将char指针传递给函数错误

Question

Hello I'm getting an error when i try to pass a char through to a function. here is my code.

variable

char *temp;

Prototype

int checkIfUniqueCourseNo(char,int);

Call

checkIfUniqueCourseNo(temp,k);

and my error

warning: improper pointer/integer combination: arg #1

Im new to C so go easy on me :)

Answer 1

Your function accepts a char ; you are trying to pass in a char* .

To fix this you need to dereference your pointer to obtain the character that it points to, so that your function receives the type of argument it expects:

checkIfUniqueCourseNo(*temp,k);

Answer 2

If the function excepts char , you should dereference the pointer:

checkIfUniqueCourseNo(*temp,k);
//                    ^ pass the char addressed by temp

Answer 3

Your variable is a char* (char pointer), but the function takes a char (not a pointer).

If you want to pass the contents of temp to the function, use checkIfUniqueCourseNo(*temp, k) . If you really do want to pass the pointer itself, declare the function as

int checkIfUniqueCourseNo(char*,int);