analysis of expression

analysis of expression

[Michel Levy]

Dans la définition de expr, on trouve :
expr := expr expr
	| expr infix-op expr
Aussi la phrase "let x = 1 in x-1" peut-être analysée de 2 manières 
1) La manière correcte (celle de Ocaml) : - est une opération infixée
entre x et 1
2) x est appliqué à -1 : d'où une erreur de types
Où est-il indiqué dans le "manuel de référence Ocaml", la solution
choisie ?

In definition of expr I read :
expr := expr expr
	| expr infix-op expr
with that syntax its possible to have two analysis of the sentence "let
x = 1 in x-1" ?
1) The correct analysis (the Ocaml anlysis) : - is an operation between
x et 1
2) x is applied to -1 and we find a type error
Where it's possible to find in the Ocaml reference the choosen solution
?

	Thank you
-- 
Michel Levy  
D106 - L.S.R.     B.P.72 - 38042 SAINT MARTIN D'HERES CEDEX - France
e.mail : Michel.Levy@imag.fr    tel :(33)476827246
http://www-lsr.imag.fr/Les.Personnes/Michel.Levy

Re: analysis of expression

[Xavier Leroy]

> In definition of expr I read :
> expr := expr expr
> 	| expr infix-op expr
> with that syntax its possible to have two analysis of the sentence "let
> x = 1 in x-1" ?
> 1) The correct analysis (the Ocaml anlysis) : - is an operation between
> x et 1
> 2) x is applied to -1 and we find a type error
> Where it's possible to find in the Ocaml reference the choosen solution ?

It's not very explicit, I agree, but the table of operator precedences
for expressions says that function application has higher precedence
than unary minus.  So, "x-1" cannot be parsed as "x applied to -1"
since this would violate the precedences.  And of course "x-1" cannot
be parsed as "(x applied to -) applied to 1" because "-" in itself is
not a valid expression.  This leaves "x binary minus 1" as the only
legal parsing.

- Xavier Leroy