Question

Say i have an array char myArrray[5] = {'T','T','T','T','T'} And i want to check how many instances of 3 pairs of T i have.

I have a forloop below that checks for all 3 instances of T. There should be 3 instances, but for some reason its not even entering the if statement that checks it.

Maybe im just lost im really sleepy.

There are 3 isntances of TTT throughout the array. Thats what we have to get the number 3 in the counter but we arent getting it. (T{T[T)T}T]

full code here: http://ideone.com/AWyOkH

Any ideas?

     for(int k = 0; k < lineInputs; k++)
{
    int counter=0;
    cout << (k+1) << " ";

    for(int u=0; u<arrayElements; u++)
    {
        //cout << myArray[u];
        if(myArray[u] == 'T' && myArray[u+1] == 'T' && myArray[u+2] == 'T')
        {
            counter++;
            cout << counter << " ";
        }
    }
}

Does the issue lay with if(myArray[u] == 'T' && myArray[u+1] == 'T' && myArray[u+2] == 'T') ?

Answer 1

Seems to work fine.

   char myArray[5] = {'T','T','T','T','T'};
    int lineInputs=1;
    for(int k = 0; k < lineInputs; k++)
    {
      int counter=0;
      cout << (k+1) << " ";
      int arrayElements=5;
      for(int u=0; u<(arrayElements-2); u++)
      {
         //cout << myArray[u];
         if(myArray[u] == 'T' && myArray[u+1] == 'T' && myArray[u+2] == 'T')
         {
             counter++;
             cout << counter << " ";
         }
      }
    }

Output I get is:

1 1 2 3