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<baader> # let d x = (x,x) ;;
<baader> val d : 'a -> 'a * 'a = <fun>
<baader> # d 2,3 ;;
<baader>
<baader> - : (int * int) * int = ((2, 2), 3)
<baader> # d (2,3) ;;
<baader>
<baader> - : (int * int) * (int * int) = ((2, 3), (2, 3))
<baader>
<baader> q: why does d 2,3 think 2 is a tuple
<mellum> why do you think d thinks 2 is a tuple? I don't see that.
<baader> d takes an x and returns a tuple x * x, right ?
<baader> i'm not exactly sure about the difference of what d returns when called with d 2,3 and d (2,3)
<baader> d (2,3) seems to make sense to me, but d 2,3 ... i'm not sure why the result is ((2,2), 3)
<mellum> It isn't. It's ((2, 3), (2, 3)).
<mellum> Which makes perfect sense to me.
<mellum> and d 2,3 is parsed as (d 2), 3, like with all infix operators
<baader> i do understand d (2,3), it returns what i'd expect it to
<mellum> Well, function application just has higher precedence than infix operators. Like in math. sin x + cos x is not sin (x + cos x), but (sin x) + (cos x)
<mellum> and therefore d 2, 3 is (d 2), 3
<mrvn> baader: I allways use () for tuples to make them more readable. Avoids all those precedence problem too.
<baader> yes
<baader> your explanation makes sense, mellum
<baader> # d 2,(5,6,7);;
<baader>
<baader> - : (int * int) * (int * int * int) = ((2, 2), (5, 6, 7))
<baader> so everything following d 2, .... is applied afterwards, right ?
<mrvn> yes
<baader> :)
<mrvn> d 2, d 2
<baader> yes, i see now
<baader> thanks
* mellum ticks "good deed for Aug, 12"
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