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references #8088
Comments
Comment author: administrator From: christophe.poucet@pandora.be
This is not a bug: the semantics of monomorphic free type variables is But the next version of the compiler is able to do backtracking in the let x = ref[];;val x : '_a list ref = {contents = []} x := (function y -> y)::!x;;
x;;
x := (function f -> f f)::!x;;This expression has type 'a -> 'b but is here used with type 'a x;;
So this non-bug is already fixed :-) Jacques |
Comment author: administrator [...]
The semantics of what you call ``monomorphic free type variables'' is [...]
So may be people will no more ask about those type variables (as Best regards, Pierre Weis INRIA, Projet Cristal, Pierre.Weis@inria.fr, http://pauillac.inria.fr/~weis/ |
Comment author: administrator Works as expected in 3.06+27 |
Original bug ID: 1622
Reporter: administrator
Status: closed
Resolution: fixed
Priority: normal
Severity: minor
Category: ~DO NOT USE (was: OCaml general)
Bug description
Full_Name: Christophe Poucet
Version: 3.06
OS: Windows
Submission from: ncasse01.telenet-ops.be (213.224.83.38)
When using ref's sometimes it will change the type specification of the
monomorphic type yet not add the given param, perhaps this will explain better:
let x = ref[];;
val x : '_a list ref = {contents = []}
x := (function y -> y)::!x;;
x;;
x := (function f -> f f)::!x;;
Characters 22-23:
x := (function f -> f f)::!x;;
^
This expression has type 'a -> 'b but is here used with type 'a
x;;
It -does- change the type of x, yet it doesn't add the function to the list....
Is this a bug, or am I missing something?
Christophe Poucet
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