Segmentation fault while taking string from user

Question

Code:

int main()
{
  char *name=NULL;
  int n;
  printf(\"\\nenter the string\\n\");
  scanf(\"%s\",name);
  n=strlen(name);
  printf(\"%d\",n);
  return 0;
}


        

Answer 1

You didn't allocate any memory for pointer to char name.

Example:

char * name = malloc( sizeof( char ) * MAX_STRING_LENGTH ) ;

Answer 2

Try this,

int main()

{
  char *name = malloc(sizeof( char ) * LENGTH); // define LENGTH as you desired

  int n;
  printf("\nenter the string\n");
  scanf("%s",name);
  n=strlen(name);
  printf("%d",n);

  free(name);
}

Problem is you did not allocate memory for pointer. So allocate memory tp pointer with malloc(BUFSIZE). Also at the end you have to free your allocated memory with free(name).

Answer 3

C is not a managed language, so you need to tell your string (char *) wihch lenght of memory are you giving it. Here comes the malloc function.

By the way, there is no GarbageCollector, so you'll need to free your char * when you'll have finish to use it.

But be careful, malloc can return null, so your char * would be unable to store any char !

int main(int argc, char **argv)
{
    char *name = null;
    // Malloc your char *
    if ((name = malloc(sizeof( char ) * LENGTH_OF_YOUR_LARGER_INPUT)) == null)
    return;
    int n;
    printf("\nenter the string\n");
    scanf("%s",name);
    n=strlen(name);
    printf("%d",n);
    // Free the allocated memory to your char *
    free(name);
}

Answer 4

Instead of char * name = malloc( sizeof( char ) * MAX_STRING_LENGTH ) ;

USE

 char * name = malloc(MAX_STRING_LENGTH+1 ); //+1 is to store Null character 

and sizeof(char)==1 so you can avoid it.