The following code is producing a warning:
const char * mystr = "\r\nHello";
void send_str(char * str);
void main(void){
send_str(mystr);
}
void send_str(char * str){
// send it
}
The error is:
Warning [359] C:\main.c; 5.15 illegal conversion between pointer types
pointer to const unsigned char -> pointer to unsigned char
How can I change the code to compile without warnings? The send_str()
function also needs to be able to accept non-const strings.
(I am compiling for the PIC16F77 with the Hi-Tech-C compiler)
Thanks
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 */
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.
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);
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 */
change the following lines
void send_str(char * str){
// send it
}
TO
void send_str(const char * str){
// send it
}
your compiler is saying that the const char pointer your sending is being converted to char pointer. changing its value in the function send_str
may lead to undefined behaviour.(Most of the cases calling and called function wont be written by the same person , someone else may use your code and call it looking at the prototype which is not right.)
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