Question

doTheMath（int [] []，int [] []，char） 将两个二维整数数组作为参数，并返回一个二维整数数组。它的第三个参数是一个字符，它告诉方法要执行哪个算术运算。它将创建一个与通过参数进入的数组大小相同的新数组。它将使用character参数来确定填充新数组时要执行的操作。例如，如果字符参数包含＆＃39; +＆＃39;符号，此方法将在每个数组中添加相应的条目，并将它们的总和放在新数组中的相同位置。它应该处理4个操作：添加（＆＃39; +＆＃39;），减去（＆＃39; - ＆＃39;），乘以（＆＃39; *＆＃39;）和除（＆＃39;） 39 /＆＃39）。如果有任何其他字符，它应该对两个数组的内容执行余数（模数）操作。

因此我目前拥有的是我的代码...

    public static int [][]doTheMath(int [][]array1, int [][]array2, char arithmetic)
{
    // Declares 2-dimensional array the same size as one in parameters
    int [][]arraySum = new int [array1.length][array2.length];

    // For loop to take each row and column to separately do arithmetic on
    for (int i = 0; i < array1.length; i++) {
        for (int j = 0; j < array1.length; j++) {
            if (arithmetic == '+') 
                arraySum [i][j] = array1 [i][j] + array2 [i][j];
            else if (arithmetic == '-') 
                arraySum [i][j] = array1 [i][j] - array2 [i][j]; 

            else if (arithmetic == '*') 
                arraySum [i][j] = array1 [i][j] * array2 [i][j];

            else if (arithmetic == '/') 
                arraySum [i][j] = array1 [i][j] / array2 [i][j];

            else 
                arraySum [i][j] = array1 [i][j] % array2 [i][j];
        }
    }

    // arraySum value is returned
    return arraySum;
}

要在我的主要方法中调用这个方法会是正确的吗？

    // 2-dimensional array created from values taken from doTheMath() method class
    int [][]sum = doTheMath(array1,array2, '+');

    // For loop to print the 2-dimensional array results after arithmetic
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < columns; j++) {
            System.out.printf("%4s", sum[i][j] + " ");
        }
        System.out.println(" ");
    }

我从这样打印得到的值似乎给了我错误的价值。

9 7 2     3 2 5     16  5  7
3 3 3  +  4 1 8  =   6  7  9
4 5 8     2 3 7      9 10 10

Array1和Array 2

    // Declared arrays
    int [][] array1;
    int [][] array2;

    // Calls getSize() to get the number for rows and then columns
    int rows = getSize("Please enter number of rows: ");
    int columns = getSize("Please enter number of columns: ");

    // Instantiate input for rows and columns into both double arrays
    array1 = new int [rows][columns];
    array2 = new int [rows][columns];

    // For loop to fill in the rows and column values with random integers
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < columns; j++) {
            array1 [i][j] = (int)(Math.random() * 9 + 1);
            array2 [i][j] = (int)(Math.random() * 9 + 1);
            System.out.printf("%4s", array1[i][j] + " ");                
            System.out.printf("%4s", array2[i][j] + " ");

        }
        System.out.println(" ");
    }

Answer 1

所以我发现如果我拿这个我认为可以用来并排打印矩阵。

    // For loop to fill in the rows and column values with random integers
for (int i = 0; i < rows; i++) {
    for (int j = 0; j < columns; j++) {
        array1 [i][j] = (int)(Math.random() * 9 + 1);
        array2 [i][j] = (int)(Math.random() * 9 + 1);
        System.out.printf("%4s", array1[i][j] + " ");                
        System.out.printf("%4s", array2[i][j] + " ");

    }
    System.out.println(" ");
}

将其更改为两个单独的循环，为我提供了正确的结果。

        // For loop to fill in the rows and column values with random integers
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < columns; j++) {
            array1 [i][j] = (int)(Math.random() * 9 + 1);
            System.out.printf("%4s", array1[i][j] + " ");                


        }
        System.out.println(" ");
    }

    // For loop to fill in the rows and column values with random integers
    for (int i = 0; i < rows; i++) {
        for (int j = 0; j < columns; j++) {
            array2 [i][j] = (int)(Math.random() * 9 + 1);               
            System.out.printf("%4s", array2[i][j] + " ");

        }
        System.out.println(" ");
    }

方法类使用（int [] []，int [] []，char）帮助“char”部分并打印错误的值

1 个答案: