在java中添加问题
我想把两个变量的输入作为整数,然后对两个变量进行加法。 ans将存储在另一个变量中。该程序将在每次添加结束后重复,并要求用户输入变量,然后再次进行添加在java中添加问题,java,addition,Java,Addition,我想把两个变量的输入作为整数,然后对两个变量进行加法。 ans将存储在另一个变量中。该程序将在每次添加结束后重复,并要求用户输入变量,然后再次进行添加 public void add() { Scanner keyboard = new Scanner(System.in); int a; int b; int total = 0; for(int i = 0; i < 2; i++) { System.out.print("\n
public void add() {
Scanner keyboard = new Scanner(System.in);
int a;
int b;
int total = 0;
for(int i = 0; i < 2; i++) {
System.out.print("\nEnter a : ");
a = keyboard.nextInt();
System.out.print("Enter b : ");
b = keyboard.nextInt();
int c = a+b;
total += c;
}
System.out.println("\nans is :"+total);
}
我的qus是taht,我如何再次添加所有ADION ans:
public void add() {
Scanner keyboard = new Scanner(System.in);
int a;
int b;
int total = 0;
for(int i = 0; i < 2; i++) {
System.out.print("\nEnter a : ");
a = keyboard.nextInt();
System.out.print("Enter b : ");
b = keyboard.nextInt();
int c = a+b;
total += c;
}
System.out.println("\nans is :"+total);
}
Exm:
public void add() {
Scanner keyboard = new Scanner(System.in);
int a;
int b;
int total = 0;
for(int i = 0; i < 2; i++) {
System.out.print("\nEnter a : ");
a = keyboard.nextInt();
System.out.print("Enter b : ");
b = keyboard.nextInt();
int c = a+b;
total += c;
}
System.out.println("\nans is :"+total);
}
Agin程序将要求
Input a= 6
Input b= 6
ans=12
public void add() {
Scanner keyboard = new Scanner(System.in);
int a;
int b;
int total = 0;
for(int i = 0; i < 2; i++) {
System.out.print("\nEnter a : ");
a = keyboard.nextInt();
System.out.print("Enter b : ");
b = keyboard.nextInt();
int c = a+b;
total += c;
}
System.out.println("\nans is :"+total);
}
现在,我如何通过程序获取所有“ans”值并添加所有“ans”
public void add() {
Scanner keyboard = new Scanner(System.in);
int a;
int b;
int total = 0;
for(int i = 0; i < 2; i++) {
System.out.print("\nEnter a : ");
a = keyboard.nextInt();
System.out.print("Enter b : ");
b = keyboard.nextInt();
int c = a+b;
total += c;
}
System.out.println("\nans is :"+total);
}
代码:
public void add() {
Scanner keyboard = new Scanner(System.in);
int a;
int b;
int total = 0;
for(int i = 0; i < 2; i++) {
System.out.print("\nEnter a : ");
a = keyboard.nextInt();
System.out.print("Enter b : ");
b = keyboard.nextInt();
int c = a+b;
total += c;
}
System.out.println("\nans is :"+total);
}
代码只是一个接一个地做加法,但我希望它能再做一个
public void add() {
Scanner keyboard = new Scanner(System.in);
int a;
int b;
int total = 0;
for(int i = 0; i < 2; i++) {
System.out.print("\nEnter a : ");
a = keyboard.nextInt();
System.out.print("Enter b : ");
b = keyboard.nextInt();
int c = a+b;
total += c;
}
System.out.println("\nans is :"+total);
}
这一切都加上了所有的理由。我该怎么做呢?只需不断地将中间和添加到另一个变量中,就可以得到最终的总数
public void add() {
Scanner keyboard = new Scanner(System.in);
int a;
int b;
int total = 0;
for(int i = 0; i < 2; i++) {
System.out.print("\nEnter a : ");
a = keyboard.nextInt();
System.out.print("Enter b : ");
b = keyboard.nextInt();
int c = a+b;
total += c;
}
System.out.println("\nans is :"+total);
}
另一种选择(虽然在简单的解决方案可用时并不可取)是将这些总和放入一个列表中,最后迭代列表并计算最终的总数
public void add() {
Scanner keyboard = new Scanner(System.in);
int a;
int b;
int total = 0;
for(int i = 0; i < 2; i++) {
System.out.print("\nEnter a : ");
a = keyboard.nextInt();
System.out.print("Enter b : ");
b = keyboard.nextInt();
int c = a+b;
total += c;
}
System.out.println("\nans is :"+total);
}
另外,不要使用ob_m.add();-只需调用add() 将每个答案存储在附加变量中。完成后,对所有答案变量求和。您可以使用循环重复操作,并将每个加法存储在一个总和中。有一个字段
GrandSum = 0;
public void add() {
Scanner keyboard = new Scanner(System.in);
int a;
int b;
int total = 0;
for(int i = 0; i < 2; i++) {
System.out.print("\nEnter a : ");
a = keyboard.nextInt();
System.out.print("Enter b : ");
b = keyboard.nextInt();
int c = a+b;
total += c;
}
System.out.println("\nans is :"+total);
}
在每次添加之后,添加ans
GrandSum += ans;
public void add() {
Scanner keyboard = new Scanner(System.in);
int a;
int b;
int total = 0;
for(int i = 0; i < 2; i++) {
System.out.print("\nEnter a : ");
a = keyboard.nextInt();
System.out.print("Enter b : ");
b = keyboard.nextInt();
int c = a+b;
total += c;
}
System.out.println("\nans is :"+total);
}
最后,Grandum会有你想要的结果
public void add() {
Scanner keyboard = new Scanner(System.in);
int a;
int b;
int total = 0;
for(int i = 0; i < 2; i++) {
System.out.print("\nEnter a : ");
a = keyboard.nextInt();
System.out.print("Enter b : ");
b = keyboard.nextInt();
int c = a+b;
total += c;
}
System.out.println("\nans is :"+total);
}
编辑:
public void add() {
Scanner keyboard = new Scanner(System.in);
int a;
int b;
int total = 0;
for(int i = 0; i < 2; i++) {
System.out.print("\nEnter a : ");
a = keyboard.nextInt();
System.out.print("Enter b : ");
b = keyboard.nextInt();
int c = a+b;
total += c;
}
System.out.println("\nans is :"+total);
}
改变
public void add() {
Scanner keyboard = new Scanner(System.in);
int a;
int b;
int total = 0;
for(int i = 0; i < 2; i++) {
System.out.print("\nEnter a : ");
a = keyboard.nextInt();
System.out.print("Enter b : ");
b = keyboard.nextInt();
int c = a+b;
total += c;
}
System.out.println("\nans is :"+total);
}
public void add(){
扫描仪键盘=新扫描仪(System.in);
INTA;
int b;
int-total=0;
对于(int i=0;i<2;i++){
系统输出打印(“\n输入a:”);
a=键盘.nextInt();
系统输出打印(“输入b:”);
b=键盘.nextInt();
INTC=a+b;
总数+=c;
}
System.out.println(“\n为:“+总计”);
}
有一个total
变量,您只需不断向其中添加c
public void add() {
Scanner keyboard = new Scanner(System.in);
int a;
int b;
int total = 0;
for(int i = 0; i < 2; i++) {
System.out.print("\nEnter a : ");
a = keyboard.nextInt();
System.out.print("Enter b : ");
b = keyboard.nextInt();
int c = a+b;
total += c;
}
System.out.println("\nans is :"+total);
}
我还将不需要的递归更改为while循环
public void add() {
Scanner keyboard = new Scanner(System.in);
int a;
int b;
int total = 0;
for(int i = 0; i < 2; i++) {
System.out.print("\nEnter a : ");
a = keyboard.nextInt();
System.out.print("Enter b : ");
b = keyboard.nextInt();
int c = a+b;
total += c;
}
System.out.println("\nans is :"+total);
}
public class math
{
public static void main(String args[])
{
int total = 0;
Scanner keyboard = new Scanner(System.in);
while (true)
{
System.out.print("\nEnter a (-999 to quit): ");
int a = keyboard.nextInt();
// terminating condition, modify appropriately
if (a == -999)
break; // break out of the while-loop
System.out.print("Enter b: ");
int b = keyboard.nextInt();
int c = a + b;
total += c;
System.out.println("\nans is: " + c);
}
System.out.println("total is: " + total);
}
}
package-farzi;
导入java.util.ArrayList;
导入java.util.Scanner;
公共类虚拟{
公共静态void main(字符串[]args){
扫描仪键盘=新扫描仪(System.in);
ArrayList list1=新的ArrayList();
ArrayList list2=新的ArrayList();
ArrayList sum=新的ArrayList();
字符串选择=”;
做{
System.out.println(“输入第一个数字”);
int a=键盘.nextInt();
System.out.println(“输入第二个数字”);
int b=keyboard.nextInt();
int tempSum=a+b;
清单1.添加(a);
清单2.添加(b);
相加(tempSum);
System.out.println(“是否继续:键入是或否”);
选择=键盘。下一步();
}while(choice.toLowerCase().charAt(0)='y');
System.out.println(“这里是带总和的输入”);
System.out.println(“num1\t num2\t sum”);
对于(int i=0;i当前您的代码只被调用一次-您可以使用一个循环-例如while循环或do while循环进行迭代,直到满足某个条件-例如,如果用户输入0作为输入,为什么您要在add方法中创建一个新的math
对象,然后再次调用add?您将永远循环添加相同的值然后一次又一次地刷新…@Katana24递归调用函数(此处-ob_m.add();
),因此它不只是调用一次。aahh-没有看到-good eyes@DukelingDefine实例变量,让我们称它为“total”。继续向其中添加sum-total+=c;就是这样。有人能举例说明如何为此任务创建循环吗?这行不通……每次添加时都要创建新对象,这样每次都会重置Grandum。多次调用同一对象上的add()将为该特定对象提供正确的Grandum。