using pointers

//CONCATENATION OF TWO STRINGS USING POINTER IN C PROGRAM

#include<stdio.h>
int main(){
  int i=0,j=0;
  char *str1,*str2,*str3;
  puts("Enter first string");
  gets(str1);
  puts("Enter second string");
  gets(str2);
  printf("Before concatenation the strings are\n");
  puts(str1);
  puts(str2);
  while(*str1){
      str3[i++]=*str1++;
  }
  while(*str2){
      str3[i++]=*str2++;
  }
  str3[i]='\0';
  printf("After concatenation the strings are\n");
  puts(str3);
  return 0;
}//LINEAR SEARCH USING C PROGRAM

#include<stdio.h>
int main(){

    int a[10],i,n,m,c=0;

    printf("Enter the size of an array: ");
    scanf("%d",&n);

    printf("Enter the elements of the array: ");
    for(i=0;i<=n-1;i++){
         scanf("%d",&a[i]);
    }

    printf("Enter the number to be search: ");
    scanf("%d",&m);
    for(i=0;i<=n-1;i++){
         if(a[i]==m){
             c=1;
             break;
         }
    }
    if(c==0)
         printf("The number is not in the list");
    else
         printf("The number is found");

    return 0;
}

Sample output:
Enter the size of an array: 5
Enter the elements of the array: 4 6 8 0 3
Enter the number to be search: 0
The number is found

//BINARY SEARCH USING C PROGRAM

#include<stdio.h>
int main(){

    int a[10],i,n,m,c=0,l,u,mid;

    printf("Enter the size of an array: ");
    scanf("%d",&n);

    printf("Enter the elements in ascending order: ");
    for(i=0;i<n;i++){
         scanf("%d",&a[i]);
    }

    printf("Enter the number to be search: ");
    scanf("%d",&m);

    l=0,u=n-1;
    while(l<=u){
         mid=(l+u)/2;
         if(m==a[mid]){
             c=1;
             break;
         }
         else if(m<a[mid]){
             u=mid-1;
         }
         else
             l=mid+1;
    }
    if(c==0)
         printf("The number is not found.");
    else
         printf("The number is found.");

    return 0;
}

Sample output:
Enter the size of an array: 5
Enter the elements in ascending order: 4 7 8 11 21
Enter the number to be search: 11
The number is found.