Question

目前，我正在尝试在我正在创建的项目中执行Caesar Cipher。但是，当我尝试将字符串传递给处理它的实例时，它似乎根本不处理它。（现在我忽略了空格和标点符号）。

import javax.swing.*;
import java.text.*;
import java.util.*;
import java.lang.*;

public class Cipher {

private String phrase; // phrase that will be encrypted 
private int shift; //number that shifts the letters


///////////////
//Constructor//
//////////////

public Cipher( int new_shift)
{

    shift = new_shift;



}//end of cipher constructor


////////////
//Accessor//
////////////

public int askShift() {


return shift;
}//end of askShift accessor

////////////
//mutators//
////////////

public void changeShift (int newShift) {

shift = newShift;

}//end of changeShift mutator

/////////////
//instances//
/////////////

public String encryptIt(String message) {

char[] charArray = message.toCharArray(); //converts to a character array
//loop that performs the encryption
for (int count = 0; count < charArray.length; count++) {
int shiftNum = 2;
charArray[count] = (char)(((charArray[count] - 'a') + shiftNum) % 26 + 'a');

} // end of for loop    

 message = new String(charArray); //converts the array to a string



return message;
}//end of encrypt instance 


//////////
///Main///
//////////
public static void main(String[] args) {

Cipher cipher = new Cipher(1); //cipher with a shift of one letter
String phrase = JOptionPane.showInputDialog(null, "Enter phrase to be messed with ");
cipher.encryptIt(phrase);
JOptionPane.showMessageDialog(null, phrase);




}//end of main function



} //end of cipher class

Answer 1

您需要再次将返回值分配给phrase。

phrase=cipher.encryptIt(phrase);

Answer 2

您的新加密字符串是返回值。您传入方法的字符串不变。试试例子

String encryption = cipher.encryptIt(phrase); 
JOptionPane.showMessageDialog(null, encryption );

Answer 3

您必须更改此行

cipher.encryptIt(phrase);

到

phrase = cipher.encryptIt(phrase);

为了更改phrase变量的值。

这是因为Java传递了所有参数的值。这意味着，当您通过方法发送变量时，您不会发送实际引用，而是发送引用的副本。

Answer 4

您应该知道的一件主要事情：方法参数是Java中的pass-by-value

简而言之：

public void foo(Bar bar) {
    bar = new Bar(999);
}
public void someMethod() {
    Bar b = new Bar(1);
    foo(b);
    // b is still pointing to Bar(1)
}

因此，您的message = new String(charArray);不会影响传递给encryptIt()

的参数

Java新手试图将字符串传递给实例java

4 个答案: