CODEWITHSUNDEEP
c
Akshat kant
Akshat kant

Reputation: 133

Why does my program exit after taking an input?

Consider:

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    int n;
    char name[100];
    int number;
    printf("Enter the value of n\n");
    scanf("%d",&n);
    printf("Enter %d values\n", n);
    for(int i=0; i<n; i++)
    {
        scanf("%[^\n]s", &name);
    }
}

Whenever I am entering the value of n, it just prints (Enter n values) and exits the program. The for loop never runs. It ran successfully for the first time, but after that it just exits the program.

There were some answers that said it will not print anything. I don’t want it to print just to take input n times. It is not doing that.

My aim is to take n as input and then take strings of names (like harry, robin, etc.) n number of times as input.

Upvotes: 1

Views: 1304

Answers (2)

Генс
Генс

Reputation: 93

Your code is a little incomplete. And there are a few errors here: scanf ("%[^\n]s", &name)

Do this and everything will be fine:

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main(void)
{
    int n;
    char name[100];
    int number;
    printf("Enter the value of n\n");
    scanf(" %d", &n);
    printf("Enter %d values\n", n);
    for(int i=0; i<n; i++)
    {
        scanf(" %99[^\n]", name);
        printf("%s\n", name);
    }
    return 0;
}

Upvotes: 2

Jabberwocky
Jabberwocky

Reputation: 50778

scanf is particularly unsuited for user input.

You probably want this:

int main() {
  int n;
  char name[100];
  int number;
  printf("Enter the value of n\n");
  scanf("%d", &n);

  printf("Enter %d values\n", n);
  for (int i = 0; i < n; i++)
  {
                                 // the space at the beginning of "%[^\n]"
                                 // gets rid of the \n which stays in the input buffer
    scanf(" %[^\n]", name);      // also there sis no 's' at the end of the "%[^\n]" specifier 
    printf("name = %s\n", name); // for testing purposes
  }
}

But this doesn't actually make much sense because the program is asking for n names, but at each run of the for loop the previous name will be overwritten with the new name.

Also be aware that scanf("%[^\n]", name); is problematic because if the user types more than 99 characters you'll get a buffer overflow.

Upvotes: 1

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