You have already displayed the Tetris bucket in your previou

You have already displayed the Tetris bucket in your previous project assignment. In this assignment, you will need to display and drop the shape. Add plenty of narrative comments. Your program must be compilable and executable.

If you need guidance, you will find some detailed instruction (below) to assist you. This is not the only approach that will work but it will present one possibility. The sample below uses Classes (a little Object oriented programming) but, if you prefer a different approach, you do not have to use classes in your project.

To represent the Tetris shapes, you can create a Class called \"TetrisShape\". In the class, declare a 2-D array of chars that is 4x4 in size (you could also use 4x2 but that would require different algorithm). Call the array \"shapeArray\". For convenience, declare it as public (but, it should not be your practice when you work in the software industry). You might need more dimensions for the array if you want to statically assign (hard code) all the shape rotations (3-d) for all the shapes (4-d). In this example, the rotation of the shapes is done by value swapping in the 4x4 array.

Now, populate or initialize the \"shapeArray\". There are three (3) options for doing this:

Option 1: Add a function in the class (call it \"populateShapeArray(int shapeType)\"), that will take an integer which will denote the type of shape.

Option 2: Use a non-default constructor. In this case, when you create an object, you will take an integer as the constructor argument, which will denote the type of shape, as shown below.

TetrisShape currentShape = new TetrisShape(shapeType);// shapeType is an int

Option 3: Use a setter \"setShape(int shapeType)\" function.

Regardless of which option you chose above, you would use a switch statement to assign the 4x4 array values for the specific shape. For example, for \'L\', it can be something like the following:

shapeArray[0][0] = \' \'; shapeArray[1][0] = \'X\'; shapeArray[2][0] = \' \'; shapeArray[3][0] = \' \';
shapeArray[0][1] = \' \'; shapeArray[1][1] = \'X\'; shapeArray[2][1] = \' \'; shapeArray[3][1] = \' \';
shapeArray[0][2] = \' \'; shapeArray[1][2] = \'X\'; shapeArray[2][2] = \'X\'; shapeArray[3][2] = \' \';
shapeArray[0][3] = \' \'; shapeArray[1][3] = \' \'; shapeArray[2][3] = \' \'; shapeArray[3][3] = \' \';

bucket[i+shapeTopLeftX][j+shapeTopLeftY] = localTetrisShape.shapeArray[i][j];

You are actually imposing the shape values on to the bucket.

Always keep record of the top left corner location (x,y dimensions) of the shape that is falling. This record will indicate where to draw the shape in the bucket after the shape moves. Initially, the location should be, (6, 0), which is the top middle of the bucket. Store the coordinates in the \"TetrisShape\" class asshapeTopLeftX and shapeTopLeftY. Note, x changes horizontally, and y changes vertically.

In this step you will generate one shape. To accomplish this, in the main() function, before the while loop (game loop), do the following:

Generate a random number from 0 to 6.

Use one of the options presented in step 2 above to create a TetrisShape object.

Create a global function called \"updatebucket(TetrisShape localTetrisShape)\" to populate the bucket with values of the \"shapeArray\". Here, you will use two nested for loops to go over the shapeArray of 4x4. You need to provide the TetrisShape object (created in step \'4b\' above) to the function as an argument in order to access its shapeArray. Inside the loop, you will do this.

Call the \"updatebucket(TetrisShape localTetrisShape)\" function from main() using the object created in step 4b.

Display the bucket.

Inside the while loop (game loop), as the shape falls, the value of y will increase. In order to accomplish this you will have \"newShapeTopLeftY = shapeTopLeftY + 1;\" so that the shape falls one cell below. Notice that you are storing the new y value in another variable because you need both the old and the new values.

Clear (assign spaces to) the current position values of the shape in the bucket using the \"old\" shapeTopLeftX and shapeTopLeftY. This will ensure that you do not leave any trail of the shape.

Call the \"updatebucket(TetrisShape localTetrisShape)\" function with the \"new\" shapeTopLeftX and shapeTopLeftY values. This will ensure the shape will be displayed in the new location (newShapeTopLeftX, newShapeTopLeftY).

Display the bucket. The shape is displayed in the new location of the bucket.

Store the new shapeTopLeftX and shapeTopLeftY values to the old shapeTopLeftX and shapeTopLeftY values to get ready for the next loop.

Sleep for a while to slow down the game.

Solution

i\'ve a 2d array of Block gadgets which became created like this:

Block[,] gameGrid;
gameGrid = new Block[10, 18];

for (int x = 0; x < gameGrid.GetLength(zero); x++)

for (int y = 0; y < gameGrid.GetLength(1); y++)

gameGrid[x, y] = new Block(false, coloration.red, posX + (blockTexture.Width * x), posY - (blockTexture.top * y), blockTexture.Width, blockTexture.top, x, y);

the first parameter of the Block constructor units the blocks to be disabled so that they do not show up. beneath is a screenshot of the grid with every block enabled to offer you an idea of what it looks as if:

What i\'m wanting to do is rotate the blocks however i\'m now not certain the way to do it. for instance, take the following scene:

The green block is the centre block which the others could be rotated around. The favored effect is the subsequent:

What I need to do is calculate which blocks want to be disabled and which blocks want to be enabled and aside from having lots and masses of if statements, I can\'t see how i\'m going to do it. I assume the calculation for each block wishes to be relative to the block to be turned around around. For the above form, each block is adjoining to the centre block. however what about the subsequent:

Which needs to come to be looking like this:

Sorry for all of the photographs, but you understand what they say? A picture\'s worth 1000 words =o)