Const 正确性 - Const 指针作为函数参数
考虑像 The给定值 at (address) source is const 这样的函数
char* strcpy (char* destination, const char* source);
，因为该函数的作者想要表明 source 的值不会被 strcpy 更改。指针本身不会被 strcpy 改变。为什么不写
char* strcpy (char* destination, const char* const source);
提前非常感谢。
consider a function like
char* strcpy (char* destination, const char* source);
The given value at (address) source is const because the author of the function wants to show that the value of source will not be changed by strcpy. The pointer itself is not changed by strcpy to. Why not to write
char* strcpy (char* destination, const char* const source);
Many thanks in advance.
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指针本身是按值传递的，所以没有意义。
The pointer itself is passed by value, so there's no point.
也可以这样写，但无论如何都不会影响调用者。
在第二种情况下，原型表明指针本身不应该被修改，但是调用者的指针无论如何都不能被修改，因为它在调用函数时被复制（按值传递）。
用 const 标记按值传递的变量仅对函数的实现者有用，作为明确其意图的一种方式。
It could as well be written like that, but it wouldn't affect the caller in any case.
In the second case the prototype says that the pointer itself should not be modified, but the caller's pointer cannot be modified anyway, because it is copied (passed by value) when calling the function.
Marking variables passed by value with
constis useful only to the implementer of the function as a way to make his intents clear.如果您尝试实现这两个功能，您会发现它们实际上是相同的。此时的 const 修饰符仅对函数体有意义，因为参数无论如何都是按值传递的。
You will see if you try to implement both functions, that they are actually the same. A
constmodifier at that point is only meaningful to the body of the function, since the argument is passed by value anyway.