Question

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）？

Answer 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

Io语言在解释器和文件执行之间的不同行为

1 个答案: