Scala 中 a += b 什么时候变成 a = a + b ？

Question

仅限 Scala 有时脱糖

a += b

到

a = a + b

但并非总是如此。例如，一些可变集合定义了一个 += 方法，而它变成了

a.+=(b)

这是行为吗

完全取决于a上是否有合适的+=方法？（包括这种行为还有其他例子吗？）
与对象是 val 还是 var 无关？

相关例子

改编自 Scala 编程，在

var s = Set("a", "b")
s += "c"

In this case, the second line of code s += "c" is essentially shorthand for:

s = s + "c"

Answer 1

When does a += b become a = a + b in Scala?

当没有适用的+=方法时，是适用的+方法，a是可赋值的（即是var 或 a_= 方法）。

或者如 the spec 所说：

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 +.

Is this behaviour

entirely determined by whether there is a suitable += method on a?

independent of whether the objects are val or var?

不完全是。如果有合适的 += 方法，它将被调用，而不管任何其他因素（例如 a 是可分配的）。但如果没有，其他因素将决定它是否已脱糖或收到错误消息。

请注意，您收到的错误消息与您从脱糖版本中收到的错误消息不同：当脱糖标准不适用时，您会收到一条错误消息，告诉您“+= is not ... 的成员”，加上无法应用脱糖的解释（例如“接收器不可分配”或如果 a + b 会产生您从 a + b 得到的类型错误类型错误）。

Answer 2

第一次回答有点仓促，抱歉。经过一些研究并更好地阅读了 @sepp2k 的回答和评论后，我得出结论，Scala 中的一些类实现了 += 方法，而其他只是 + 方法，我玩了一些 Scala 代码，例如：

  //Set, Int, Double, String implements the "+" method,
  //and then "+=" is syntactic sugar of a = a + b 
  var set = Set("a", "b")
  set += "c"

  var num = 3
  num += 2

  var str = "43"
  str += 5

  var l = List()
  l += "someString"

  // As you mention, MutableList implements "+=" method, and when you do
  // mutL += 4, is the same as call the method mutL.+=
  var mutL = new mutable.MutableList[Int]
  mutL += 4

Scala 中 a += b 什么时候变成 a = a + b ？

When does a += b become a = a + b in Scala?

scala

assignment-operator

syntactic-sugar