C++ 有人可以解释这些 for 循环在说什么吗？

Question

所以这段代码是我从网上复制过来的一个猜谜游戏的基本大纲。 来源: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;
}

下面的代码是有问题的代码，它是上面代码的一部分。

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)

我想知道是否有人可以将此代码转换为不涉及使用 ?然后。我确实知道它的简单部分，例如“？”表示返回，“:”表示下一行，但我迷失了它在 for 循环中使用的事实以及它看起来像的事实

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

我哪里做错了？

Answer 1

？ : block 只是一个看起来很奇怪的 if 语句。如果您愿意，它是一个内联。

这是格式

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

举个例子

1<2 ? "Hurray" : "boo"

将评估为 "Hurray" 因为 1<2 为真。但是，如果我们将其切换为 1>2，它将计算为 "boo"。