我如何申报使用malloc是在C挥发创建数组++ [英] How do I declare an array created using malloc to be volatile in c++

问题描述

我presume了下面就给我10挥发性整数

 挥发性INT富[10];
 

不过，我不认为下面会做同样的事情。

 挥发为int * foo的;
富=的malloc（sizeof的（INT）* 10）;
 

请纠正我，如果我错了这件事，我怎么能有使用malloc项目的挥发性数组。

感谢。

解决方案

  INT挥发性* foo的;
 

从右读向左foo是一个指向挥发性INT

所以无论你诠释美孚通过访问，整型会挥发。

P.S。

 为int * foo的波动; //foo是挥发性指向int
 

==

 挥发为int * foo的; // foo是一个指向一个int，挥发性
 

含义foo是不稳定。第二种情况是真的只是一个普通从右到左规则剩。
在使用习惯要吸取的教训是获得

 字符常量* foo的;
 

而不是更常见的

 为const char * foo的;
 

如果你想喜欢更复杂的东西指针函数返回指向int的指针，以任何意义。

P.S，这是一个biggy（和主要的原因我添加一个答案）：

我注意到，你列入多线程作为标记。你是否意识到，挥发性做的好一点的/没有关于多线程？

I presume that the following will give me 10 volatile ints

volatile int foo[10];

However, I don't think the following will do the same thing.

volatile int* foo;
foo = malloc(sizeof(int)*10);

Please correct me if I am wrong about this and how I can have a volatile array of items using malloc.

Thanks.

解决方案

int volatile * foo;

read from right to left "foo is a pointer to a volatile int"

so whatever int you access through foo, the int will be volatile.

P.S.

int * volatile foo; // "foo is a volatile pointer to an int"

==

volatile int * foo; // foo is a pointer to an int, volatile

Meaning foo is volatile. The second case is really just a leftover of the general right-to-left rule. The lesson to be learned is get in the habit of using

char const * foo;

instead of the more common

const char * foo;

If you want more complicated things like "pointer to function returning pointer to int" to make any sense.

P.S., and this is a biggy (and the main reason I'm adding an answer):

I note that you included "multithreading" as a tag. Do you realize that volatile does little/nothing of good with respect to multithreading?

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