scala中int的+=方法在哪里
Where is += method located for int in scala
在 Scala 中，+=（或任何赋值运算符）是 Int 类型的方法。
例如，
var x=5
x+=1
这里只有当它是一个变量时我才能使用+=方法
我做不到，
5+=1
scala 编译器是否将此方法视为特例？
为什么在 scala.Int class 中不可用？
没有+=方法，编译器扩展为x = x + 1。这在 specification:
中有详细说明
6.12.4 赋值运算符
Let's consider an assignment operator such as += in an infix operation
l += r, where l, r
are expressions. This operation can be re-interpreted as an operation
which corresponds to the assignment
l = l + r
except that the operation's left-hand-side l is evaluated only once.
The re-interpretation occurs if the following two conditions are
fulfilled.
The left-hand-side l does not have a member named +=, and also cannot be converted by an implicit conversion to a value with a member
named +=.
The assignment l = l + r is type-correct. In particular this implies that l refers to a variable or object that can be assigned to,
and that is convertible to a value with a member named +.
在 Scala 中，+=（或任何赋值运算符）是 Int 类型的方法。
例如，
var x=5
x+=1
这里只有当它是一个变量时我才能使用+=方法
我做不到，
5+=1
scala 编译器是否将此方法视为特例？
为什么在 scala.Int class 中不可用？
没有+=方法，编译器扩展为x = x + 1。这在 specification:
6.12.4 赋值运算符
Let's consider an assignment operator such as += in an infix operation l += r, where l, r
are expressions. This operation can be re-interpreted as an operation which corresponds to the assignment
l = l + rexcept that the operation's left-hand-side l is evaluated only once.
The re-interpretation occurs if the following two conditions are fulfilled.
The left-hand-side l does not have a member named +=, and also cannot be converted by an implicit conversion to a value with a member named +=.
The assignment l = l + r is type-correct. In particular this implies that l refers to a variable or object that can be assigned to, and that is convertible to a value with a member named +.