问题描述
代码片段如下:
public static void main (String[]arg)
{
char ca = 'a' ;
char cb = 'b' ;
System.out.println (ca + cb) ;
}
输出是:
195
为什么会这样?我认为 'a' + 'b'
将是 "ab"
、 "12"
或 3代码>.
Why is this the case? I would think that 'a' + 'b'
would be either "ab"
, "12"
, or 3
.
这是怎么回事?
推荐答案
+
的两个 char
是算术加法,而不是字符串连接.你必须做类似 ""+ ca + cb
,或者使用String.valueOf
和Character.toString
方法保证+
是一个String
,用于操作符进行字符串连接.
+
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 isString
, 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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