英文:
Java += operator
问题
我理解的是，x+=1 相当于 x=x+1，但为什么在字符串中不起作用？
String str = "";
str = str + null + true; // 没问题
str += null + true; // 错误信息：对于参数类型 null、boolean，未定义运算符 +
英文:
as far as I understand x+=1 works as x=x+1, but why it doesn't work in String?
String str = "";
str = str + null + true; // fine
str += null + true; // Error message: The operator + is undefined for the argument type(s) null, boolean
答案1
得分: 5
在Java中，表达式从左到右进行求值。因此，
str = str + null + true;
与
str = (str + null) + true;
是等价的，并且null和true会被隐式转换为字符串。这是因为在str + null中，编译器知道str是一个字符串，并将null转换为字符串。这是可能的，因为在Java中任何值都可以转换为字符串。按照同样的论证，编译器知道(str + null)是一个字符串，因此将true转换为字符串。
另一方面，
str += null + boolean;
等同于
str = str + (null + boolean);
因此，首先会对null + boolean进行求值。由于类型null、boolean的操作符+未定义，因此会生成编译器错误。
英文:
In Java, expression are evaluated from left to right. Thus
str = str + null + true;
is the same as
str = (str + null) + true;
and null and true are implicitly converted to Strings. This works because in str + null, the compiler knows that str is a String and converts null to a String. This is possible because every value can be converted to a String in Java. By the same line of argumentation, the compiler knows that (str + null) is a String and thus coverts true to a String.
On the other hand,
str += null + boolean;
is equivalent to
str = str + (null + boolean);
hence, null + boolean is evaluated first. Since the operator + is not defined for types null, boolean, a compiler-error is generated.
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