Arithmetic on char in java

问题:

I am learning Java.

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.

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');
        }
}

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

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

回答1:

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

ch = (char)(ch + 32)

Notes:

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.

2. 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."

3. 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.

回答2:

Type casting is required, try this code:

if (ch >= 97 & ch <= 122)
    ch = (char) (ch - 32);
if (ch >= 65 & ch <= 90)
    ch = (char) (ch + 32);

回答3:

The result of arithmetic between a char and an int is an int and you cannot store an int in a char without explicit typecasting - by which you tell to compiler that I know what I'm doing, now do it for me

So, you need to do a typecast: -

    char ch = 'a';
    if (ch >= 97 & ch <= 122) {
       ch = (char)(ch - 32);
    } 
    System.out.println(ch);  // Prints `A`

But, you have already method in Character class, that will do it for you: -

char ch = 'a';
ch = Character.toUpperCase(ch);
System.out.println(ch);  // Prints `A`

回答4:

Try

ch = (char) (ch + 32);

if you're sure that ch + 32 won't be bigger than what a char can hold.

回答5:

That is because you trying to assign an int value (Ascii value) to the char. But you need to explicitly convert the result value int to char. Just cast int to char.

ch= (char) (ch+32);
ch= (char) (ch-32);

It will work with out any Exception.