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07:12
<
baader >
# let d x = (x,x) ;;
07:12
<
baader >
val d : 'a -> 'a * 'a = <fun>
07:12
<
baader >
# d 2,3 ;;
07:12
<
baader >
- : (int * int) * int = ((2, 2), 3)
07:12
<
baader >
# d (2,3) ;;
07:12
<
baader >
- : (int * int) * (int * int) = ((2, 3), (2, 3))
07:13
<
baader >
q: why does d 2,3 think 2 is a tuple
07:53
<
mellum >
why do you think d thinks 2 is a tuple? I don't see that.
07:57
<
baader >
d takes an x and returns a tuple x * x, right ?
07:59
<
baader >
i'm not exactly sure about the difference of what d returns when called with d 2,3 and d (2,3)
07:59
<
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)
08:05
<
mellum >
It isn't. It's ((2, 3), (2, 3)).
08:05
<
mellum >
Which makes perfect sense to me.
08:06
<
mellum >
and d 2,3 is parsed as (d 2), 3, like with all infix operators
08:07
<
baader >
i do understand d (2,3), it returns what i'd expect it to
08:08
<
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)
08:08
<
mellum >
and therefore d 2, 3 is (d 2), 3
08:08
<
mrvn >
baader: I allways use () for tuples to make them more readable. Avoids all those precedence problem too.
08:13
<
baader >
your explanation makes sense, mellum
08:15
<
baader >
# d 2,(5,6,7);;
08:15
<
baader >
- : (int * int) * (int * int * int) = ((2, 2), (5, 6, 7))
08:15
<
baader >
so everything following d 2, .... is applied afterwards, right ?
08:16
<
baader >
yes, i see now
08:18
* mellum
ticks "good deed for Aug, 12"
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