ruby 意外返回（LocalJumpError）
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/17800629/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Unexpected Return (LocalJumpError)
提问by Eli Rose -- REINSTATE MONICA
What is the problem with this Ruby 2.0 code?
这个 Ruby 2.0 代码有什么问题？
p (1..8).collect{|denom|
(1...denom).collect{|num|
r = Rational(num, denom)
if r > Rational(1, 3) and r < Rational(1, 2)
return 1
else
return 0
end
}
}.flatten
The error is in block (2 levels) in <main>': unexpected return (LocalJumpError). I want to create a flat list containing nones (and the rest zeroes) where nis the number of rational numbers with denominators below 8 which are between 1/3 and 1/2. (it's a Project Euler problem). So I'm trying to return from the inner block.
错误在block (2 levels) in <main>': unexpected return (LocalJumpError). 我想创建一个包含n个（以及其余的零）的平面列表，其中n是分母低于 8 且介于 1/3 和 1/2 之间的有理数的个数。（这是一个 Project Euler 问题）。所以我试图从内部块返回。
回答by sarahhodne
You can't returninside a block in Ruby*. The last statement becomes the return value, so you can just remove the return statements in your case:
您不能return在 Ruby * 中的块内。最后一条语句成为返回值，因此您可以删除案例中的返回语句：
p (1..8).collect{|denom|
(1...denom).collect{|num|
r = Rational(num, denom)
if r > Rational(1, 3) and r < Rational(1, 2)
1
else
0
end
}
}.flatten
*: You can inside lambdablocks: lambda { return "foo" }.call # => "foo". It has to do with scoping and all that, and this is one of the main differences between lambdablocks and procblocks. "Normal" blocks you pass to methods are more like procblocks.
*：您可以在lambda块内：lambda { return "foo" }.call # => "foo". 它与范围和所有这些有关，这是lambda块和proc块之间的主要区别之一。您传递给方法的“普通”块更像是proc块。
