这有什么错铸像（无效**）及device_array？ [英] What's wrong with casting like (void**)&device_array?
问题描述
有 cudaMalloc（（无效的另外一个问题这个答案 ** ）及device_array，NUM_BYTES），它使用无效** 作为输出参数，而不是通过一个无效* 像标准的返回值的malloc 。
There is this answer on another question about the use of cudaMalloc((void**)&device_array, num_bytes), which uses void** as output argument instead of passing a void* as return value like the standard malloc.
它批评NVIDIA的API，并指出：
It criticizes NVIDIA's API and states :
铸造，如（无效**）及device_array，是无效的C和结果
未定义的行为。
Casting, as in (void**)&device_array, is invalid C and results in undefined behavior.
，并已upvoted几次（8截至目前），所以我认为有一定的道理在里面。
and has been upvoted several times (8 as of now), so I assume there is some truth in it.
我不明白这有什么错铸那里。
I don't understand what's wrong with casting there.
- 什么是无效这儿用？
- 在什么情况下会这导致不确定的行为？
我所知道的是，它编译没有警告，与我预期的行为运行。但我不掌握相关C到标准规范的水平。
All I know is that it compiles without warning and runs with the intended behavior for me. But I am not knowledgeable with C up to standard specification level.
推荐答案
的问题是，无效* 有一个在C特殊的含义，具有特殊的规则（1）。它是唯一指针类型/从中你可以安全地将任何其他指针类型。然而，这些特殊规则不递归适用无效** 。
The problem is that void* has a special meaning in C, with special rules (1). It is the only pointer type to/from which you can safely convert any other pointer type. However, these special rules do not apply recursively to void**.
意思就是说code像为int * PTR =的malloc（X）; 是完全没有问题的，但
Meaning that code like int* ptr = malloc(x); is perfectly fine, but
int* ptr;
cudaMalloc(&ptr, x); // bad
不精！从 INT ** A指针转换为无效** 是不明确。理论上，这可能会导致不确定的行为和不对（2）。
is not fine! A pointer conversion from int** to void** is not well-defined. In theory this could cause undefined behavior and misalignment (2).
在另外，也可能是与指针锯齿问题。编译器是免费的假设的无效** 从未通过 INT访问** 和内容可以因此，优化以意想不到的方式在code，导致对违反严格别名规则（6.5）的未定义的行为。
In addition, there might also be problems with pointer aliasing. The compiler is free to assume that the contents of a void** is never accessed through a int** and could therefore optimize the code in unexpected ways, leading to undefined behavior for violation of the strict aliasing rule (6.5).
这意味着你将不得不写code这样才能安全地使用功能：
Which means you will have to write code like this in order to safely use the function:
void* vptr;
int* iptr;
cudaMalloc(&vptr, x);
iptr = vptr;
（1）C11 6.3.2.3/1：
(1) C11 6.3.2.3/1:
要无效的指针可被转换成或从一个指针到任何物体
类型。的指针的任何对象类型可被转换成一个指针
作废，然后再返回;结果应比较等于原
指针。
A pointer to void may be converted to or from a pointer to any object type. A pointer to any object type may be converted to a pointer to void and back again; the result shall compare equal to the original pointer.
（2）C11 6.3.2.3/7：
(2) C11 6.3.2.3/7:
一个指针对象类型可被转换成一个指针
不同的对象类型。如果所得到的指针不正确
为引用类型一致，行为是不确定的。
A pointer to an object type may be converted to a pointer to a different object type. If the resulting pointer is not correctly aligned for the referenced type, the behavior is undefined.
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