Java-如何将Char []数组与字符串进行比较？

Question

I have a question regarding the comparison of the characters within an Array with a specific input which is a String.

I seem to have a problem at line where it states: String[] oneRow = alphabet[i];

package MemoryGame;

import java.util.Scanner;

public class Memory {

/*
 * array list of the alphabet.
 */

char[] alphabet = {'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};

/*
 * Constructor
 */
public Memory(){

    /*
     * Scanner initiation for input end-user.
     */
    System.out.println("What is the first letter of you name? ");
    Scanner scan = new Scanner(System.in);
    String letter = scan.nextLine();
    String s = scan.nextLine();

    /*
     * Comparing Input with array
     */
    String[] matchedRow;
    for(int i=0; i<alphabet.length; i++)
    {           
        String[] oneRow = alphabet[i];

        if(oneRow[0].equals(letter))
        {
            matchedRow = oneRow;
            break;
        }
    }

    for(int i=0; i<matchedRow.length; i++)
    {
        System.out.println(matchedRow[i]);

    }
}

public static void main(String[] args) {
Memory memory = new Memory();   
}

Eclipse says:

Type mismatch: cannot convert from char to String[]

Thanks for your help!

M

Answer 1

You can't assign a single character to a String array. You couldn't assign a single String to an array, either; that part has nothing to do with chars.

Of course, you can't assign a char to a String directly, either, so you probably want something like this:

String onerow = String.valueOf(alphabet[i]);

if(oneRow.equals(letter))

Answer 2

The assignment of a character to a String is invalid. You can use a concatenation trick to do it anyway, but that borders on a hack:

String[] oneLetter = ""+alphabet[i];

A better approach would be to check that letter has the length of 1 , and that its only character matches alphabet[i] :

if (letter.length() == 1 && letter.charAt(0) == alphabet[i]) {
    matchedLetter = alphabet[i];
    break;
}

Answer 3

char is incompatible with String and so String[] is not compatible with char[] .

alphabet[i] will return a char , you can transform it into a String by constructing a new one, for example:

char c = aplhabet[i];
String oneRow = String.valueOf(c);
if (oneRow.equals(letter)) { ... }

Or, you can compare the first char of the input String with the char you took from alphabet :

char c = aplhabet[i];
if (letter.chartAt(0) == c) { ... }