Help me understand lambda expression precedence

Question

Looking at the Operator Precedence table, I'm confused by where lambda expressions fit in to all of this. Numbering the table 1 to 25 (lowest to highest precedence), I see the two key tokens used for composing lambda expressions: fun is non-associative 6, and -> is right-associative 8. That is, fun ... -> .... This might be my first mistake in understanding, since I'm not even sure if precedence plays a role in the body of a lambda expression (the right hand side of ->) I can only think of precedence playing a role in the entire expression, so I'll move on to there. According to this table, the sequence operator, ;, has lower precedence, 4, than the fun "operator". Which I believe says that

fun (x:int) -> x + 3 ; 2;;

should be equivalent to

((fun (x:int) -> (x + 3)) ; 2);;

yet it is actually equivalent to

(fun (x:int) -> (x + 3 ; 2));;

I appreciate any help clarifying my interpretation of the rules of precedence given by the referenced table vs. the actual rules of precedence for lambda expressions! Thanks!

Answer 1

MU

(If you are reading an operator precedence table, you've already lost. The table probably has bugs, but who cares, don't read the table. And don't use the sequence operator (;). And if you do use the sequence operator next to a lambda, then put the lambda in parentheses. Or just put every lambda in parentheses. I touch the code for the F# parser all the time, and I have no clue what the relative precedence of ; versus fun versus -> is, I don't even know what it means for -> to have a precedence (it is not an operator in expressions (only in types, where it is indeed right-associative)... I cannot imagine any satisfying answer to this question or any followup. You should have no interest in any of this unless you're writing your own parser/compiler for F#. These are not the droids you're looking for.)