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Home/ Questions/Q 8801485
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T00:50:44+00:00 2026-06-14T00:50:44+00:00

So this code is the base outline for a boggle game from online that

  • 0

So this code is the base outline for a boggle game from online that I copied over.
SOURCE: http://www.codingfriends.com/index.php/2010/06/10/boggle/

bool findUsersWord(string findThis, Grid<char> &theBoard, Vector<cell> &theRoute, string alreadyFound, int placeY, int placeX)
{  
  // need to find the findThis  base case
  if (findThis == alreadyFound)
    return true;
  // need to find the first letter within the board and then progress around that.
  if (alreadyFound.empty())
  {
    for (int rows = 0; rows < theBoard.numRows(); rows++)
      for (int cols = 0; cols < theBoard.numCols(); cols++)
        // find the each character within the 
        if (theBoard[rows][cols] == findThis[0])
        {
          alreadyFound = findThis[0];
          cell newR;
          newR.row = rows;
          newR.col = cols;
          theRoute.add(newR);
          if (findUsersWord(findThis, theBoard, theRoute, alreadyFound, rows, cols))
            return true;
          else
            // clear out the found Board 
            theRoute.clear();
        }
  }
  else
  {
    // try and find the next letters within the area around the base letter
    // spin around the letter 3 * 3 grid
    for (int y= (placeY > 0 ? placeY-1: placeY); y <=(placeY == (theBoard.numRows()-1) ? placeY : placeY+1);y++)
      for (int x=(placeX > 0 ? placeX-1: placeX); x<=(placeX == (theBoard.numCols()-1) ? placeX : placeX+1); x++)
        if ((theBoard[y][x] == findThis[alreadyFound.length()]) && (!(y==placeY && x ==placeX)))
          // already used letter
          if (!placeAlreadyUsed(y,x,theRoute))
          {
            alreadyFound += findThis[alreadyFound.length()];
            cell newR;
            newR.row = y;
            newR.col = x;
            theRoute.add(newR);
            if (findUsersWord(findThis, theBoard,theRoute, alreadyFound, y, x))
              return true;
            else
            {
              if (alreadyFound.length() > 1)
                alreadyFound = alreadyFound.substr(0, alreadyFound.length()-1);
              theRoute.removeAt(theRoute.size()-1);
            }
          }
    return false;
  }
  return false;
}

The code below is the code in question which is part of the code above.

for (int y= (placeY > 0 ? placeY-1: placeY); y <=(placeY == (theBoard.numRows()-1) ? placeY : placeY+1);y++)
  for (int x=(placeX > 0 ? placeX-1: placeX); x<=(placeX == (theBoard.numCols()-1) ? placeX : placeX+1)

I am wondering if someone could turn this code into more simple code that doesn’t involve the using of ? and that. I do know the simple parts of it such as the “?” means return and “:” means next line, but I am lost in the fact that it is being used in a for loop and the fact that it would just look like

if(placeY > 0)
 return playceY-1
placeY;

Where have I gone wrong?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-14T00:50:45+00:00Added an answer on June 14, 2026 at 12:50 am

    The ? : block is just a strange looking if statement. It’s an inline if, if you will.

    Here’s the format

    argument ? result evaluated to if true : result evaluated to if false
    

    Here’s an example

    1<2 ? "Hurray" : "boo"
    

    Will evaluate to "Hurray" because 1<2 is true. However, if we switch it to 1>2 it will evaluate to "boo".

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