编写一个程序来读取两个数字，第一个小于第二个数字。该计划将写入所有号码的两个数字

Question

//所以我写这之间：（输出那里）

import java.util.Scanner; 
public class E7L6{ 
public static void main(String[]args){ 
int num1, num2; 
Scanner keyboard= new Scanner(System.in); 
System.out.print("Type two numbers:"); 
num1= keyboard.nextInt(); 
num2= keyboard.nextInt(); 

if(num1<num2){ 
    while(num1<=num2){ 
    int counter=num1; 
    System.out.print(counter+" "+num2); 
    counter=counter+1; 
    }} 
else{ 
    System.out.print("Error: the first number must be smaller than the second"); 
} 
    }} 

Output: 
----jGRASP exec: java E7L6 
Type two numbers:4 124 
4 4 4 4 4 4 4 4 4 
----jGRASP: process ended by user. 

//这么多数量的4重复谁能告诉我在哪里，我错了？谢谢！！

Answer 1

while(num1<=num2){ 
    int counter=num1; 
    System.out.print(counter+" "+num2); 
    counter=counter+1; 
    } 

这将以无限循环结束。 while循环的条件需要修改。

//修改后的代码 - 完整的代码。

import java.util.Scanner; 

public class E7L6 { 
    public static void main(String[] args) { 
     int num1, num2; 
     Scanner keyboard = new Scanner(System.in); 
     System.out.print("Type two numbers:"); 
     num1 = keyboard.nextInt(); 
     num2 = keyboard.nextInt(); 

     if (num1 < num2) { 
      int counter = num1; 
      while (counter <= num2) { 
       System.out.print(counter + " "); 
       counter = counter + 1; 
      } 
     } else { 
      System.out 
        .print("Error: the first number must be smaller than the second"); 
     } 
    } 
} 

Answer 2

int counter=num1; 

您创建了一个新的counter变量循环的每个迭代。
因此，它们都具有相同的价值。

它永远运行，因为您的循环条件永远不会是错误的（您永远不会更改num1或num2）。

Answer 3

作为一个新手程序员，我会用循环来代替。它会更简单，更轻松：

for(ctr=num1;ctr<=num2;ctr++) 
    s.o.p(ctr) 

Answer 4

在这里你去：

System.out.print("First:"); 
    int first = Integer.parseInt(reader.nextLine()); 
    System.out.print("Second:"); 
    int second = Integer.parseInt(reader.nextLine()); 
    while (first <= second) { 
     System.out.println(first); 
     first++; 
    }