java中char的算法 [英] Arithmetic on char in java

问题描述

我正在学习Java。

我应该编写一个程序，将所有大写字母转换为小写，全部小写转换为大写。它在书中说我只需要从大写中减去32并将32添加到小写。

I am supposed to write a program that converts all uppercase letters to lowercase and all lowercase to uppercase. It said in the book I just need to subtract 32 from uppercase and add 32 to lowercase.

这是我的代码......

Here is my code...

class Caseconv {
    public static void main(String args[])
        throws java.io.IOException {
            char ch;

            do {
            ch = (char) System.in.read();

            if (ch >= 97 & ch <= 122) ch = ch - 32;
            if (ch >= 65 & ch <= 90) ch = ch + 32;
            System.out.print(ch);
            } while (ch != '\n');
        }
}

但编译器不想这样做，我收到此错误。

But the compiler doesn't want to do this, I get this error.

Caseconv.java:13: error: possible loss of precision
            if (ch >= 97 & ch <= 122) ch = ch - 32;
                                              ^
  required: char
  found:    int
Caseconv.java:14: error: possible loss of precision
            if (ch >= 65 & ch <= 90) ch = ch + 32;
                                             ^
  required: char
  found:    int
2 errors

我应该做什么来减去char？

What am I supposed to be doing to subtract from the char?

推荐答案

你需要添加一个类型转换为将表达式的结果转换为 char 。例如。

You need to add a type cast to convert the result of the expression to char. For example.

ch = (char)(ch + 32)

注意：

1. 这是必要的原因是因为 32 是一个 int 字面值，并添加 char 使用 int 算术执行 int ，并给出 int 结果。

1. The reason this is necessary is because 32 is an int literal, and the addition of a char and an int is performed using int arithmetic, and gives an int result.

将 int 分配给 char 可能导致截断。有效地添加类型转换对编译器说：是的。我知道。没关系。就这样做。

Assigning an int to a char potentially results in truncation. Adding the type cast effectively says to the compiler: "Yes. I know. It is OK. Just do it."

+ 子表达式周围的括号是必要的，因为type-cast的优先级高于 + 。如果你把它们排除在外，那么类型转换没有区别，因为它将 char 强制转换为 char 。

The parentheses around the + subexpression are necessary because type-cast has higher precedence than +. If you leave them out, the type-cast makes no difference because it "casts" a char to a char.

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