挥发的char *＆QUOT;类型＆QUOT的说法;与类型和QUOT的参数不符;为const char *＆QUOT; [英] Argument of type "volatile char *" is incompatible with parameter of type "const char *"

问题描述

我有一个函数，其原型如下：

I have a function whose prototype is as follows:

void foo(const char * data);

在别处在我的code，我已经声明为一个全局变量如下

Elsewhere in my code, I have a global variable declared as follows

volatile char var[100];

每当我试着这样做：

Whenever I try to do this:

foo(var);

编译器抛出了以下错误消息：

The compiler throws up the following error message:

键入挥发性的char *的参数是与类型为const char *的参数不符

为什么会这样呢？据我了解，在我的函数变量是不允许改变指针或其内容。据我所知，因为我的全局变量是挥发性的，它可能在任何时候改变，但看到因为它是完全合法的具有挥发性常量变量，我不明白为什么我收到此编译器错误。

Why is that the case? As I understand it, the variable in my function is not allowed to change the pointer or its contents. I understand that because my global variable is volatile, it could potentially change at any time, but seeing as it is perfectly legal to have a volatile const variable, I don't see why I am getting this compiler error.

感谢

- 阿姆鲁

推荐答案

这是因为隐含的类型转换的添加的预选赛指针类型的目标，而不是将其删除。所以，如果你想你的函数能够接受挥发性和/或常量合格的指针，你必须声明它既：

It's because implicit conversions can add qualifiers to the target of pointer types, but not remove them. So if you want your function to be able to accept volatile and/or const qualified pointers, you must declare it with both:

void foo(const volatile char * data);

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