If statement operator as a variable in ruby

Question

I am trying to do something like this in ruby:

x = 0
y = 1
z = "!="

if x #{z} y
  puts "True"
end
#True
#=> nil

x = 1
if x #{z} y
  puts "True"
end
#True
#=> nil

Using a operator as a variable is not evaluating the expression. Any way to do this.

`if x #{z} y` is the same as `if x` because `#` introduces an end-of-line comment in Ruby. — axiac, Nov 14 '17 at 09:47
Can you please remove the irb-prompts (the `1.9.3-p429 :091 etc) as it is of no use and only makes your code harder to read and does not allow users to work with it using only a copy/paste. — Sagar Pandya, Nov 14 '17 at 09:49
You can try to use [`Kernel#eval`](https://ruby-doc.org/core-2.4.2/Kernel.html#method-i-eval). — axiac, Nov 14 '17 at 09:50
@DheerajGupta I'm curious, what are you trying to do? Why do you have the operator as a string? — Stefan, Nov 14 '17 at 10:03
@Stefan: implementing simple calculator / formula evaluator? ¯\\_(ツ)_/¯ — Sergio Tulentsev, Nov 14 '17 at 10:10
Note you can only interpolate within a double-quoted string or regex. — Sagar Pandya, Nov 14 '17 at 10:28
It might be overkill but you could try creating an [infix operator](https://stackoverflow.com/a/43429202/7324619) — Asone Tuhid, Nov 14 '17 at 13:01

Ursus · Accepted Answer · 2017-11-14T10:08:40.783

7

Try this one

x = 0
y = 1
z = "!="

x.public_send(z, y)
=> true

The trick here is know that 3 >= 5 is syntactic sugar for 3.>=(5)

edited Nov 14 '17 at 10:08

answered Nov 14 '17 at 09:51

Ursus

29,643
3
33
50

1

Nice, though it may be prudent to recommend `public_send` instead. – Sagar Pandya Nov 14 '17 at 09:55
Also worth noting that `x.public_send(z, y)` alone returns `true ` so if only the boolean is required then the if-expression is not. – Sagar Pandya Nov 14 '17 at 09:59
Sure, I just wanted recreate the user example – Ursus Nov 14 '17 at 10:08

Tom Lord · Answer 2 · 2017-11-14T12:06:55.813

4

To reiterate the answer above, I'd suggest using send.

However, it is possible to write the code in your original style - you just need to eval (i.e. execute) it:

x = 0
y = 1
z = "!="

if eval("x #{z} y")
  puts "True"
end

Generally speaking, the use of eval is strongly discouraged. It can lead to major security issues (e.g. if you're evaluating arbitrary user input!), and is quite a slow operation.

edited Nov 14 '17 at 12:06

answered Nov 14 '17 at 10:24

Tom Lord

27,404
4
50
77

1

btw, no need to interpolate the x and y into the string. :) – Sergio Tulentsev Nov 14 '17 at 11:53
heh... I did it that way originally, then doubted myself and "corrected" it. I'll revert that change :) – Tom Lord Nov 14 '17 at 12:06

If statement operator as a variable in ruby

2 Answers2