C语言 C 中不兼容的指针类型
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incompatible pointer type in C
提问by Nathan
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.
所以我试图将一个类型传递double *给一个接受void **作为参数之一的函数。这是我收到的警告。
incompatible pointer type passing 'double **' to parameter of type 'void **'
Here is a snippet of my code.
这是我的代码片段。
int main( void )
{
// Local Declaration
double *target;
// Statement
success = dequeue(queueIn, &target);
}
Here's the prototype declaration of the function.
这是函数的原型声明。
int dequeue ( QUEUE *queue, void **dataOutPtr );
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?
我认为如果我将目标作为两级指针传递，它会起作用，但我想我错了。有人可以向我解释我怎么会收到这个警告吗？
回答by caf
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.
尽管所有其他指针类型都可以在void *不丢失信息的情况下相互转换，但void **其他指针到指针类型并非如此；如果您取消引用一个void **指针，它需要指向一个真正的void *对象1。
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:
在这种情况下，假设dequeue()通过提供的指针存储它来返回单个指针值，要正式正确，您需要执行以下操作：
int main( void )
{
void *p;
double *target;
success = dequeue(queueIn, &p);
target = p;
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).
当您像这样编写它时，从void *to的转换double *是显式的，这允许编译器执行任何必要的魔术（即使在绝大多数情况下，根本没有魔术）。
1. ...or a
char *, unsigned char *or signed char *object, because there's a special rule for those.1. ......或者char *，unsigned char *或signed char *对象，因为有一个特殊的规则相同。回答by Mani
In your prototype declaration , you said second argument as void**,so you have to type cast double**to void**.
Instead of this line success = dequeue(queueIn, &target);.
在你的原型声明中，你说第二个参数 as void**，所以你必须输入 cast double**to void**。而不是这条线success = dequeue(queueIn, &target);。
Call like this success = dequeue(queueIn,(void**) &target);
像这样打电话 success = dequeue(queueIn,(void**) &target);
回答by anshul garg
int main( void )
{
// Local Declaration
double *target;
// Statement
success = dequeue(queueIn, (void**)&target);
}
Use it like this.
像这样使用它。
