查找a和b含之间所有奇数的总和，其中a和b均为用户输入我的set_1有效，但set_2无效-为什么？

如何解决查找a和b含之间所有奇数的总和，其中a和b均为用户输入我的set_1有效，但set_2无效-为什么？

注意：两组扫描仪均已导入

在主要方法中：

代码集_1的开始...

    Scanner input = new Scanner(system.in); //scanner for user input
    System.out.print("Enter number,a: "); //prompt for user to enter number for variable,a
    int numA = input.nextInt(); //variable for 1st number,a
    int a = numA; //variable created to track 1st number entered - will be used to print later

    System.out.print("Enter number,b: "); //prompt for user to enter number for variable,b
    int numB = input.nextInt(); //variable for 2nd number,b


    int oddSum = 0; //variable to hold calculation of total odd sum
    //to loop between two numbers
    while (numA <= numB){
        //> 1 for odd #s to enter conditional
        if ((numA % 2) > 0){
            oddSum += numA; //variable,oddSum will store all added numbers between the 2 entered #s
        }
        numA++; //numA is counted +1
    }
    System.out.println("The sum of all odd numbers between " + a + " and " + numB + " is " + oddSum);

...代码set_1的结尾==>可以100％工作

代码集_2的开头...

    Scanner input = new Scanner(system.in); //scanner for user input
    System.out.print("Enter number,a                             
    System.out.print("Enter number,b


    int oddSum = 0; //variable to hold calculation of total odd sum
    int numerBetweenAandB = numA; //variable created to track 1st # entered and count up
    //to loop between two numbers
    while (numA <= numB){
        //> 1 for odd #s to enter conditional
        if ((numA % 2) > 0){
            oddSum += numberBetweenAandB; //variable,will store all added numbers between the 2 #s
        }
        numberBetweenAandB++; // counted +1
    }
    System.out.println("The sum of all odd numbers between " + numA + " and " + numB + " is " + oddSum);

...代码set_2的结尾==>这会陷入无限循环。为什么？

在这里我的逻辑有什么问题？据我了解-第一个数字存储在numberbetweenAandB中，并且使用++计数器进行递增，直到循环结束为止，此时，时间变量sumSum已将a和b之间的所有＃都加了。

感谢您为此花费的时间和精力。我还在学习。

解决方法

在您的代码中，您具有：

while (numA <= numB){
  // a bunch of code; and none of it changes numA or numB
}

显然，如果循环进入一次，它将永远进入。 numA和numB不会更改，因此，如果一开始是正确的，那么它将永远是真实的。