char[][]数组在for循环中被分配单个数组[i]时不保存值

xqkwcwgp  于 2021-07-08  发布在  Java
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我注解了我的代码,以帮助您理解我的逻辑,我基本上收到一个边值,我必须画一个三角形使用x的视觉显示其周长(各方都是平等的)
因此,在创建将填充包含图形的矩阵的单个数组时,随着迭代的继续,一些赋值会丢失,只保留最后一次迭代的值,并用这些值填充整个矩阵,这使我失去了所有的前一行,组成一个漂亮的三角形绘图,我无法打印成一个txt文件(我可以通过在循环中打印来在控制台中打印它,因为它生成了对应于一条绘图线的每个数组,但这只是一个暂时的幻觉,因为我没有将矩阵和x保持在适当的位置)。
如果有人能帮我不失去这个价值,并且能把它储存起来,我会很高兴的。
注意:父类figura只提供一个名为lado的int。

public class Triangulo extends Figura {

    char[] lineaI;
    char[] lineaT;
    char[][] matT;

    public char[][] matDibujo(int lado) {

        this.linea = new char[(lado+lado-1)]; //side int value to determine an eq. triangle base length
        lineaI = new char[linea.length]; //multiple intermediate single arrays that fill triangle diagonally
        lineaT = new char[linea.length]; //triangle base array
        matT = new char[lado][linea.length]; // matrix containing {{},{},{},{}} n individual arrays representing a row in the drawing (top to bottom)

        //Create triangle base line array 
        for (int i = 0; i < linea.length; i++) {
            if (i == 0 ) {
                linea[i] = 'X';
            } else if (i%2 == 0) {
                linea[i] = 'X';
            } else {
                linea[i] = ' ';
            }
            //Create triangle top point array
            if (i == (lado-1)) {
                lineaT[i] = 'X'; 
            } else {
                lineaT[i] = ' ';
            }
        }

        //Fill matrix with first array (top axis)
        matT[0] = lineaT;
        //Fill matrix last array (triangle base) 
        matT[lado-1] = linea;
        //System.out.println(matT[lado-1]);

        //THIS IS THE LOOP NOT SAVING CORRESPONDING lineaI full single array to matT[j], 
        //instead it's replacing every matT[j] with the last value of lineaI.

        //Create multiple arrays to fill n-sided triangle from top axis to base diagonally (parting from top axis)  
        for (int j = 1; j < (lado-1); j++) {
            for (int i = 0; i < linea.length; i++) {
                if (i == (lado-1-j) ) {
                    lineaI[i] = 'X';
                } else if (i == (lado-1+j) ) {
                    lineaI[i] = 'X';
                } else {
                    lineaI[i] = ' ';
                }
            }
            matT[j] = lineaI;
        }

        //return the matrix so we can print it
        return matT;

    }

    //Dibujar en txt
    public void dibuja(char[][] matriz) {

        for (int i = 0; i < matriz.length; i++) {
            System.out.println(matriz[i]);      
        }

    }

    public static void main(String[] args) {

        char[][] mat;
        Triangulo t = new Triangulo();
        mat = t.matDibujo(4);
        t.dibuja(mat);

        //This is the outcome, since we lost matT[1] and got it replaced by the last lineaI contents, which should
        //only belong to matT[2]
        /*      X                     The outcome should be             X                       
              X   X                                                    X X     ---> this line is lost and
              X   X                                                   X   X    ----> replaced by this one
             X X X X                                                 X X X X                                   */

    }

}
u4dcyp6a

u4dcyp6a1#

这里的问题是lineai数组值一旦计算出来,就将它赋给matt[j],然后再次用新值覆盖同一lineai引用,并为每个j值附加到matt。
这将导致值被覆盖,并显示所有间歇行的最终数组值。
要解决这个问题,只需在每个j值的迭代开始时,将lineai变量重新初始化为一个新数组。

for (int j = 1; j < (lado-1); j++) {
        //initialize the lineaI variable for start of new j value in loop
        lineaI = new char[linea.length]; // <----- new change
        for (int i = 0; i < linea.length; i++) {
            if (i == (lado-1-j) ) {
                lineaI[i] = 'X';
            } else if (i == (lado-1+j) ) {
                lineaI[i] = 'X';
            } else {
                lineaI[i] = ' ';
            }

        }
        matT[j] = lineaI;

    }

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