警告：格式“%s”需要类型“char *”，但参数 2 的类型为“char (*)”

Question

I am trying to run a simple C program but I am getting this error:

warning: format '%s' expects type 'char *', but argument 2 has type 'char (*)[20]'

Running Mac OSX Mountain Lion, compiling in terminal using gcc 4.2.1

#include <stdio.h>

int main() {
    char me[20];

    printf("What is your name?");
    scanf("%s", &me);
    printf("Darn glad to meet you, %s!\n", me);

    return (0);
}

Answer 1

scanf("%s",&me);

should be

scanf("%s",me);

Explaination:

"%s" means that scanf is expecting a pointer to the first element of a char array. me is an object array and could evaluated as pointer. So that's why you can use me directly without adding & . Adding & to me will be evaluated to 'char (*)[20]' and your scanf is waiting char *

Code critic:

Using "%s" could cause a buffer overflow if the user input string with length > 20. So change it to "%19s" :

scanf("%19s",me);

Answer 2

Except when it is the operand of the sizeof , _Alignof , or unary & operators, or is a string literal being used to initialize an array in a declaration, an expression of type "N-element array of T " will be converted ("decay") to an expression of type "pointer to T ", and it will evaluate to the address of the first element in the array.

The array me is declared as a 20-element array of char ; normally, when the expression me appears in your code, it will be treated as an expression of type "pointer to char ". If you had written

scanf("%s", me);

then you wouldn't have gotten the error; the expression me would have been converted to an expression of the correct type.

By using the & operator, however, you've bypassed that rule; instead of a pointer to char , you're passing a pointer to an array of char ( char (*)[20] ), which is not what scanf expects for the %s conversion specifier, hence the diagnostic.

Answer 3

Another way you could fix this issue is by doing this:

scanf("%s",&me[0]);

You actually have to give the array's starting point (which in most languages is 0).