ruby 如何将布尔值转换为整数？
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How do I convert boolean values to integers?
提问by marcamillion
I have a boolean value to check if it is true, then set a local variable. How do I refactor this so it is more Ruby-ish?
我有一个布尔值来检查它是否为真，然后设置一个局部变量。我如何重构它以使其更像 Ruby-ish？
if firm.inflection_point
inflection_point = 1
else
inflection_point = 0
end
回答by rudolph9
inflection_point = (firm.inflection_point ? 1 : 0)
回答by sawa
If you just have that at one point, then rudolph9's answeris good, but if you are having a similar kind of logic all over the place, then maybe it might make sense with general use in mind to monkey patch:
如果您只是在某一时刻拥有它，那么rudolph9 的答案是好的，但是如果您到处都有类似的逻辑，那么考虑到猴子补丁的一般用途，这可能是有意义的：
class FalseClass; def to_i; 0 end end
class TrueClass; def to_i; 1 end end
inflection_point = firm.inflection_point.to_i
Within Ruby, you should keep all of your logic dealing with truth values rather than 0and 1, but I guess you are dealing with some inputs or outputs from/to some external system that deals with 0and 1. Then, doing like this will make sense.
在 Ruby 中，您应该让所有逻辑处理真值而不是0and 1，但我猜您正在处理来自/到某个处理0and 的外部系统的一些输入或输出1。那么，这样做是有意义的。
回答by Andre Figueiredo
Another alternative is use of short-circuit operators:
另一种选择是使用短路运算符：
inflection_point && 1 || 0
irb(main):001:0> true && 1 || 0
=> 1
irb(main):002:0> false && 1 || 0
=> 0
回答by J?rg W Mittag
In Ruby, ifis an expression. There's no need to assign to a variable inside the thenand elsebranches, just return the value you want and assign the variable to the result of the if expression:
在 Ruby 中，if是一个表达式。不需要在thenandelse分支内部分配变量，只需返回您想要的值并将变量分配给 的结果if expression：
inflection_point = if firm.inflection_point
1
else
0
end
In simple cases like this, it's more readable to write the entire expression on a single line:
在像这样的简单情况下，将整个表达式写在一行上会更易读：
inflection_point = if firm.inflection_point then 1 else 0 end
You can also use the conditional operator, which I personally find to be much less readable:
您还可以使用条件运算符，我个人认为它的可读性要差得多：
inflection_point = firm.inflection_point ? 1 : 0
回答by thebugfinder
What you need is a conditional operation that is known as Ternary Operator It's used in almost every language and it uses the symbols ? and :
您需要的是一种称为三元运算符的条件运算。它几乎在所有语言中都有使用，并且使用符号？和 ：
inflection_point = firm.inflection_point ? 1 : 0
basically means, if the first condition evaluates to true (firm.inflection_point), return the value after "?" (1) otherwise, return the value after ":" (0)
基本上意味着，如果第一个条件评估为真（firm.inflection_point），则返回“？”之后的值。(1) 否则，返回“:”后的值 (0)
回答by Darth Egregious
Here's another method:
这是另一种方法：
5 - bool.to_s.length
5 - bool.to_s.length
This takes advantage of the fact that 'true'has four characters, while 'false'has 5.
这利用了'true'有四个字符而'false'有 5个字符的事实。
回答by Miko?aj Wawrzyniak
It is not pure ruby solution but, You can use ActiveRecord::Type::Integer.new.cast(true)
它不是纯红宝石解决方案，但是，您可以使用 ActiveRecord::Type::Integer.new.cast(true)
