您好我正在尝试创建一个checkArg方法，该方法检查用户是否应该键入一个整数参数时键入一个字符串，如果用户输入的参数超过3个，那么告诉他们“请输入正确的值”，如果这些条件不符合，然后再次询问并继续该计划。我的代码底部有一个粗略的轮廓。
我尝试在底部的main中调用checkArg方法，比较参数'size'。它编译，但不会告诉我“请输入正确的值”，然后继续该程序。
我的代码：
import java.util.*;
public class trominoZ2 {
//create a drawing panel of width=400px and height=400px
private int[][] grid;
private int currentNum;
// Pre-condition: size must be a perfect power of 2 and 0<=x<size, 0<=y<size
// Post-condition: creates an empty tromino object with dimensions size x size.
public void tromino(int size, int x, int y) {
int actualsize = 1;
while (actualsize < size) actualsize*=2; //actualsize = actualsize * 2
// Make sure the grid size is a perfect power of 2.
grid = new int[actualsize][actualsize];
currentNum = 1;
// Fill in the grid with all empty squares.
for (int i=0; i<actualsize; i++) {
for (int j=0; j<actualsize; j++) {
grid[i][j] = 0;
}
}
// This represents the original hole in the tromino.
grid[x][y] = -1;
}
// Wrapper call for recursive method.
public void tile() {
tileRec(grid.length, 0, 0);
}
private void tileRec(int size, int topx, int topy) {
// No recursive case needed here, just fill in your one tromino...
if (size == 2) {
// Fill in the one necessary tromino. The hole is identified by a
// non-zero number, so don't fill in that one square.
for (int i=0; i<size; i++)
for (int j=0; j<size; j++)
if (grid[topx+i][topy+j] == 0)
grid[topx+i][topy+j] = currentNum;
// Advance to the next tromino.
currentNum++;
}
// Recursive case...
else {
// Find coordinates of missing hole
int savex=topx, savey=topy;
for (int x=topx; x<topx+size; x++)
for (int y=topy; y<topy+size; y++)
if (grid[x][y] != 0) {
savex = x;
savey = y;
}
// Hole in upper left quadrant.
if (savex < topx + size/2 && savey < topy + size/2) {
// Recursively tile upper left quadrant.
tileRec(size/2, topx, topy);
// Fill in middle tromino
grid[topx+size/2][topy+size/2-1] = currentNum;
grid[topx+size/2][topy+size/2] = currentNum;
grid[topx+size/2-1][topy+size/2] = currentNum;
// Advance to the next tromino
currentNum++;
// Now we can make our three other recursive calls.
tileRec(size/2, topx, topy+size/2);
tileRec(size/2, topx+size/2, topy);
tileRec(size/2, topx+size/2, topy+size/2);
}
// Hole in upper right quadrant
else if (savex < topx + size/2 && savey >= topy + size/2) {
// Recursively tile upper right quadrant.
tileRec(size/2, topx, topy+size/2);
// Fill in middle tromino
grid[topx+size/2][topy+size/2-1] = currentNum;
grid[topx+size/2][topy+size/2] = currentNum;
grid[topx+size/2-1][topy+size/2-1] = currentNum;
// Advance to the next tromino
currentNum++;
// Now we can make our three other recursive calls.
tileRec(size/2, topx, topy);
tileRec(size/2, topx+size/2, topy);
tileRec(size/2, topx+size/2, topy+size/2);
}
// Hole in bottom left quadrant
else if (savex >= topx + size/2 && savey < topy + size/2) {
// Recursively tile bottom left quadrant.
tileRec(size/2, topx+size/2, topy);
// Fill in middle tromino
grid[topx+size/2-1][topy+size/2] = currentNum;
grid[topx+size/2][topy+size/2] = currentNum;
grid[topx+size/2-1][topy+size/2-1] = currentNum;
// Advance to the next tromino
currentNum++;
// Now we can make our three other recursive calls.
tileRec(size/2, topx, topy);
tileRec(size/2, topx, topy+size/2);
tileRec(size/2, topx+size/2, topy+size/2);
}
else {
// Recursively tile bottom right quadrant.
tileRec(size/2, topx+size/2, topy+size/2);
// Fill in middle tromino
grid[topx+size/2-1][topy+size/2] = currentNum;
grid[topx+size/2][topy+size/2-1] = currentNum;
grid[topx+size/2-1][topy+size/2-1] = currentNum;
// Advance to the next tromino
currentNum++;
// Now we can make our three other recursive calls.
tileRec(size/2, topx+size/2, topy);
tileRec(size/2, topx, topy+size/2);
tileRec(size/2, topx, topy);
}
} // end large if-else
} // end tileRec
// Prints out the current object.
public void print() {
for (int i=0; i<grid.length; i++) {
for (int j=0; j<grid[i].length; j++)
System.out.print(grid[i][j] + "\t");
System.out.println();
}
}
public static int checkArg(int ch){
String str = "";
char[] all = str.toCharArray();
for(int i = 0; i < all.length;i++) {
if(!Character.isDigit(all[i])) {
System.out.println("Please retype a correct value");
}
}
return ch;
}
public static void main(String[] args) {
Scanner stdin = new Scanner(System.in);
// Get user input...
int size = stdin.nextInt();
int x = stdin.nextInt();
int y = stdin.nextInt();
if(checkArg(size) != size){
checkArg(size);
} else if (x != x){
System.out.println("Please retype a correct value");
} else if (y != y){
System.out.println("Please retype a correct value");
}
// Create our object and tile it!
tromino thisguy = new tromino(size, x, y);
//tile grid with trominos
thisguy.tile();
// Print romino grid.
thisguy.print();
}
}
我如何获得checkArg方法来检查用户是否输入了3个以上的参数，或者不是是否为整数参数？
答案 0 :(得分：0)
你可以这样做。让该方法返回一个布尔值。然后检查main中的args以获取boolean响应。并保持循环，直到方法返回true
public static boolean checkArg(String line){
String[] args = line.split("\\s+");
if (args.length != 3) {
return false;
}
try {
int num1 = Integer.parseInt(args[0].trim());
int num1 = Integer.parseInt(arss[1].trim());
int num3 = Integer.parseInt(args[2].trim());
} catch (Exception ex) {
System.out.println("Must enter 3 integers");
return false;
}
return true;
}
在main
public static void main(String[] args) {
Scanner stdin = new Scanner(System.in);
String input = stdin.nextLine();
while (!checkArgs(input)) {
input = stdin.nextLine();
}
String[] args = input.split("\\s+");
int size = Integer.parseInt(args[0].trim());
int x = Integer.parseInt(arss[1].trim());
int y = Integer.parseInt(args[2].trim());
// Create our object and tile it!
tromino thisguy = new tromino(size, x, y);
//tile grid with trominos
thisguy.tile();
// Print romino grid.
thisguy.print();
}