C:指针类型之间的非法转换:指向const unsigned char的指针->指向无符号字符的指针 [英] C: Illegal conversion between pointer types: pointer to const unsigned char -> pointer to unsigned char

问题描述

以下代码产生警告:

const char * mystr = "\r\nHello";
void send_str(char * str);

void main(void){
    send_str(mystr);
}
void send_str(char * str){
    // send it
}

错误是:

Warning [359] C:\main.c; 5.15 illegal conversion between pointer types
pointer to const unsigned char -> pointer to unsigned char

如何更改代码以在没有警告的情况下进行编译? send_str()函数还需要能够接受非常量字符串.

How can I change the code to compile without warnings? The send_str() function also needs to be able to accept non-const strings.

(我正在使用Hi-Tech-C编译器为PIC16F77进行编译)

(I am compiling for the PIC16F77 with the Hi-Tech-C compiler)

谢谢

推荐答案

您需要添加强制类型转换，因为您要将常量数据传递给表示我可能会更改"的函数:

You need to add a cast, since you're passing constant data to a function that says "I might change this":

send_str((char *) mystr);  /* cast away the const */

当然，如果函数确实决定更改实际上应该是常量的数据(例如字符串文字)，则会出现未定义的行为.

Of course, if the function does decide to change the data that is in reality supposed to be constant (such as a string literal), you will get undefined behavior.

不过，也许我误会了你.如果send_str()永远不需要更改其输入，但是可能会在调用者的上下文中使用非恒定数据来调用，那么您应该仅使用参数const，因为这只是说不会改变这一点":

Perhaps I mis-understood you, though. If send_str() never needs to change its input, but might get called with data that is non-constant in the caller's context, then you should just make the argument const since that just say "I won't change this":

void send_str(const char *str);

可以安全地使用常量和非常量数据来调用它:

This can safely be called with both constant and non-constant data:

char modifiable[32] = "hello";
const char *constant = "world";

send_str(modifiable);  /* no warning */
send_str(constant);    /* no warning */

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