In the process of figuring out blocks and yields I came across these comma-separated assignments:
def fibit
n,m =1,1
loop do |variable|
yield n
n,m = m,n+m # this line
puts "n is #{n} m is #{m}"
end
end
fibit do |num|
puts "Next : #{num}"
break if num > 100
end
Why does the m get assigned first in this scenario?
Does the last one always go first? If so why?
This was also seen as only e has the value of 1 meaning e went first?
e,r=1
puts r
puts "-------"
puts e
Also, does anyone have an idea why the code-block versions just executes, where if I write the same code with no code block I actually need to call the method for it to run?
def fibit
n,m =1,1
loop do |variable|
puts "Next : #{n}"
break if n > 100
n,m = m,n+m
end
end
fibit
If I didn't have the last line it wouldn't run. Where in the first one I don't actually call the fibit method? Or is the block kicking it off?
m does not get assigned first. When using multiple assignments, all right hand side calculations are done before any assignment to the left hand side.
That's how this code works:
a = 1
b = 3
a, b = b, a
a
# => 3
b
# => 1
This would not be possible if the assignment was done serially, since you would get that both would be either equal 1 or 3.
To further prove my point, simply swap the assignment of n and m in your code, and you'll find that the result is the same:
def fibit
n,m =1,1
loop do |variable|
yield n
m,n = n+m,m # this line
puts "n is #{n} m is #{m}"
end
end
fibit do |num|
puts "Next : #{num}"
break if num > 100
end