Problem scanning strings with spaces in C

Question

I have to scan a string with spaces so I can do somethings with it in this case moving the string n chars to the right or to the left. For example:

If I give n = 1, the string

house

becomes:

ehous

But the input can also be a string with spaces like, n being the same:

yellow house

becomes:

eyellow hous

So to scan the string im doing this:

char text[166];
scanf("%d %[^\n]%*c", &n, &text);

And everything works great, but now the I submitted the program I get this error:

src.c: In function 'main':
src.c:30:26: error: format '%[^
 ' expects argument of type 'char *', but argument 3 has type 'char (*)[166]' [-Werror=format=]
 30 |             scanf("%d %[^\n]%*c", &nVezes, &texto);
    |                       ~~~^~                ~~~~~~
    |                          |                 |
    |                          char *            char (*)[166]

What can I solve this? Also I can't use these libraries string.h, strings.h, and stdlib.h.

Every bit of help is appreciated.

Answer 1

scanf() does not truly scan a string but stdin .
Instead OP want to read a line of user input into an int and string .

Save headaches - ditch scanf() with its many weaknesses.

//         v---------------------- missing width limit
//                        v------- do not use here
scanf("%d %[^\n]%*c", &n, &text);
//     ^ ^ ----------------------- possible to read more than 1 line
// return value not checked

---

Use fgets() @David C. Rankin

char text[166];
char line[sizeof text + 20];
if (fgets(line, sizeof line, stdin)) {
  if (sscanf(line, "%d %165[^\n]", &n, text) == 2) {
    Success();  // OP's code here
  }
}

Answer 2

You don't need to pass the char array as a pointer(*). when you pass an array in C as an argument to a function, the name of the array acts as a pointer to the start of the array.

So, you should pass the array as follows:

scanf("%d %[^\n]%*c", &n, text);