Question

嘿，我真的需要帮助我的java编程类的程序。我将介绍我到目前为止的说明和代码。任何帮助，将不胜感激！提前致谢！！说明：编写一个名为Box（Box.java）的程序，它将打印/显示一个空心盒子形状使用星号（*）。该程序将以2到24的范围内的偶数读取指定框中的行数/列数。显示错误并重新提示输入错误值时的数字。然后该程序将显示一个空心适当的大小。提示：在循环中使用循环。基本上它应该是一个方框，所以如果你给它数字boxSize = 5输出是一个尺寸为5x5的盒子。轮廓由星号组成，但内部是空的

到目前为止我的代码是什么

import java.util.Scanner;

public class Box
{

    public static void main(String[]args)
    {
        //numrows and numcols are equal however the spacing is differnt
        Scanner input = new Scanner(System.in);
        System.out.print("Enter an even number (2-24):  ");
            int boxSize = input.nextInt();

        int numRows = boxSize;
        int numCols = numRows;
        // This program demonstrates compound decisions with the logical and operator &&
        //asks if the number is less than or equal to 24 and greater than or equal to 2 and that the remainder is 0 
        //when divided by 2, checks if its an even number like it should be

        if(boxSize >= 2 && boxSize <= 24 && boxSize%2 == 0)
        {   
            //nested loops are used to print out the asterisk in the correct pattern
            for(int r = 0; r<numRows; r++)
            {
                    System.out.println("*");
                for(int c = 0; c<numCols; c++)      
                {   
                            System.out.print("*");    
                }
                    }
        }
    }

            //This program demonstrates compound decisions with the logical or ( || ) operator
        //checks if any of the following are true
        //if one or more is true then that means that it is an incorrect number
        //then reprompts the user to put in a new number then checks again
        if(boxSize<2||boxSize>24||boxSize% 2 != 0)
        {
            System.out.println("Value must be an even number from 2-24");
        }

基本上我的问题是我不知道在循环中放什么以及在哪里获得形状。如果数字是奇数或不在2到24之间，我也不知道如何再次使用它作为boxSize值，并且还需要显示错误消息，该值必须介于2和24之间，甚至等等。 / p>

Answer 1

使用嵌套循环的方法基本上没问题。只是里面的东西不是你想要的......你想要的更像是这样：

for(int r = 0; r<numRows; r++) {
    for(int c = 0; c<numCols; c++) {
        // print a "*" ONLY IF on border, and " " otherwise 
    }
    // here you finished printing the "*"/" ", so print just a newline
}

对于proming和reprompting我将使用do {} while(yourIfCondition)循环来重新显示提示。

Answer 2

下面。我把所有这些都放在一个方法中。请注意，有很多ifs。如果您已经在课程中达到该部分，则可以使用三元运算符进行优化。

public static void main(String...args){
        int boxSize = 0;
        Scanner input = new Scanner(System.in);

        do {
            System.out.print("Enter box size [-1 to quit] >> ");
            boxSize = input.nextInt();

            if(boxSize == -1){
                System.exit(0);
            }

            /* check if number is valid */
            if(boxSize < 2 || boxSize > 24 || boxSize % 2 != 0){
                System.err.println("--Error: please enter a valid number");
                continue; // prompt again
            }

            // draw the box
            for (int col = 0; col < boxSize; col++) {
                for (int row = 0; row < boxSize; row++) {
                    /* First or last row ? */
                    if (row == 0 || row == boxSize - 1) {
                        System.out.print("*");
                        if (row == boxSize - 1) {
                            System.out.println(); // border reached start a new line
                        }
                    } else { /* Last or first column ? */
                        if (col == boxSize - 1 || col == 0) {
                            System.out.print("*");
                            if (row == boxSize - 1) {
                                System.out.println();
                            }
                        } else {
                            System.out.print(" ");
                            if (row == boxSize - 1) {
                                System.out.println();
                            }
                        }
                    }
                }
            }

        }while (true);
    }

如何用java中的星号制作盒子？嵌套循环？

2 个答案: