英文:
java.util.InputMismatchException scanner issue?
问题
我正在尝试通过键盘扫描器收集用户输入，并验证输入是否与存储的值匹配。因此，如果输入是 2927942074l，应该显示行 Correct，但实际上我收到了以下错误。此外，如果输入与 PIN 不匹配，则会显示行 Wrong。不确定我做错了什么。
import java.util.Scanner;class app {public static void main(String[] args){long pin = 2927942074l;System.out.println("Please enter your pin.");Scanner keyboard = new Scanner(System.in);long input = keyboard.nextLong();if (input != pin)System.out.println("Wrong");if (input == pin)System.out.println("Correct");}}
英文:
I am trying to collect user input via the keyboard scanner and verifying if the input matches the stored value. So if the input is 2927942074l the line Correct should appear however I get the following error. Also if the input doesn't match with the pin the line Wrong appears. Not sure what I am doing wrong.
Exception in thread "main" java.util.InputMismatchExceptionat java.base/java.util.Scanner.throwFor(Scanner.java:939)at java.base/java.util.Scanner.next(Scanner.java:1594)at java.base/java.util.Scanner.nextLong(Scanner.java:2373)at java.base/java.util.Scanner.nextLong(Scanner.java:2328)at app.main(app.java:11)
import java.util.Scanner;class app {public static void main(String[] args){long pin = 2927942074l;System.out.println("Please enter your pin.");Scanner keyboard = new Scanner(System.in);long input = keyboard.nextLong();if (input != pin)System.out.println("Wrong");if (input == pin)System.out.println("Correct");}}
答案1
得分: 0
你的代码是正确的，并且按照要求工作。
你需要输入的是 2927942074，而不是 2927942074l。
英文:
Your code is correct and works as per requirement.
The thing that you need to input is 2927942074 not 2927942074l
答案2
得分: 0
Type just the number `2927942074` without trailing 'l'. Else it will treat it as a String.Also you may want to add a condition using keyboard.hasNextLong() if you dont want to take String inputs.if (keyboard.hasNextLong()) {long input = keyboard.nextLong();if (input != pin)System.out.println("Wrong");if (input == pin)System.out.println("Correct");} else {System.out.println("Enter a valid pin");}
英文:
Type just the number 2927942074 without trailing 'l'. Else it will treat it as a String.
Also you may want to add a condition using keyboard.hasNextLong() if you dont want to take String inputs.
if (keyboard.hasNextLong()) {long input = keyboard.nextLong();if (input != pin)System.out.println("Wrong");if (input == pin)System.out.println("Correct");} else {System.out.println("Enter a valid pin");}
答案3
得分: 0
你必须输入
2927942074
而不是
2927942074l
“L” 只是为了让 Java 编译器理解这些数字表示的是长整型而不是整型（整型是默认的）。
英文:
You have to input
2927942074
not
2927942074l
The "L" is only for the Java Compiler to understand, that die digits resemble a long and not an int (which would be default).
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