How to store string value in specific index of array using while loop

Question

I wanted to store string values at a particular index of the character array using a while loop.

Condition for while loop termination: 'q' should be pressed to stop taking input

My Code so far

  char s[100],in;
  int count = 0;
  printf("Enter individual names: \n");
  scanf("%c",&in);
   
  while (in != 'q')
  {
    s[count++] = in;
    scanf("%c", &in);
  }
  printf("%d", count);
  printf("%s" , s);

Input:

    sam 

    tam 

    q

Output:

    9����

I don't understand how can I store strings at an individual index of array and why is count giving me wrong value when it should have been 2.

Is there any other method to store string using while loop?

Kyle Burns · Accepted Answer · 2021-02-01 18:56:16Z

1

The problem is you're scanning a single character instead of a string. tam and sam are 3 characters, not one. You would need to modify your code to something like this which reads the input string into an input buffer and then copies it to the s buffer you have.

Edit: Sorry I misunderstood your question. This should do what you're wanting. Let me know if you have any questions about it.

#include <stdio.h> // scanf, printf
#include <stdlib.h> // malloc, free
#include <string.h> // strlen, strcpy

#define MAX_NUM_INPUT_STRINGS 20

int main(int argc, char **argv) { // Look no further than our friend argv!
    char* s[MAX_NUM_INPUT_STRINGS], in[100]; // change in to buffer here, change s to char pointer array
    size_t count = 0, len;
    printf("Enter individual names: \n");
    do {
        scanf("%s",in);
        size_t len = strlen(in);
        if (in[0] == 'q' && len == 1) {
            break;
        }
        // allocate memory for string
        s[count] = malloc(len + 1); // length of string plus 1 for null terminating char
        s[count][len] = '\0'; // Must add null terminator to string.
        strcpy(s[count++], in); // copy input string to c string array
    } while (count < MAX_NUM_INPUT_STRINGS); // allows user to enter single char other than 'q'
    printf("Count: %lu\n", count);
    for (size_t i = 0; i < count; i++) {
        printf("%s\n", s[i]);
    }

    // free allocated memory
    for (size_t i = 0; i < count; i++) {
        free(s[i]);
    }
    return 1;
}

edited Feb 1, 2021 at 18:56

answered Jan 27, 2021 at 18:21

Kyle Burns

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2 Comments

Satyam thassu Over a year ago

Thank you so so much for the answer! This was exactly what i was looking for . And would you be kind enough to please answer why you have used malloc[len + 1] Why was 1 added? Code works fine without it too. Thank you again.

Kyle Burns Over a year ago

@Satyamthassu Because c-strings are supposed to be terminated by a zero character so functions know when the string ends. The strlen function returns the length not including that last zero. For example, if you had a string "bob", the length is 3. The compiler automatically inserts the '\0' for you. But when you're reading in strings using scanf, they won't have the '\0' and you have to add it yourself. I actually forgot to do that in the code. I just added it.

vmp · Accepted Answer · 2021-01-31 17:31:45Z

1

C-strings need a '\0' at the end and you not adding one to s.

printf("%d", count);
s[count] = '\0'; // ADD THIS LINE
printf("%s" , s);

But you could alternatively read as follows:

char s[100];
scanf("%99[^q]" s); // Will read up to 99 chars that are not a 'q'
printf("%s\n", s);
int count = strlen(s);

answered Jan 31, 2021 at 17:31

vmp

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1 Comment

Алексей Неудачин Over a year ago

Will read up to 99 chars until q isn't met

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