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Closed 10 years ago.
Today I had a really hard time programming in ruby. I came upon this question, and I was really stumped.
Print a sequence of number pairs "x,y" such that x ranges from 0 to 3 and y from 5 to 7 *
I already submitted my answer, but I'm not quite sure if I got it right. I'd just like my question answered now for the sake of understanding.
Something simple like
p [*0..3].product([*5..7])
0.upto 3 do |x|
5.upto 7 {|y| puts x, y}
end
Related
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Closed 10 years ago.
I am experimenting on arrays in Ruby. I cannot access the maximum nested array count. Example
experiment is shown below:
array = [[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[[]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]]
So, do you know a limit of dimensions of array in Ruby?
The limit is when you run out of memory.
Per commenter #HunterMcMillen's idea, you could test it like so:
# Warning: be prepared for a sad computer!
depth=0; arr=[]; a=arr
begin
while true do
b = []
a.push(b)
a = b
depth += 1
end
rescue => e
puts "OK: depth=#{depth}, error=#{e} (#{e.class})"
end
Presumably at some point it will raise some error (e.g. out of memory) and you'll see how many arrays you were able to nest.
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Closed 10 years ago.
I wrote this code:
def sumeven (i)
result = 0
while i < 100
if i % 2 == 0
result += i
end
i += 1
end
result
end
How do I give the i for the equation? When I run the equation in the terminal (mac), I get no output because I'm unable to figure out how to substitute the i for a number. Substituting the "i" with a number in the text editor still gives no output.
Try:
def sumeven(i)
# method implementation...
end
# call the method, passing it an argument
sumeven(4)
If this actually answers your question, may I suggest taking a tiny step back and reading a book on ruby before you get too much farther? You are asking why the car doesn't go before you know how to use the gas pedal.
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Closed 10 years ago.
Is there any method available in ruby 1.8.6 for calculating n values n power?
Like 2 has 5 power then answer is 32 and 5 has 3 power then answer is 125?
2**5 will give you the result 32.
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Closed 10 years ago.
I have this regex:
diff.\*\n.\*\n.\*\n.\*\n.\*\n
I want to match everything from the word "diff" to the fifth new line. Can someone simplify it for me?
You need grouping:
diff(?:.*\n){5}
I think you are asking for this:
/diff(.*\n){5}/
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Closed 11 years ago.
Mathematica 8.0.1 on Mac OS X 10.7.2 does this: Quit the kernel and execute
MathieuS[MathieuCharacteristicA[ 1, -(1/4)], -0.25`, 15.707963267948966`]
MathieuS[MathieuCharacteristicA[ 1, -(1/4)], -0.25`, 15.707963267948966`]
(*
5.10119 10^-15
MathieuS[MathieuCharacteristicA[1, -(1/4)], -0.25, 15.708]
*)
That is, the first time it numerically evaluates the expression, while the second time it returns the unevaluated form.
Why? Or have I just spent too long staring at my screen and am doing something stupid?