我正在尝试用Io语言创建一个新的运算符,但是当我用解释器检查我的工作与文件执行时,我注意到了不同的行为:
## Executing nand.io
##########################
OperatorTable addOperator("nand", 10)
true nand := method(bool, if(bool, false, true))
false nand := method(bool, true)
true nand false println # => false
true nand true println # => true
false nand false println # => false
false nand true println # => true
# Why does this print 'true' (it should be 'false')?
(true nand true) println # => true
# I noticed that the argument to nand is always printed
true nand "not" println # => not
true nand "what" println # => what
false nand "I" println # => I
false nand "expected" println # => expected
# Another sanity check. I thought this might have to do with println behaving
# oddly, but this conditional really does return true
if(true nand true, "This should not print" println, "This should print" println)
# => This should not print
# More unexpected behavior
false nand := method(arg, "foo")
false nand "bar" println # => "bar"
true nand := method(arg, "foo")
true nand "bar" println # => "bar"
false nand := "Why is this not returned?"
false nand "aaaaahhhh" println # => "aaaaahhhh"
## Ran in the io interpreter
###############################
OperatorTable addOperator("nand", 10)
true nand := method(bool, if(bool, false, true))
false nand := method(bool, true)
# Same as file execution when sent to println
true nand false println # => false
true nand true println # => true
false nand false println # => false
false nand true println # => true
# It acually does what's expected here
true nand false # => true
true nand true # => false -- It works here!!!
false nand false # => true
false nand true # => true
if(true nand true, "This should not print" println, "This should print" println)
# => This should print
也许这需要编译与解释?这实际上是预期的行为,也许我错过了一些关于语言的基本信息(我今天才开始学习Io)?
答案 0 :(得分:1)
好的,这与VM分析运营商的方式有关。文件的第一次传递将通过并重写它已知道的任何操作符,重写与作为优先级提供的数字相关的消息。然后第二遍将执行该修改后的消息树。因此,当您将一个新运算符添加到稍后通过VM的文件中的运算符表时,新运算符将仅对您之后加载的任何文件或模块可用。对于从该文件生成的消息树的生命周期,您的运算符不存在。
基本上,这里的经验法则是维护一个单独的operators.io
文件,并将主io文件分开;然后确保在main.io之前加载operator.io,你就可以了。
CLI没有此问题的原因是它逐个评估消息。也就是说,您输入的下一条消息将具有新的运算符st