用C兼容的指针类型 [英] incompatible pointer type in C

问题描述

所以我想传递一个类型双* 来接受无效** 作为一个功能的参数。这是我收到的警告。

so I'm trying to pass a type double * to a function that accepts void ** as one of the parameters. This is the warning that I am getting.

incompatible pointer type passing 'double **' to parameter of type 'void **'

下面是我的code片段。

Here is a snippet of my code.

int main( void )
{
    //  Local Declaration
    double *target;

   //   Statement
   success = dequeue(queueIn, &target);
}

下面是函数的原型声明。

Here's the prototype declaration of the function.

int    dequeue     ( QUEUE *queue, void **dataOutPtr );

我想，如果我通过目标作为一个二级指针，它会的工作，但我想我错了。可有人请向我解释怎么来的我得到这个警告？

I thought that if I passed target as a two level pointer that it would work, but I guess I'm wrong. Can someone please explain to me how come i'm getting this warning?

推荐答案

即使所有其他指针类型可以转换为从无效* 不会丢失信息，在同样不无效** 和其他指针到指针类型的真实;如果取消引用无效** 指针，它需要在一个真正的无效* 对象 ^1是指向。

Even though all other pointer types can be converted to and from void * without loss of information, the same is not true of void ** and other pointer-to-pointer types; if you dereference a void ** pointer, it needs to be pointing at a genuine void * object¹.

在这种情况下，presuming的出列（）通过存储它通过提供的指针，被正式纠正你需要返回一个指针值这样做：

In this case, presuming that dequeue() is returning a single pointer value by storing it through the provided pointer, to be formally correct you would need to do:

int main( void )
{
    void *p;
    double *target;

    success = dequeue(queueIn, &p);
    target = p;

当你写的这个样子，从无效* 来转换双* 是明确的，这使得编译器做任何魔法必要（尽管在绝大多数常见的情况，有没有在所有魔法）。

When you write it like this, the conversion from void * to double * is explicit, which allows the compiler to do any magic that's necessary (even though in the overwhelmingly common case, there's no magic at all).

---

^{1。 ......或的char * ，无符号字符* 或符号字符* 对象，因为有那些特殊的规则。}

---

^{1. ...or a char *, unsigned char * or signed char * object, because there's a special rule for those.}

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