00:14 <discocaml> <uberpyro181> Can anyone confirm if OCaml's implementation of first-class polymorphism in record types is based on these [TACS97](<https://link.springer.com/chapter/10.1007/BFb0014546>) and [IC99](<https://www.sciencedirect.com/science/article/pii/S0890540199928304>) papers?

00:14 <discocaml> <uberpyro181> i.e., this system where the nominal data definition doubles as an annotation for the quanitier, and then there's no additional annotations required beyond that

00:16 <discocaml> <uberpyro181> i.e., this system where the nominal data definition doubles as an annotation for the quantifier, and then there's no additional annotations required beyond that

18:15 <discocaml> <octachron> Yes, with some simplifications if my memory is correct.

23:33 <discocaml> <dubious245> Ah so `Float.pi` exist but `Float.e` doesn't, that is interesting.

23:33 <discocaml> <Kali> `Float.exp 1.`

23:35 <discocaml> <Kali> like many programming languages, rather than providing a constant e ocaml provides a function that calculates e^x, so to get the value of e you can do e^1 = `Float.exp 1.`

23:35 <discocaml> <dubious245> Oh look at that it worked.

23:35 <discocaml> <dubious245> Thank you.

23:36 <discocaml> <Kali> like many programming languages, rather than providing a constant e ocaml provides a function that calculates e^x ("the exponential function"), so to get the value of e you can do e^1 = `Float.exp 1.`

23:36 <discocaml> <Kali> like many programming languages, rather than providing a constant e ocaml provides a function that calculates e^x ("the exponential function", called exp() for short), so to get the value of e you can do e^1 = `Float.exp 1.`

23:37 <discocaml> <Kali> like many programming languages, rather than providing a constant e ocaml provides a function that calculates e^x ("the exponential function", called exp() for short in math), so to get the value of e you can do e^1 = `Float.exp 1.`

23:38 <discocaml> <Kali> oops, forgot what channel i was in, sorry about the edits

23:38 <discocaml> <Kali> irc users

23:39 <discocaml> <dubious245> Oh yeah wasn't asking in beginners. I originally solved it by copy and pasted the number in from google as a literal.