allocating dynamic memory using malloc

Question

Here is the sample of my code. I want to create dynamic character array for storing string.

Here is my code:

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

    int main(void)
    {
      int i , j , n;
      char *ptr;
      printf("enter number of elements \n");
      scanf("%d",&n);
      ptr  = (char *) malloc((n + 1)*sizeof(char));
      for (i = 0;i< n;i++)
      {
        scanf("%c",&ptr[i]);
      }
      for ( i = 0;i <n; i++)
      {
        printf("at %d is %c\n",i,*(ptr + i));
      }
      free(ptr);
    }

But when I try to compile and run this code, no character is assigned to memory which is pointed by pointer p.

Here is the output of my program:

jharvard@appliance (~/c): ./test2
enter number of elements 
8
asdfghjk
at 0 is 
at 1 is a
at 2 is s
at 3 is d
at 4 is f
at 5 is g
at 6 is h
at 7 is j

Please [don't cast the result of malloc](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc?lq=1). — Martin, Aug 28 '15 at 11:11
Note that `sizeof(char)` is always 1, so you can replace allocation with `ptr = malloc(n + 1);`. — user694733, Aug 28 '15 at 11:15
@WayneSudo wrong. Since C99 the return value of `main()` is implicitly `0`. — RamblingMad, Aug 28 '15 at 11:24
After `scanf("%d",&n)`, call `scanf` one more time in order to "get rid" of the newline (enter) character that was scanned during the first call to this function. — barak manos, Aug 28 '15 at 11:26
Don't cast malloc() and "ptr = (char *) malloc((n + 1)*sizeof(char));" is the same as malloc((n+1) * 1), this is what you need ? — Michi, Aug 28 '15 at 11:43
Make sure to null terminate the array of characters after reading them, so that you an use them as a C string. — Lundin, Aug 28 '15 at 12:38

ameyCU · Answer 1 · 2015-08-28T11:32:53.840

3

Leave a space before %c in scanf-

scanf(" %c",&ptr[i]);

because there will be '\n' left in the buffer after the first scanf forn as you press ENTER after giving value of n.

And you don't need to cast result of malloc.

As what Matt McNabb Sir said in his comment ,for that you can do this -

 scanf("%*c%c",&ptr[i]);

%*c will take care of '\n' and wont even skip if only space is hit.

edited Aug 28 '15 at 11:32

answered Aug 28 '15 at 11:12

ameyCU

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He might want to enter spaces and have them count – M.M Aug 28 '15 at 11:25
@MattMcNabb Please see the edit Sir .I tied to cover your point. – ameyCU Aug 28 '15 at 11:28
1

Actually, that entire loop should be replaced by `fgets(ptr, n + 1, stdin);`. – The Paramagnetic Croissant Aug 28 '15 at 11:49
@TheParamagneticCroissant Actually that is right but then also he will have to take care of `'\n'` . – ameyCU Aug 28 '15 at 11:51
@ameyCU yes, no question about that! it's just that `fgets()` makes reading the line simpler. – The Paramagnetic Croissant Aug 28 '15 at 12:01
@TheParamagneticCroissant Yes I totally agree with you Sir !! – ameyCU Aug 28 '15 at 12:02

allocating dynamic memory using malloc

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