Question

See the following program:

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

main(void){
  printf("Array concatenateor\n-------------------------\n");

  //declarations
  char s1[50],s2[50],s3[50];
  short int n=0;

  //Array initialisation
  printf("Array 1: ");
  gets(s1);
  printf("Array 2: ");
  gets(s2);

  strcpy(s3, s1); //asure initial form of s1 in s3
  strcat(s1, s2); //concatenate s1 with s2  
  //at this point s1 is in the concatenation form and s3 is s1's initial form

  printf("Arrays concatenated with STRCPY: \"%s\"\n", s1); //print concatenation, s3 ok
  printf("Number of characters to concatenate: "); //scan number, s3 ok
  scanf("%d",&n); //beyond this point s3 becomes null... peculiar
  printf("S3: %s\n",s3);    //this is the proof of s3 being null
  strncat(s3,s2,n); //s3 concatenates with n chars of s2
  printf("Arrays concatenated with STRNCAT(%d chars): %s\n", n, s3);  //print s3
  system("PAUSE");
  return 0;
}

It is peculiar how that specific scanf erases the s3 array without even being implied. How come this happens?

Answer 1

Try making “n” an “int” rather than a “short int”, run it, and see if that fixes it. If so, I’ll explain why.

As Soren points out, scanf expects to read in 32 bits of data; when the buffer is only 16 bits wide, the extra memory is shuffled somewhere it doesn’t belong, essentially zeroing out memory (assuming the user enters a small enough number).