import java.util.Scanner;
class Main {
static Scanner userInput = new Scanner(System.in);
public static void main(String[] args) {
int testNumber = userInput.nextInt();
do{
System.out.println(newNumber * 2);
newNumber++;
}while( testNumber < 1000000);
}
}
答案 0 :(得分:3)
将数字乘以2后,您需要更新数字。
newNumber = newNumber * 2;
System.out.println(newNumber);
您还使用newNumber
和testNumber
,而newNumber
似乎没有在任何地方定义...
}while( ***testNumber***newNumber*** < 1000000);
您需要选择一个,因为如果要更新newNumber
但在循环中比较testNumber
,则会创建一个无限循环。
除非您在发布后遗漏了某些内容,否则显示的代码不应编译。
答案 1 :(得分:1)
您对循环有正确的想法,但是变量有多个问题。
<
,但是(错误地)您正在操纵一个完全不同变量-testNumber
。 / li>
您可能希望循环如下:
newNumber
答案 2 :(得分:1)
您还可以为其创建递归方法。
public int reachMillion(int num) {
if(num<=0)
return -1; // indicating it is not possible.
if(num>=1000000) // Base Condition denoting we have reached 1 million
return num;
return reachMillion(num*2); // recursive part to multiply by 2 until we reach 1 million
}
答案 3 :(得分:1)
class Main {
private static Scanner userInput = new Scanner(System.in);
public static void main(String[] args) {
int newNumber = 0;
do{
System.out.println("Enter a positive number: ");
try{
newNumber = userInput.nextInt();
}catch(Exception ignored){ }
System.out.println("");
}while(newNumber <= 0);
System.out.println("----- " + newNumber + " multiply by 2 ------");
while(newNumber <= 1_000_000){
System.out.print("2 * " + newNumber +" = ");
newNumber <<= 1;//in some compilers left shift is faster than multiply
System.out.println(newNumber);
}
}
答案 4 :(得分:1)
@brso05 has done well描述这里出了什么问题。我想提供一个完整的示例:
import java.util.Scanner;
public class Main {
private static Scanner userInputScanner = new Scanner(System.in);
public static void main(String[] args) {
System.out.print("Please input a number: ");
int userInputNumber = userInputScanner.nextInt();
System.out.println();
int newNumber = userInputNumber;
while (newNumber < 1_000_000) {
newNumber *= 2; // Take the variable on the left, multiply it by the number on the right, and save it in the variable on the left
System.out.println(newNumber);
}
}
}
当心!!该代码不会处理任何错误的用户输入。例如,如果给它0
,它将永远循环;如果给它foo
,它将崩溃。如果您想处理用户输入的所有极端情况,可以这样做:
import java.util.*;
public class Main {
private static Scanner userInputScanner = new Scanner(System.in);
public static void main(String[] args) {
int userInputNumber;
//
while(true) {
System.out.print("Please input a number: ");
if (userInputScanner.hasNext()) {
// The user gave us something, but we don't know if it's a number
String rawUserInput = userInputScanner.next();
try {
userInputNumber = Integer.parseInt(rawUserInput);
// If that previous line runs, the user has given us an integer!
System.out.println();
if (userInputNumber > 0) {
// The user has given a valid number. Break out of the loop and multiply it!
break;
}
else {
// The user has given a bad number. Tell them why and ask again.
System.out.println("The number has to be greater than 0.");
}
}
catch (NumberFormatException exception) {
// The user has given us something, but it wasn't an integer
System.out.println();
System.out.println("That is not a number: " + exception.getMessage());
}
}
else {
// There is no input, so we can't do anything.
return;
}
}
// Done looping through user input
int newNumber = userInputNumber;
while (newNumber < 1_000_000) {
newNumber *= 2; // Take the variable on the left, multiply it by the number on the right, and save it in the variable on the left
System.out.println(newNumber);
}
}
}
答案 5 :(得分:0)
do-while
循环中有一个棘手的部分。在这种类型的循环中,首先执行do
部分。对于下面的示例,尽管输入已经大于1000000,但它会显示1000001。
public void doWhileLoop() {
int num = 1000001;
do {
System.out.println(num);
num *= 2;
} while (num < 1000000);
}
因此,在do-while循环中执行某些操作之前,最好使用一些保护性子句(也称为if-statements
)。喜欢,
public void doWhileLoop() {
int num = 1000001;
if(num >= 1000000) {
return;
}
do {
System.out.println(num);
num *= 2;
} while (num < 1000000);
}