Question

这个程序的目的是创建一个比原始数组放大10倍的更大字节数组。例如，[0] [0]中的1应该是新数组中1x的10x10平方。我提供代码和输出，它似乎在较大数组的填充期间正常工作，但随后打印不同的值。我目前正在试验这些行，以便限制我在测试期间处理的变量数量。任何人都可以想到这种情况发生的原因吗？

public class Test 
{
static byte[][] byteArray =
{{1, 0},
 {0, 1}};

public static void main(String[] args)
{
    byte newarray[][] = converter();
    for(int i = 0; i < 20; i++)
    {
        System.out.println(newarray[i][0]);
    }
}

private static byte[][] converter()
{
    byte[][] b = new byte[20][20];

    for(int r = 0; r < 2; r++)
    {
        for(int i = 0; i < 10; i++)
        {
            b[r+i][0] = byteArray[r][0];
            System.out.println(byteArray[r][0]);
            System.out.println(b[r+i][0]);
        }
    }

    return b;
}

}

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Answer 1

为什么不直接使用截断整数除法：

static void printMat(byte[][] mat) 
// just a utility function to print a matrix
{ 
    for(byte[] row : mat)
    {
        System.out.println(Arrays.toString(row));
    }
}

private static byte[][] stretch(byte[][] bytes, int rfactor, int cfactor)
// stretch the matrix in 'bytes'
//stretch the rows by 'rfactor' and the columns by 'cfactor'
{ 
    // create an empty matrix:
    int rows = bytes.length*rfactor; // rows in the new matrix
    int cols = bytes[0].length*cfactor; // columns in the new matrix
    byte[][] out = new byte[rows][cols]; // our new, stretched matrix

    // loop through the rows and columns of the *new* matrix:
    for(int r = 0; r < rows; r++)
    {
        for(int c = 0; c < cols; c++)
        {
            // Divide the row and column indices by the 
            // appropriate factors to find the correct value
            // in the original matrix.
            // Integer division just drops any remainder,
            // which is what we want.
            out[r][c] = bytes[r/rfactor][c/cfactor];
        }
    }
    return out;
}

public static void main(String[] args) throws Exception 
{
    // your example:
    byte[][] byteArray =
        {{1, 0},
         {0, 1}};
    byte[][] newarray = stretch(byteArray, 10, 10);
    printMat(newarray);

    System.out.println();

    // can stretch any matrix by any dimensions:
    byte[][] byteArray2 =
        {{1, 2, 3},
         {4, 5, 6}};
    byte[][] newarray2 = stretch(byteArray2, 3, 2);
    printMat(newarray2);

}

输出：

[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]

[1, 1, 2, 2, 3, 3]
[1, 1, 2, 2, 3, 3]
[1, 1, 2, 2, 3, 3]
[4, 4, 5, 5, 6, 6]
[4, 4, 5, 5, 6, 6]
[4, 4, 5, 5, 6, 6]

缩放数组（矩阵）

1 个答案: