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vrom911
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chapter1
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| let digit1 = mod (abs n) 10 | ||
| digit2 = mod (div (abs n) 10) 10 |
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That is a wonderful solution! 👏🏼 You correctly noticed that it is the div and mod, cool 😎
One hint to make your solution even shorter: you can see that you use both:
mod m 10
div m 10The standard library has the divMod function, that actually combines inside both div and mod. And this is exactly what you use!.
So you could write it this way:
(x, y) = divMod m 10You can see how we could pattern match on the pair 🙂
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Hmm...
I fail to see how this would be simpler - I need to apply the modulo-division by 10 to the result of the first division whereas divMod only applies it to the original value albeit as div and as mod at the same time.
| firstDigit n = error "firstDigit: Not implemented!" | ||
| firstDigit :: Int -> Int | ||
| firstDigit n | ||
| | abs n < 10 = n |
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In case of n being negative, this would return negative result. as you are using n here 🙂
But you are on the right way! 💪🏼
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Ouch! I ran make test-chapter1-basic only, and hence missed the failure - should have caught this :-/
Co-authored-by: Veronika Romashkina <vrom911@gmail.com>
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Thank you very much for the elaborate feedback - you are putting a lot of thoughts and work into this 🙇 🙏 |
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Solutions for Chapter 1
cc @vrom911 @chshersh
This was real fun - thx a lot 🙇♂️