字符串和字符的连接 [英] Concatenation of Strings and characters

问题描述

以下陈述，

String string = "string";   

string = string +((char)65) + 5;
System.out.println(string);

产生输出 stringA5 。

但是，

String string = "string";

string += ((char)65) + 5;
System.out.println(string);

生成 string70 。

区别在哪里？

推荐答案

您看到这种行为是因为这种结合运算符优先级和字符串转换。

You see this behavior as a result of the combination of operator precedence and string conversion.

JLS 15.18.1 陈述：

如果只有一个操作数表达式是String类型，然后在另一个操作数上执行字符串转换（第5.1.11节）以在运行时生成一个字符串。

If only one operand expression is of type String, then string conversion (§5.1.11) is performed on the other operand to produce a string at run time.

因此，第一个表达式中的右手操作数被隐式转换为字符串： string = string +（（char）65）+ 5;

Therefore the right hand operands in your first expression are implicitly converted to string: string = string + ((char)65) + 5;

对于第二个表达式但是 string + =（（char）65）+ 5; + = 复合赋值运算符必须与 + 一起考虑。由于 + = 弱于 + ， + 运算符首先评估。我们有一个 char 和一个 int ，它会产生二进制数字促销到 int 。只评估 + = ，但此时已经评估了涉及 + 运算符的表达式的结果。

For the second expression however string += ((char)65) + 5; the += compound assignment operator has to be considered along with +. Since += is weaker than +, the + operator is evaluated first. There we have a char and an int which results in a binary numeric promotion to int. Only then += is evaluated, but at this time the result of the expression involving the + operator has already been evaluated.

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