sml - 为什么使用这些重载运算符定义函数时不会出错？

Question

我正在阅读 Colin Myers、Chris Clack 和 Ellen Poon 的Programming with Standard ML。在 § 2.4.1 受限多态函数中:

Unlike the operators = and <>, the other comparison operators (>, < etc.) are overloaded; they are not restricted polymorphic and therefore the following definition fails:

- fun wrong_ordered (x, y, z)
     = (x < y) andalso (y < z);
Error: overloaded variable "<" cannot be resolved

This failure occurs because SML does not know the types of x, y and z.

当我将该定义输入 SML/NJ 110.99.2 和 Poly/ML 5.9 时，我没有收到任何错误消息。定义成功:

- fun wrong_ordered (x, y, z) = (x < y) andalso (y < z);
val wrong_ordered = fn : int * int * int -> bool

为什么我没有得到作者说我会得到的错误？

Answer 1

这本书早于 SML'97。使用该语言版本，重载运算符的类型解析得到了改进。特别是，当重载类型无法通过上下文解析时，它们现在默认为 int。