Code syntax comma usage

ruby-talk
1

That I can see; Pragmatic Programming Ruby's, index, has reference to
all sorts of symbols, but I can't find a explanation for a ',' seperator
such as used in the following:

var1 = 1; p var1 #=> 1
var2 = 2; p var2 #=> 2
var3 = 3; p var3 #=> 3

var1, var2, var3 = 5

p var1 #=> 5
p var2 #=> nil
p var3 #=> nil

What's happening here? I read it as:
set var1 to 5 and reset var2 and var3 to nil.
Is that all the comma does?

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2

Jcm Mz wrote:

That I can see; Pragmatic Programming Ruby's, index, has reference to
all sorts of symbols, but I can't find a explanation for a ',' seperator
such as used in the following:

var1 = 1; p var1 #=> 1
var2 = 2; p var2 #=> 2
var3 = 3; p var3 #=> 3

var1, var2, var3 = 5

p var1 #=> 5
p var2 #=> nil
p var3 #=> nil

What's happening here? I read it as:
set var1 to 5 and reset var2 and var3 to nil.
Is that all the comma does?

Consider:

irb(main):005:0> a,b,c = 5,6,7
=> [5, 6, 7]

irb(main):006:0> a
=> 5

irb(main):007:0> b
=> 6

irb(main):008:0> c
=> 7

Multiple assignment from array parameters is very useful as well...

ra_parm = [1,2,3]

def meth(ra_parm)
p1, p2, p3 = ra
end

Regards

···

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3

set var1 to 5 and reset var2 and var3 to nil.
Is that all the comma does?

It can be used for parallel assignments:

a,b,c = [1,2,3]
a #=> 1
b #=> 2
c #=> 3

When you enter "var1, var2, var3 = 5" in irb, you'll get [5] as return
value. It seems the 5 gets silently converted into an array here.
Interesting.

4

You've stumbled on one of Ruby's most elegant bits of syntactical
niceness: multiple assignment. It isn't the comma that's interesting
here so much as the equal-sign and how it distributes assignments based
on lists. Try these:

arr = [1, 2, 3]

# now look at what a, b, and c are after each of the following:

a, b, c = 1, 2, 3
a, b, c = [1, 2, 3]
a = b, c
a, b = b, a
a, b, c = arr
a, b, c = arr, 4
a, b, c = 4, arr
a, b, c = 4, *arr

m.

···

Jcm Mz <menzies.main@gmail.com> wrote:

That I can see; Pragmatic Programming Ruby's, index, has reference to
all sorts of symbols, but I can't find a explanation for a ',' seperator
such as used in the following:

var1, var2, var3 = 5

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5

Jcm Mz wrote:

That I can see; Pragmatic Programming Ruby's, index, has reference to
all sorts of symbols, but I can't find a explanation for a ',' seperator
such as used in the following:

var1 = 1; p var1 #=> 1
var2 = 2; p var2 #=> 2
var3 = 3; p var3 #=> 3

var1, var2, var3 = 5

p var1 #=> 5
p var2 #=> nil
p var3 #=> nil

What's happening here?

It's called "parallel assignment". Here is the basic case:

x, y = 10, 20
puts x #10
puts y #20

Then there are rules for dealing with the cases when the number of
variables on the left side do not equal the number of values on the
right side. You are seeing the rule: if there are more "lvalues" than
"rvalues", the excess lvalues will have nil assigned to them.

One of the little tricks you can do with parallel assignment is switch
values in two variables without using a third variable:

str = "hello"
num = 100
str, num = num, str

puts str #100
puts num #hello

Normally, if there are more rvalues than lvalues then the excess rvalues
are discarded, for example:

x, y = 10, 20, 100, 200
puts x #10
puts y #20

But you can also do this:

x, y, *remaining = 10, 20, 100, 200

puts x #10
puts y #20
p remaining #[100, 200]

And this is also another way to do parallel assignment:

x, y, z = [10, 20, 30]
puts x #10
puts y #20
puts z #30

In pickaxe2, a detailed description of parallel assignment is on p. 340.

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6

Jcm Mz wrote:

What's happening here? I read it as:
set var1 to 5 and reset var2 and var3 to nil.
Is that all the comma does?

The answer I was looking for was in fact the multiple assignment comma
operation.

The original code I was looking at was:

class Timer
  private
  def dispatch
     now = Time.now.to_f
     ready, @queue = @queue.partition{|time, proc| time <= now }
     ready.each {|time, proc| proc.call((time)}
  end
end

p24 Practical Ruby Projects

···

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7

Multiple assignment from array parameters is very useful as well...

ra_parm = [1,2,3]

def meth(ra_parm)
p1, p2, p3 = ra
end

Regards

Make that...

def meth(ra_parm)
   p1, p2, p3 = ra_parm
end

···

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8

It seems the 5 gets silently converted into an array here.

The 5 doesn't get converted of course and [5] is the value of the
assignment. At times, I find ruby's permissiveness slightly
surprising.

9

Hi --

···

On Sun, 15 Feb 2009, Tom Link wrote:

It seems the 5 gets silently converted into an array here.

The 5 doesn't get converted of course and [5] is the value of the
assignment. At times, I find ruby's permissiveness slightly
surprising.

That's been fixed in 1.9, so that a,b,c = 5 returns 5.

David

--
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Ruby/Rails consulting & training: http://www.rubypal.com
Coming in 2009: The Well-Grounded Rubyist (http://manning.com/black2\)

http://www.wishsight.com => Independent, social wishlist management!

10

Tom Link wrote:

When you enter "var1, var2, var3 = 5" in irb, you'll get [5] as return
value. It seems the 5 gets silently converted into an array here.
Interesting.

The 5 doesn't get converted of course and [5] is the value of the
assignment. At times, I find ruby's permissiveness slightly
surprising.

The value of the assignment is written as 5, so if you end up with [5]
then it seems logical that something got converted.

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11

Jeff Moore wrote:

Multiple assignment from array parameters is very useful as well...

ra_parm = [1,2,3]

def meth(ra_parm)
p1, p2, p3 = ra
end

Regards
    
Make that...

def meth(ra_parm)
   p1, p2, p3 = ra_parm
end
  
Don't you need a splat for that?

def meth(ra_parm)
  p1, p2, p3 = *ra_parm
end

12

Hi --

Jeff Moore wrote:

Multiple assignment from array parameters is very useful as well...

ra_parm = [1,2,3]

def meth(ra_parm)
p1, p2, p3 = ra
end

Regards

Make that...

def meth(ra_parm)
   p1, p2, p3 = ra_parm
end

Don't you need a splat for that?

def meth(ra_parm)
p1, p2, p3 = *ra_parm
end

No. If ra_parm is an array, it will assign in parallel to the
variables on the left.

irb(main):010:0> a,b,c = [1,2,3]
=> [1, 2, 3]
irb(main):011:0> a
=> 1
irb(main):012:0> b
=> 2
irb(main):013:0> c
=> 3

David

···

On Sun, 15 Feb 2009, David Masover wrote:

--
David A. Black / Ruby Power and Light, LLC
Ruby/Rails consulting & training: http://www.rubypal.com
Coming in 2009: The Well-Grounded Rubyist (http://manning.com/black2\)

http://www.wishsight.com => Independent, social wishlist management!

13

David Masover wrote:

Jeff Moore wrote:

Make that...

def meth(ra_parm)
   p1, p2, p3 = ra_parm
end
  
Don't you need a splat for that?

def meth(ra_parm)
  p1, p2, p3 = *ra_parm
end

1) You could test that yourself in less than 30 seconds.

2) You could actually read the thread and discover the answer: there
are two posts containing the answer.

···

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