为什么在输出过程中字符值会发生变化？

Question

我正在制作我的程序，为JavaSE的学校作业扔一个骰子（如骰子）。用户可以将一个字符作为标准输入，因此用户选择的字符将代表骰子的眼睛。有时当我打印结果时，它会显示一个完全不同的字符。

package ThrowADie;
import java.util.Scanner;

public class ThrowADie {

public static void main(String[] args) {

    //Ask user for the char in which the dices eyes should be printed in.
    System.out.print("Which character should I use for the eye: ");

    //Allow user to place input in the eye variable
    Scanner input = new Scanner(System.in); //Make the stdinput object
    char eye = input.next().charAt(0);

    //Time to throw the die. Place result in dieResult
    int dieResult = throwDie();

    //Reveal of the result
    printDieResult(dieResult, eye);

}


    /*
    * Method name: throwDie()
    * Purpose: Picks a number from 1 to 6 randomly, like a die does
    * Parameters: N/A
    * Returns: Integer number from 1 to 6    
    */
    public static int throwDie(){
        int range = (6 - 1) + 1;     
        return (int)(Math.random() * range) + 1;
    }


    /*
    * Method name: printDieResult()
    * Purpose: Generate result of the die throw in ASCII art
    * Parameters: numberOfEyes, typeOfEyes
    * Returns: N/A
    */
    public static void printDieResult(int numberOfEyes, char typeOfEyes){
        if (numberOfEyes == 1){
            //Print art
            System.out.println(
                    " " + " " + " \n"
                  + " " + typeOfEyes + " \n"
                  + " " + " " + " ");
        } else if (numberOfEyes == 2){
            //Print art
            System.out.println(
                    typeOfEyes + " " + " \n"
                  + " " + " " + " \n"
                  + " " + " " + typeOfEyes);
        } else if (numberOfEyes == 3){
            //Print art
            System.out.println(
                    typeOfEyes + " " + " \n"
                  + " " + typeOfEyes + " \n"
                  + " " + " " + typeOfEyes);
        } else if (numberOfEyes == 4){
            //Print art
            System.out.println(
                    typeOfEyes + " " + typeOfEyes + "\n"
                  + " " + " " + " \n"
                  + typeOfEyes + " " + typeOfEyes);
        } else if (numberOfEyes == 5){
            //Print art
            System.out.println(
                    typeOfEyes + " " + typeOfEyes + "\n"
                  + " " + typeOfEyes + " \n"
                  + typeOfEyes + " " + typeOfEyes);
        } else {
            //Print art
            //Accidentally written down 9 eye representation
            System.out.println(
                    typeOfEyes + typeOfEyes + typeOfEyes + "\n"
                  + typeOfEyes + typeOfEyes + typeOfEyes + "\n"
                  + typeOfEyes + typeOfEyes + typeOfEyes);
        }
    }
}

输出此程序将生成正确的结果。但偶尔输入的字符（代表模具的眼睛）会转换为数字。

在以下情况下，程序应打印9个“@”字符。相反，它在第一行打印192。（我知道骰子有6只眼睛，但我在无意中打印了9只眼睛时遇到了这个奇怪的输出）

run:
Which character should I use for the eyes: @
192
@@@
@@@

我找不到原因，有人能看出我做错了什么吗？

Answer 1

int test = (int) (typeOfEyes + typeOfEyes + typeOfEyes);

        System.out.println("\n" + test + "\n"
                + typeOfEyes + "" + typeOfEyes + "" + typeOfEyes + "\n"
                + typeOfEyes + "" + typeOfEyes + "" + typeOfEyes + "\n"
                + typeOfEyes + "" + typeOfEyes + "" + typeOfEyes);

Which character should I use for the eye: @

192
@@@
@@@
@@@

Answer 2

字符算术！

请参阅此表<代码>@是字符<代码>64

typeofees-typeofees-typeofees“\n”

第一行实际上是将字符（64 64 64) = 192的值相加，然后用换行符将其相加，因此我们得到192\n。

编译器选择将它们相加而不是创建字符串。解决这个问题的简单方法是在前面加上一个空字符串："typeOfEyes...

基本上，编译器是“愚蠢的”。当我们将整数添加到String时，"foo"123编译器可以将其解释为fo123很好，因为它将第一个元素识别为String。但是，我们定义了一个char，它是表示字符的数字类型。所以编译器会对其进行数学运算。即使我们不应该这样做。添加字符串文字告诉它我们实际上需要文本。