为什么输出过程中 char 值会发生变化？答案

【问题标题】：Why is the char value transforming during the output?为什么输出过程中 char 值会发生变化？
【发布时间】：2014-11-21 21:53:21
【问题描述】：

我正在编写我的程序，以便在 Java SE 中为学校作业掷骰子（如掷骰子）。用户可以放置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);
        }
    }
}

输出该程序将产生适当的结果。但是偶尔输入的字符，代表骰子的眼睛，会转换成一个数字。

在下面的例子中，程序应该打印 9 个“@”字符。相反，它在第一行打印 192。（我知道骰子有 6 只眼睛，但我在不小心打印 9 只眼睛时碰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，谁能看到我在这里做错了什么？

【问题讨论】：

标签： java

【解决方案1】：

字符算术！

请咨询table。 @ 是字符64

typeOfEyes + typeOfEyes + typeOfEyes + "\n"

第一行实际上是将字符 (64 + 64 + 64) = 192 的值相加，然后加上换行符，所以我们得到 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字符串时，"foo" + 123 编译器可以将其解释为 foo123，因为它将第一个元素识别为字符串。然而，我们已经定义了一个char，它是一个代表一个字符的numeric 类型。所以编译器用它做数学运算。即使我们不应该。添加字符串字面量告诉它我们实际上需要文本。

【讨论】：

基本上，这些总和之间缺少一个字符串。
精彩的解释。我觉得有义务说：谢谢！

【解决方案2】：

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
@@@
@@@
@@@

【讨论】：