问题描述
因此,您可以执行以下操作:
So you can do this:
void foo(const int * const pIntArray, const unsigned int size);
其中说来的指针是只读的,指向的整数是只读的
Which says that the pointer coming is read-only and the integer's it is pointing to are read-only.
您可以在函数内部访问此代码,如下所示:
You can access this inside the function like so:
blah = pIntArray[0]
您还可以执行以下声明:
You can also do the following declaration:
void foo(const int intArray[], const unsigned int size);
这几乎是相同的,但是您可以这样做:
It is pretty much the same but you could do this:
intArray = &intArray[1];
我可以这样写吗:
void foo(const int const intArray[], const unsigned int size);
对吗?
推荐答案
否,您的最后一个变体不正确。您正在尝试通过以下新语法在C99中实现
No, your last variant is not correct. What you are trying to do is achieved in C99 by the following new syntax
void foo(const int intArray[const], const unsigned int size);
等效于
void foo(const int *const intArray, const unsigned int size);
[const]
语法特定于C99。在C89 / 90中无效。
That [const]
syntax is specific to C99. It is not valid in C89/90.
请记住,有些人认为函数参数的顶级cv限定词无用,因为它们限定了复制实际参数。我认为它们根本没有用,但就我个人而言,在现实生活中我不会遇到太多使用它们的原因。
Keep in mind that some people consider top-level cv-qualifiers on function parameters "useless", since they qualify a copy of the actual argument. I don't consider them useless at all, but personally I don't encounter too many reasons to use them in real life.
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