Reading a string without wasting memory

Question

I have to read a string from stdin, allocating memory dinamically without wasting it. I've done this, but i'm not convinced about it,because in this way i think i waste memory!

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

char *alloc_memory(int n)
{
    char *p;
    p=malloc(n*sizeof(char));
    if(p==NULL)
    {
        fprintf(stderr,"Error in malloc\n");
        exit(EXIT_FAILURE);
    }
    return p;
}


int main(int argc, char *argv[])
{
    if(argc != 1)
    {
        fprintf(stderr,"Usage: %s \n",argv[0]);
        return EXIT_FAILURE;
    }

    char string[64];
    int lung;
    char *p,*s,*w;

    printf("Insert string: \n");

    p=fgets(string,63,stdin);
    if(p==NULL)
    {
        fprintf(stderr,"Error in fgets\n");
        exit(EXIT_FAILURE);
    }

    printf("You've inserted: %s", string);

    lung=strlen(p);

    s = alloc_memory(lung+1);

    w=strncpy(s,p,lung);

    printf("Final string:%s", w);   

    return EXIT_SUCCESS;

}

any idea? Should i read one character at a time?

What memory do you think you're wasting? (Apart from the obvious leak.) — Mat
– Mat, Commented Sep 1, 2013 at 10:44
One character at a time will be much slower (and user will get irritated :P) than your approach. You are not wasting any memory, other than the fact that your program is allocating dynamic memory continuously without ever releasing it (which is bad^infinity). — MD Sayem Ahmed
– MD Sayem Ahmed, Commented Sep 1, 2013 at 10:47
For this? Warning: If there is no null byte among the first n bytes of src, the string placed in dest will not be null-terminated. — SegFault
– SegFault, Commented Sep 1, 2013 at 11:53
@Butterfly: Exactly that's the dangerous "feature" of strncpy(). — alk
– alk, Commented Sep 1, 2013 at 12:08

alk · Accepted Answer · 2013-09-01 13:18:28Z

2

To have char str[64] (string is not a good name for a variable, it might lead to ambiguities) declared only temporarily just put it in a local context:

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

char * alloc_memory(size_t n)
{
  char * p = malloc(n); /* * sizeof(char) is always 1 */

  if (p == NULL)
  {
    fprintf(stderr, "Error in malloc() when trying to allocate %zu bytes.\n", n);
    exit(EXIT_FAILURE);
  }

  memset(p, 0, n); /* Avoid having strncpy() choke .. later down in this example. */

  return p;
}

int main(int argc, char * argv[])
{
  if (argc != 1)
  {
    fprintf(stderr, "Usage: %s \n", argv[0]);
    return EXIT_FAILURE;
  }

  {
    char * w = NULL;

    printf("Insert string: ");

    {
      char str[64]; /* here str is allocated */
      char * p = fgets(str, 63, stdin);
      if (p == NULL)
      {
        fprintf(stderr, "Error in fgets().\n");
        exit(EXIT_FAILURE);
      }

      printf("You've inserted: '%s'\n", str);

      {
        size_t lung = strlen(p);
        char * s = alloc_memory(lung + 1);

        w = strncpy(s, p, lung);
      }
    } /* here "str" is deallocated */

    printf("Final string: '%s'\n", w);
  }

  return EXIT_SUCCESS;
}

edited Sep 1, 2013 at 13:18

answered Sep 1, 2013 at 10:51

alk

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4 Comments

Nikos C. Over a year ago

Or you can malloc() string and then free() it once the data is copied to a new buffer.

alk Over a year ago

@Butterfly: Please see my comment (on the usage of strncpy()) to your question.

Nikos C. Over a year ago

Btw, a local scope is sometimes a hint that it should be split into its own function. In this case, something like char* read_alloc_line(FILE* f);

Joshua Green Over a year ago

In alloc_memory, why not use calloc rather than malloc and save the memset?

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