Why isn't this simple code waiting for student name (entered by user)?

Question

This C code is not waiting for student name. It directly prints total student number. But when I comment out the first printf-scanf statement(or enter number of students:), then code is waiting for the user to enter student name.

#include <stdio.h>

int main()
{

 char name[10];
 int count;

 printf("ENTER NUMBER OF STUDENTS:\n");
 scanf("%d", &count);

 printf("ENTER STUDENT NAME:\n");
 scanf("%[^\n]%*c", &name);

 printf("Total_Students: %d\n", count);
 printf("NAME: %s\n", name);
 return (0);
 }

I think your second scanf is reading newline from first scanf — Rohan Kumar, Mar 03 '21 at 18:15
Try changing the format in the first `scanf` call to "%d\n" and see what happens. — Bob Jarvis - Слава Україні, Mar 03 '21 at 18:16
https://stackoverflow.com/questions/5240789/scanf-leaves-the-new-line-char-in-the-buffer — ilkkachu, Mar 03 '21 at 18:17
What are you expecting `scanf()` to do with `%*c` ? Also, do you really want `&name` (`name` is a pointer by itself) ? — donjuedo, Mar 03 '21 at 18:18
One of the *many* duplicates of this problem [here](https://stackoverflow.com/questions/36281871/scanf-skip-scanning-character). — WhozCraig, Mar 03 '21 at 18:25

score 0 · Accepted Answer · answered Mar 03 '21 at 18:20

The second scanf is skipped because the newline character is being interpreted from the first scanf.

For instance, if you entered 2 for number of student, what is being entered is 2\n. The first scanf reads the number 2 and leaves the \n in the buffer which is being interpreted by the 2nd scanf.

You can simply add a space in the second scanf to get past this issue

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

score -1 · Answer 2 · answered Mar 03 '21 at 18:25

-1

change the second scanf() argument to %s and it directly makes the last entry as NULL .

answered Mar 03 '21 at 18:25

Nitin Shrinivas

1

Why isn't this simple code waiting for student name (entered by user)?

2 Answers2