译器，但它们是不可移植的(甚至可能无法执行您想要的操作；另请参阅问题13.9).如果void **指向的指针不是void *，并且其大小或表示形式与void *不同，则编译器将无法正确访问它.

In other words, any void ** value you play with must be the address of an actual void * value somewhere; casts like (void **)&dp, though they may shut the compiler up, are nonportable (and may not even do what you want; see also question 13.9). If the pointer that the void ** points to is not a void *, and if it has a different size or representation than a void *, then the compiler isn't going to be able to access it correctly.