I have a variable x which is set to 10. And I want to write a while loop that increments that. I know you can easily just use a for-loop for this, but easy isn't fun. The code I have is:
def add(a)
g = a + 1
puts g
end
def loop(d)
x = 0
while x <= 4
x += 1
add(d)
end
end
loop(9)
When ran I get 9, four times. How can I get this code to have an output of 9, 10, 11, 12?
Your problem is that you say add(d) and d is the parameter of your loop, loop(d). Ruby does blindly what you tell him : loop(9), so here d=9 and remains equal to 9. You need to increment the value of d. To do that add will now return the incremented value, and we assign the returned value to d (in loop).
To solve your problem you will want to do something like :
def add(a)
g = a + 1
puts g
g
end
def loop(d)
x = 0
while x <= 4
x += 1
d = add(d)
end
end
loop(9)
BUT and that's a huge but, your code is not the ruby way at all.
If I were to do it I would do it like this:
def loop(start_number, repeat_number, increment)
repeat_number.times do
start_number += increment
p start_number
end
end
loop(9, 4, 1)
If you rly wanna to use while and your expected outpit is Array numbers 9 up to 12.
Simply define the array variable before while loop and return it like this:
def loop(number)
x, a = 0, []
while x <= 3
a << number + x
x += 1
end
a
end
loop(9)
# => [9, 10, 11, 12]
BETTER WAY
But better way is use some ruby functions like times and map
in this case we use times and map
def loop(d)
4.times.map{|i| d + i}
end
p loop(9)
# => [9, 10, 11, 12]
maybe this will helps.
xwas never set to10, what contradicts the preface,) that there is no way to fix it save for completely rewrite from the scratch after reading a book on ruby language.xis set to0, not10and your code outputs10four times, not9. Please fix either your code or your question.arrayof numbers 9 to 12 ?