ruby-core

Issue #13815 has been updated by phluid61 (Matthew Kerwin).


opti (Andreas Opti) wrote:
> >If you want the same behavior, you need to use parentheses:
> >3.times { p((x,y = x+1,y+1)) }
> 
> # ok, but also with ((...)) [most people] would assume its meaning is (x,  y=x+1,  y+1)...
> 

Most people wouldn't write it in the first place.

You *could*, if you really, really wanted to do a multiple assignment **and** capture the result into an array **and** `p` the array, all in one step.

I don't think it's a parser issue, though; the parser does a logical thing at each step.  It's the author who's doing something weird.

> 
> *Other example:
> p(x+=1,x+=1)  # ok
> p (x+=1,x+=1)  # error

Well, yeah, but this is well known, well defined ruby.  `p(x)` is unambiguous function call; `p (x)` is a parenthesised `(x)` being passed as the first positional argument to `p`.

It might help illustrate by replacing the `p` function call with, say, an assignment:

~~~
#p(x)
a = x

#p((x))
a = (x)

#p x
a = x

#p (x)
a = (x)

#p (x+=1,x+=1)
a = (x+=1,x+=1) #???

#p ((x,y)=[x+1,y+1])
a = ((x,y)=[x+1,y+1])
# Note: multiple assignment returns the whole array, so a == [x,y]
# This can also be written: a = (x,y=[x+1,y+1])  or:  a = (x,y=x+1,y+1)

#p((x,y)=[x+1,y+1])
a = (x,y)=[x+1,y+1] #???  This part doesn't make sense: a = (x,y)
~~~

> 
> So the parsing of p(Args) might be improved...

I don't like saying this, but I think this is a case where you have to get familiar with ruby's syntax.  It all makes sense once you understand how the parser sees what you've written.

----------------------------------------
Bug #13815: p args
https://bugs.ruby-lang.org/issues/13815#change-66208

* Author: opti (Andreas Opti)
* Status: Open
* Priority: Normal
* Assignee: 
* Target version: 
* ruby -v: 2.5.x
* Backport: 2.2: UNKNOWN, 2.3: UNKNOWN, 2.4: UNKNOWN
----------------------------------------
x=y=0  # outside defined vars!
3.times {|i| x,y=x+1,y+1 }  # as expected: x==3
3.times {|i| p(x,y=x+1,y+1) }  # NOT as expected: x not changed!
  # x,y are the same vars as above. p shoudn't have any effect on scope






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