Question

我到处都看到这个片段，它有效！为什么呢？

while gets
    print if /start/../end/
end

如果没有Lvalue，ruby如何评估/start/？我希望我们首先必须在某处存储'gets'的值，然后再进行

gets_result =~ /start/.. gets_result =~ /end/

那么为什么代码片段有效？

让我澄清一下。

ruby如何知道比较正则表达式与获取在上面的代码段中，我没有指定ruby将reg exp与gets进行比较但它只是知道。问题是怎么样的？

Answer 1

Kernel#gets不仅会返回下一行，还会将值指定给$_。

如果没有争论，

Kernel#print打印$_。

触发器操作符（/start/../end/）也在$_上运行。

Answer 2

好问题。

请记住：在条件语句中使用Range运算符（..或...）时，它会执行某些操作完全出乎意料：它没有创建一个Range对象。相反，它充当“触发器”运算符。

下面的代码实际上是

while gets
    print if /start/../end/
end

默认为

while gets
    # gets_input I put just to make the code more expressive
    # actually the input taken using gets method applied here implicitly.
    print if  /start/ =~ gets_input .. /end/ =~ gets_input
end

让我来证明你。我接受了Ruby Tracer类的帮助。

trace = TracePoint.new do |tp|
  p [tp.lineno, tp.event, tp.defined_class,tp.method_id]
end
trace.enable do
  while gets
    # when you type start in your console, 11 will be output.
    print 11 if /start/../end/
  end
end

让我运行此代码，向您展示我的上述代码作为证明以及Ruby Flip-Flop 功能：

(arup~>Ruby)$ ruby test.rb
test.rb:6: warning: regex literal in condition
test.rb:6: warning: regex literal in condition
[4, :b_call, nil, nil]
[5, :line, nil, nil]
[5, :c_call, Kernel, :gets]
[5, :c_call, ARGF.class, :gets]
end # I presses **end** here.
[5, :c_return, ARGF.class, :gets]
[5, :c_return, Kernel, :gets]
[6, :line, nil, nil]
# Regexp#=~ call begin happened for /end/ =~ gets_input
[6, :c_call, Regexp, :=~]
# Regexp#=~ call end happened for /end/ =~ gets_input
[6, :c_return, Regexp, :=~] 
[5, :c_call, Kernel, :gets]
[5, :c_call, ARGF.class, :gets]
start # I presses **start** here.
[5, :c_return, ARGF.class, :gets]
[5, :c_return, Kernel, :gets]
[6, :line, nil, nil]
# Regexp#=~ call begin happened for /start/ =~ gets_input
[6, :c_call, Regexp, :=~] 
# Regexp#=~ call end happened for /start/ =~ gets_input
[6, :c_return, Regexp, :=~]
# Regexp#=~ call begin happened for /end/ =~ gets_input
[6, :c_call, Regexp, :=~] 
# Regexp#=~ call end happened for /end/ =~ gets_input
[6, :c_return, Regexp, :=~] 
[6, :c_call, Kernel, :print]
[6, :c_call, IO, :write]
[6, :c_call, Fixnum, :to_s]
[6, :c_return, Fixnum, :to_s]
# As there is a match so **if** clause true, thus 11 printed
11[6, :c_return, IO, :write] 
[6, :c_return, Kernel, :print]
[5, :c_call, Kernel, :gets]
[5, :c_call, ARGF.class, :gets]
end  
[5, :c_return, ARGF.class, :gets]
[5, :c_return, Kernel, :gets]
[6, :line, nil, nil]
[6, :c_call, Regexp, :=~]
[6, :c_return, Regexp, :=~]
[6, :c_call, Kernel, :print]
[6, :c_call, IO, :write]
[6, :c_call, Fixnum, :to_s]
[6, :c_return, Fixnum, :to_s]
11[6, :c_return, IO, :write]
[6, :c_return, Kernel, :print]
[5, :c_call, Kernel, :gets]
[5, :c_call, ARGF.class, :gets]

Answer 3

在这种情况下，..是触发器操作符。给定'start'时，条件的计算结果为true，并继续true直到end为止。

如何评估孤立的正则表达式

3 个答案: