ASCII values

I am currently trying to get the program to search for more than one character. This has been successful up to a certain stage.

For some reason the program is reading the count for the first character and the second the same. I have looked through the code a few times and still stuck. Any help would be much appreciated

Here is the code.

Code:

#include<stdio.h> 
int main(void)
{
 
 
   int array[100], search, c, n, count = 0;
   int             search2, c1, n1, count2 = 0;
 
   printf("Enter the number of elements in array\n");
   scanf("%d",&n);
 
   printf("Enter %d numbers\n", n);
 
   for ( c = 0 ; c < n ; c++ )
      scanf("%d",&array[c]);
   for ( c1 = 0 ; c1 < n ; c1++)
   
   printf("Enter the number to search\n");
   scanf("%d",&search);
   printf("enter second number to search\n");
   scanf("%d", &search2);
 
   for ( c = 0 ; c < n ; c++ )
   for ( c1 = 0; c1< n ; c1++)
   {
      if ( array[c] == search )
      if ( array[c1] == search2)     
      {
         printf("%d is present at location %d.\n", search, c+1);
    count++;
         printf("%d is present at location22i3rwefkhc %d.\n", search2, c1+1);
    count2++;     
      }
   }
   if ( count2 == 0 )
      printf("%d is not present in array.\n", search2);
    else
      printf("%d is present %d times in array.\n", search2, count2);   
      
      
      
      
      
   if ( count == 0 )
      printf("%d is not present in array.\n", search);
   else
      printf("%d is present %d times in array.\n", search, count);
    
     
         
  


      
   
      
   return 0;
   
 }

Originally Posted by SDH

I am currently trying to get the program to search for more than one character. This has been successful up to a certain stage.

This might be related to your problem

Code:

for ( c = 0 ; c < n ; c++ )
      for ( c1 = 0; c1 < n ; c1++)
      {
         if ( array[c] == search )
            if ( array[c1] == search2)     
            {
               printf("%d is present at location %d.\n", search, c+1);
               count++;
               printf("%d is present at location22i3rwefkhc %d.\n", search2, c1+1);
               count2++;     
            }
      }

This is a nested for loop. For example, if n == 16, then the loop will run 256 times. Is that what you want? If you want to go through each array, and each array is n==16 elements, then you probably want two loops, one after the other.

Also, don't write if(FOO) if (BAR). If you want to say that both FOO and BAR must be true, the logical way (clearer to read and understand) is "if (FOO && BAR)"

This is a nested for loop. For example, if n == 16, then the loop will run 256 times. Is that what you want? If you want to go through each array, and each array is n==16 elements, then you probably want two loops, one after the other.

I think I have now sorted out the problem, by running the first loop to search for the first number and how many times it occurs. Then start the second one after that is complete. I am now having problems with the second number to search for though. It for some reason thinks that I am asking it to search for an 8. Any ideas on why this is occuring would be a big help, thanks.

Code:

 #include<stdio.h> 
int main(void)
{
 
 
   int array[100], search, c, n, count = 0;
 
   printf("Enter the number of elements in array.");
   scanf("%d",&n);
   printf ("There is %d numbers in the array.\n", n);
 
   printf("Enter %d numbers\n", n);
 
   for ( c = 0 ; c < n ; c++ )
      scanf("%d",&array[c]);
 
   printf("Enter the number to search\n");
   scanf("%d",&search);
 
   for ( c = 0 ; c < n ; c++ )
   {
      if ( array[c] == search )    
      {
         printf("%d is present at location %d.\n", search, c+1);
    count++;
      }
   }
  
   if ( count == 0 )
      printf("%d is not present in array.\n", search);
   else
      printf("%d is present %d times in array.\n", search, count);
   if ( count == 2)
      printf("One Pair.\n"); 
   if ( count == 3)
      printf("Three of a kind.\n");
   ///////////////////////////////////////////////////////////
   ///////////////////////////////////////////////////////////   
      
   int search2, c2, count2 = 0;


 
   for ( c2 = 0 ; c2 < n ; c2++ )
      scanf("%d",&array[c2]);
 
   printf("Enter the number to search\n");
   scanf("%d",&search2);
 
   for ( c2 = 0 ; c2 < n ; c2++ )
   {
      if ( array[c2] == search2 )    
      {
         printf("%d is present at location %d.\n", search2, c2+1);
    count2++;
      }
   }
  
   if ( count2 == 0 )
      printf("%d is not present in array.\n", search2);
   else
      printf("%d is present %d times in array.\n", search2, count2);
   if ( count2 == 2)
      printf("One Pair.\n"); 
   if ( count2 == 3)
      printf("Three of a kind.\n");   
      
      
         
      
     return 0;
   }

Sorry, have just found what the problem was. I was asking it look for a input to then say the array has that many values. So I have just removed this and it has sorted the problem.

How would I modify this program to either read a number or letter, the letters represent j,q,k,a?
thanks