为什么一个 char + 另一个 char = 一个奇怪的数字

+ of two char is arithmetic addition, not string concatenation. You have to do something like "" + ca + cb, or use String.valueOf and Character.toString methods to ensure that at least one of the operands of + is a String for the operator to be string concatenation.

如果 + 运算符的任一操作数的类型为 String，则该操作为字符串连接.

If the type of either operand of a + operator is String, then the operation is string concatenation.

否则，+ 运算符的每个操作数的类型必须是可转换为原始数值类型的类型，否则会出现编译时错误.

Otherwise, the type of each of the operands of the + operator must be a type that is convertible to a primitive numeric type, or a compile-time error occurs.

至于为什么你得到 195，那是因为在 ASCII 中，'a' = 97 和 'b' = 98，以及 97 + 98= 195.

As to why you're getting 195, it's because in ASCII, 'a' = 97 and 'b' = 98, and 97 + 98 = 195.

这执行基本的 int 和 char 转换.

This performs basic int and char casting.

 char ch = 'a';
 int i = (int) ch;   
 System.out.println(i);   // prints "97"
 ch = (char) 99;
 System.out.println(ch);  // prints "c"

这忽略了字符编码方案的问题(初学者不应该担心......但是！).

This ignores the issue of character encoding schemes (which a beginner should not worry about... yet!).

作为注释，Josh Bloch 指出，很遗憾 + 对字符串连接和整数加法都进行了重载(对于字符串连接重载 + 运算符可能是一个错误." -- Java Puzzlers，谜题 11:最后的笑声).通过使用不同的字符串连接标记可以轻松避免很多此类混淆.

As a note, Josh Bloch noted that it is rather unfortunate that + is overloaded for both string concatenation and integer addition ("It may have been a mistake to overload the + operator for string concatenation." -- Java Puzzlers, Puzzle 11: The Last Laugh). A lot of this kinds of confusion could've been easily avoided by having a different token for string concatenation.

• 是与空字符串连接进行字符串转换真的那么糟糕吗?

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