Question

我需要使用方法

    public static String readString() { 

    return input.nextLine(); 
}

为了使用Scanner方法next.Line（），它只是项目的一个要求，那就是当我使用静态方法ProjectUtils.readString（）将用户输入收集为字符串时它只是抛出我犯了很多错误，我无法理解为什么，我这样使用它

       public static void encryptAString() {

        ProjectUtils.println("Enter the phrase and the key");
        String p = ProjectUtils.readString();
        ProjectUtils.println("Enter key");
        int k = ProjectUtils.readInteger();
        ProjectUtils.println(EncryptString.encryptString(p,k ));

但总是会终止并抛出这个

Exception in thread "main" java.util.InputMismatchException
at java.util.Scanner.throwFor(Unknown Source)
at java.util.Scanner.next(Unknown Source)
at java.util.Scanner.nextInt(Unknown Source)
at java.util.Scanner.nextInt(Unknown Source)
at ProjectUtils.readInteger(ProjectUtils.java:24)
at ProjectUtils.encryptAString(ProjectUtils.java:88)
at StringMenuManager.run(StringMenuManager.java:37)
at ProjectUtils.operationsOnStrings(ProjectUtils.java:69)
at MainMenuManager.run(MainMenuManager.java:47)
at P1Main.main(P1Main.java:9)

提前致谢。

MCVE

主要课程

public class Main {

    public static void main(String[] args) {
        MCVE.encryptAString();
    }
}

原始项目中的MCVE方法或ProjectUtils类

import java.util.Scanner;





public class MCVE {

private static final Scanner input = new Scanner(System.in);

public static void println(String s) { 
    System.out.println(s); 
}
public static int readInteger() { 
    // for the moment, just assume that the input is a
    // valid integer.... but eventually we want to be 
    // more robust and explicitly read and test first if
    // the input was really an integer or not....
    return input.nextInt(); 
}
public static String readString() { 
    // for the moment, just assume that the input is a
    // valid String.... but eventually we want to be 
    // more robust and explicitly read and test first if
    // the input was really an integer or not....
    return input.nextLine(); 
}

public static void encryptAString() {

    MCVE.println("Enter the phrase and the key");
    String p = MCVE.readString();
    int k = MCVE.readInteger();
    MCVE.println(EncryptString.encryptString(p,k ));

     }
   }

加密类

 public class EncryptString {
 private static final int ALENGTH = 26; 

/**
 * Encrypt a string using a given key value. 
 * @param s the String to be encrypted
 * @param key the key 
 * @return the encrypted string
 */
public static String encryptString(String s, int key) {
    // if the key value is not in the accepted range (-25..25)
    // the encrypted string is the same as the input string
    if (key < -25 || key > 25) 
        return s; 

    // the key is valid, construct the encrypted string as 
    // described in P1 specs...
    String newString = "";
    for(int x = 0; x<s.length(); x++)
    {
        if(((int)s.charAt(x)>64&&(int)s.charAt(x)<123))
            newString = newString + encryptChar(s.charAt(x), key);
        else
            newString = newString + s.charAt(x);

    }
    return newString;
}


/**
 * Encrypt a particular character.
 * @param ch the character to encrypt - assumed to be a letter ‘a’..’z’ or ‘A’..’Z’
 * @param key the key to be used. It is assumed to be a value in range (-25..25)
 * @return the new character after encryption
 */
private static char encryptChar(char ch, int key) {
    // PRE: ch is a letter 'A'..'Z' or 'a'..'z'
    // PRE: key is an integer in the range -25..25

    int base; 
    if (Character.isUpperCase(ch))
        base = (int) 'A'; 
    else 
        base = (int) 'a'; 
    return (char) (Math.abs((((int) ch - base) + key + ALENGTH) 
                                % ALENGTH) + base); 
}

}

Answer 1

虽然我不确定确切的潜在问题，但您的输入中似乎有一个新行不应该存在。它导致第一个typedef void(^blockTakingInt)(int); // Returns a block blockTakingInt f() { return ^(int i) { printf("i = %d\n", i); }; } // Accepts a block as a parameter void g(int i, blockTakingInt b) { b(i); } // Store the block returned by the function f in b, and pass it to g void (^b)(int) = f(); g(4, b);读取（大概）空行。

临时解决方法是通过调用input.nextLine()并忽略返回值来使用此换行符，然后通过另一个readString()调用读取实际的String输入。

调用Scanner的nextLine

1 个答案: