C语言 初始化丢弃来自指针目标类型的限定符
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/2316387/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Initialization discards qualifiers from pointer target type
提问by Crystal
I'm trying to print the list of a singly linked list that I referred to in link text. It works, but I do get the compiler warnings:
我正在尝试打印我在link text 中引用的单链表的列表。它有效，但我确实收到了编译器警告：
Initialization discards qualifiers from pointer target type
Initialization discards qualifiers from pointer target type
(on declaration of start = head) and
（在声明 start = head 时）和
return discards qualifiers from pointer target type
return discards qualifiers from pointer target type
(on return statement) in this code:
（在返回语句中）在此代码中：
/* Prints singly linked list and returns head pointer */
LIST *PrintList(const LIST *head)
{
LIST *start = head;
for (; start != NULL; start = start->next)
printf("%15s %d ea\n", head->str, head->count);
return head;
}
I am using XCode. Any thoughts?
我正在使用 XCode。有什么想法吗？
回答by GManNickG
It's this part:
这是这部分：
LIST *start = head;
The parameter for the function is a pointer to a constant, const LIST *head; this means you cannot change what it is pointing to. However, the pointer above is to non-const; you could dereference it and change it.
该函数的参数是一个指向一个常数，const LIST *head; 这意味着您无法更改它所指向的内容。但是，上面的指针是指向非常量的；你可以取消引用它并改变它。
It needs to be constas well:
它也需要const：
const LIST *start = head;
The same applies to your return type.
这同样适用于您的返回类型。
All the compiler is saying is: "Hey, you said to the caller 'I won't change anything', but you're opening up opportunities for that."
编译器只是说：“嘿，你对调用者说‘我不会改变任何东西’，但你正在为此打开机会。”
回答by Eric Wang
In following function, would get the warning that you encountered with.
在以下函数中，将收到您遇到的警告。
void test(const char *str) {
char *s = str;
}
There are 3 choices:
有3种选择：
Remove the const modifier of param:
void test(char *str) { char *s = str; }Declare the target variable also as const:
void test(const char *str) { const char *s = str; }Use a type convert:
void test(const char *str) { char *s = (char *)str; }
移除 param 的 const 修饰符：
void test(char *str) { char *s = str; }将目标变量也声明为 const：
void test(const char *str) { const char *s = str; }使用类型转换：
void test(const char *str) { char *s = (char *)str; }
