c program to find a substring

Replies

• 

  reachrkata

  Using For Loops and Character comparison.
  
  -Karthik
  😁
• 

  samaira

  that's quite obvious but i m not able to get the output.To be frank i want the code of the program.Plz help if u can
• 

  Saandeep Sreerambatla

  Please post clear question, post what you have tried so far such taht we can guide you..
• 

  samaira

  #include
  #include
  int main()
  {
  int i=0,j=0;
  char a[10],b[10];
  clrscr();
  printf("enter the main string: ");
  scanf("%s",a);
  printf("\n enter the second string: ");
  scanf("%s",b);
  while(a!='\0')
  {
  if(a==b[j])
  {
  i++;
  j++;}
  
  else
  {
  j=0;
  if(a==b[j])
  j++;
  i++;
  }
  else
  {
  printf("the given string is not substring");
  }
  printf("\n the given substring is present");
  }
  getch();
  }
  just check the program and tell me where i m going wrong
• 

  Saandeep Sreerambatla

  This is my understanding..
  
  Before the first if loop you have initialized i=0 and j=0.
  
  If that condition fails again you have initialized j=0.
  
  So if the first IF loop fails ..
  Then the If loop in the else part will also fail for j=0.
  
  This functionality is very easy by using for loops.
  
  First take the main string into a[] , then substring into b[]
  
  Use two for loops and you will get the answer.
• 

  samaira

  i have initialized j=0 again after the if loop fails coz the string might appear at any position of the second string like e.g we have to find whether the string "comp" is a substring of "mech & comp".so after the condition fails the second string is again ready to be checked from first while the counter of second goes on increasing.
• 

  shalini_goel14

  Hi samaira,
  
  

  int blen=0, k=0;
  
  //assign length of be to blen
  blen =strlen(b); //not sure
  
   while(a[i]!='\0')
        {
         if(a[i]==b[j])
            {
             i++;
             j++;
  if(j==blen){
  j=0;
  }
  
  }
  
          else
          {
              j=0;
            //if(a[i]==b[j])
             //j++;
          i++; 
             }
          else
  {
          printf("the given string is not substring");
  }
  printf("\n the given substring is present");
  }
  getch();
  }
  

  Try this including b length's check in your program, if it works. Also, I think you can include a separate variable k in your program and initialize it to 0 and set its value as 1 when your check in if loop works and then at the end while printing message you can check if it is 1 then "matched" else "unmatched"
  
  Thanks !
• 

  samaira

  hey shalini thanks for ur help.:smile:
• 

  shalini_goel14

  @Samaira If you program works then don't forget to post the complete working program here also.
  
  Thanks !
• 

  yogesh_dhiman

  //Without using Functions.
  
  #include(stdio.h) // place this '<' & '>' instead of '(' & ')' before stdio.h
  #include(string.h)
  void main( )
  {
  char st[20],sub[10],temp[10];
  int pos, i, j;
  clrscr( );
  printf("Enter the main string:");
  gets(st);
  printf("\nEnter the substring to insert:");
  gets(sub);
  printf("Enter the index position:");
  scanf("%d",&pos);
  if(pos<=strlen(st)) {
  for(i=0;i<=strlen(st);i++) {
  if(i==pos)
  {
  for(j=0;st!='\0';j++) // to store the 'st' to 'temp' from given position.
  {
  temp[j]=st;
  i++;
  }
  temp[j]='\0';
  i=pos;
  for(j=0;sub[j]!='\0';j++) // to insert a sub-str to main string.
  {
  st=sub[j];
  i++;
  }
  for(j=0;temp[j]!='\0';j++) // Lastly to insert the 'temp' to 'st' after sub-str.
  {
  st=temp[j];
  i++;
  }
  st='\0';
  }}
  printf("\nAfter adding the sub-string: %s",st);
  }
  else
  printf("\nSorry, it is not possible to insert a substring in that position.");
  getch();
  }
  
• 

  jkr0103

  [HASHTAG]#include[/HASHTAG] 
  using namespace std;
  int main()
  {
      char a[] = {"Hanuman"};
      char b[] = {"Hanuu"};
      int i = 0, j = 0;
      while(a[I]!='\0')
      {
          if(a[I]==b[j])
          {
              j++;
              if (b[j] == '\0')
                  break;
          }
          else
          {
              i = i-j;
              j = 0;
          }
          i++;
      }
      if(b[j] == '\0')
      {
          cout << "substring exists\n";
      }
      else
      {
          cout << "substring does not exists\n";
      }
      return 0;
  } 

  [/I][/I]