blocks|key|1948759|text|在下面的示例中，lambda仍然是匿名的，但它有一个引用。(这算是匿名吗？)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1948760|(l+=+lambda+{+l.call+}).call|code-block|syntax|javascript|1948761|(感谢Niklas+B.指出了我原始答案中的错误；我只在IRB中测试了它，它在那里工作)。|1948762|当然，这以SystemStackError:+stack+level+too+deep错误结束，但它演示了目的。|offset|length|style|CODE|1948763|entityMap^0|0|0|0|5|12|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

In the following example, the lambda is still anonymous, but it has a reference. (Does that pass for anonymous?)

<pre><code>(l = lambda { l.call }).call
</code></pre>

(Thanks to Niklas B. for pointing out the error in my original answer; I had only tested it in IRB and it worked there).

This of course ends in a <code>SystemStackError: stack level too deep</code> error, but it demonstrates the purpose.

blocks|key|1735392|text|除了KL-7's+comment之外，这里还有一个Y组合子解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1735393|lambda+{+%7Cf%7C
++lambda+{+%7Cx%7C+x.call(x)+}.call(
++lambda+{+%7Cx%7C+f.call(+lambda+{+%7Cv%7C+x.call(x).call(v)+}+)+}+)
}.call(
++lambda+{+%7Cf%7C
++++lambda+{+%7Cn%7C+n+==+0+?+1+:+n+*+f.call(n+-+1)+}
++}
).call(5)+#=>+120|code-block|syntax|javascript|1735394|您通常会将以下内容拆分：|1735395|y+=+lambda+{+%7Cf%7C
++lambda+{+%7Cx%7C+x.call(x)+}.call(
++lambda+{+%7Cx%7C+f.call(+lambda+{+%7Cv%7C+x.call(x).call(v)+}+)+}+)
}

fac+=+y.call(
++lambda+{+%7Cf%7C+lambda+{+%7Cn%7C+n+==+0+?+1+:+n+*+f.call(n+-+1)+}+}
)

fac.call(5)+#=>+120|1735396|请注意，尽管分配了fac，但它不会在lambda中使用。|style|CODE|1735397|我会使用Ruby语言的->语法和.()，而不是.call()|1735398|y+=+->(f)+{
++->(x)+{+x.(x)+}.(
++->(x)+{+f.(->(v)+{+x.(x).(v)+})+}+)
}

fac+=+y.(->(f)+{
++->(n)+{+n+==+0+?+1+:+n+*+f.(n+-+1)+}
})

fac.(5)+#=>+120|1735399|通过使用curry，可以稍微简化y调用|1735400|y+=+->(f)+{
++->(x)+{+x.(x)+}.(
++->(x)+{+f.curry.(->(v)+{+x.(x).(v)+})+}+)
}

fac+=+y.(
++->(f,+n)+{+n+==+0+?+1+:+n+*+f.(n+-+1)+}
)

fac.(5)+#=>+120|1735401|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/9516061/can-i-reference-a-lambda-from-within-itself-using-ruby#comment12052922_9516061|1|http://ruby-doc.org/core-2.2.3/Proc.html#method-i-curry^0|2|E|0|0|0|0|0|9|3|0|B|2|G|3|N|7|0|0|4|5|G|1|4|5|1|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@$A|19|B|1A|1|1B]]|C|$]]|$1|D|3|E|5|F|7|1C|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|1D|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|1E|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|1F|8|@$A|1G|B|1H|O|P]]|9|@]|C|$]]|$1|Q|3|R|5|6|7|1I|8|@$A|1J|B|1K|O|P]|$A|1L|B|1M|O|P]|$A|1N|B|1O|O|P]]|9|@]|C|$]]|$1|S|3|T|5|F|7|1P|8|@]|9|@]|C|$G|H]]|$1|U|3|V|5|6|7|1Q|8|@$A|1R|B|1S|O|P]|$A|1T|B|1U|O|P]]|9|@$A|1V|B|1W|1|1X]]|C|$]]|$1|W|3|X|5|F|7|1Y|8|@]|9|@]|C|$G|H]]|$1|Y|3|-4|5|6|7|1Z|8|@]|9|@]|C|$]]]|Z|$10|$5|11|12|13|C|$14|15]]|16|$5|11|12|13|C|$14|17]]]]

In addition to <a href="https://stackoverflow.com/questions/9516061/can-i-reference-a-lambda-from-within-itself-using-ruby#comment12052922_9516061">KL-7's comment</a>, here's a Y combinator solution:

<pre><code>lambda { |f|
 lambda { |x| x.call(x) }.call(
 lambda { |x| f.call( lambda { |v| x.call(x).call(v) } ) } )
}.call(
 lambda { |f|
 lambda { |n| n == 0 ? 1 : n * f.call(n - 1) }
 }
).call(5) #=&gt; 120
</code></pre>

You would normally split these:

<pre><code>y = lambda { |f|
 lambda { |x| x.call(x) }.call(
 lambda { |x| f.call( lambda { |v| x.call(x).call(v) } ) } )
}

fac = y.call(
 lambda { |f| lambda { |n| n == 0 ? 1 : n * f.call(n - 1) } }
)

fac.call(5) #=&gt; 120
</code></pre>

Note that although <code>fac</code> is being assigned, it is not used within the lambda.

I'd use Ruby's <code>-&gt;</code> syntax and <code>.()</code> instead of <code>.call()</code>:

<pre><code>y = -&gt;(f) {
 -&gt;(x) { x.(x) }.(
 -&gt;(x) { f.(-&gt;(v) { x.(x).(v) }) } )
}

fac = y.(-&gt;(f) {
 -&gt;(n) { n == 0 ? 1 : n * f.(n - 1) }
})

fac.(5) #=&gt; 120
</code></pre>

The <code>y</code> invocation can be simplified a bit by using <a href="http://ruby-doc.org/core-2.2.3/Proc.html#method-i-curry" rel="nofollow noreferrer"><code>curry</code></a>:

<pre><code>y = -&gt;(f) {
 -&gt;(x) { x.(x) }.(
 -&gt;(x) { f.curry.(-&gt;(v) { x.(x).(v) }) } )
}

fac = y.(
 -&gt;(f, n) { n == 0 ? 1 : n * f.(n - 1) }
)

fac.(5) #=&gt; 120
</code></pre>

blocks|key|1735438|text|fact+=+->+(x){+x+<+2+?+1+:+x*fact.(x-1)}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1735439|极小函数|unstyled|1735440|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>fact = -&gt; (x){ x &lt; 2 ? 1 : x*fact.(x-1)}
</code></pre>

minimal function

I want to be able to call an anonymous lambda from within itself using Ruby. Consider the following recursive block (returns a factorial). I know I can assign it to a variable, and that variable is within the scope of the lambda:

<pre><code>fac = lambda { |n| n == 1 ? 1 : n * fac.call(n - 1) }
fac.call(5)
</code></pre>

But, I want to be able to do the following (for no practical reason as of yet, I'm just interested in exploring the language some more):

<pre><code>(lambda { |n| n == 1 ? 1 : n * self.call(n - 1) }).call(5)
</code></pre>

I know that won't work, because <code>self</code> is the <code>main</code> object. Am I doing it wrong? Am I trying to do something that's not possible -- and if not, is this because of some theoretical restriction or is it simply not implemented in Ruby?

Can I reference a lambda from within itself using Ruby?

Ruby语言

我希望能够使用Ruby从其内部调用匿名lambda。考虑下面的递归块(返回阶乘)。我知道我可以将它赋给一个变量，而该变量在lambda的作用域内：fac = lambda { |n| n == 1 ? 1 : n * fac.call(n - 1) }fac.call(5)但是，我希望能够做到以下几点(到目前为止还没有...

问我可以使用Ruby从lambda内部引用它吗？
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问我可以使用Ruby从lambda内部引用它吗？
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