Ocaml - matching two lists
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I am trying to write a shuffle function in OCaml, but there is a problem with type inference. Merlin tells me that l1 and l2 are of type 'a list list, which is not true because they are only 'a list. Why does it claims that?
let shuffle l1 l2 =
let rec scan l1 l2 acc =
match (l1, l2) with
| , -> acc
| (,h2::t2) -> scan t2 h2::acc
| (h1::t1, ) -> scan t1 h1::acc
| (h1::t1,h2::t2) -> scan t1 t2 h1::h2::acc
in scan l1 l2
;;
compiler-errors ocaml type-inference operator-precedence
edited Nov 9 at 19:51
glennsl
8,823102545
asked Nov 9 at 19:42
mvxxx
31
add a comment |
up vote
0
down vote
favorite
I am trying to write a shuffle function in OCaml, but there is a problem with type inference. Merlin tells me that l1 and l2 are of type 'a list list, which is not true because they are only 'a list. Why does it claims that?
let shuffle l1 l2 =
let rec scan l1 l2 acc =
match (l1, l2) with
| , -> acc
| (,h2::t2) -> scan t2 h2::acc
| (h1::t1, ) -> scan t1 h1::acc
| (h1::t1,h2::t2) -> scan t1 t2 h1::h2::acc
in scan l1 l2
;;
compiler-errors ocaml type-inference operator-precedence
edited Nov 9 at 19:51
glennsl
8,823102545
asked Nov 9 at 19:42
mvxxx
31
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to write a shuffle function in OCaml, but there is a problem with type inference. Merlin tells me that l1 and l2 are of type 'a list list, which is not true because they are only 'a list. Why does it claims that?
let shuffle l1 l2 =
let rec scan l1 l2 acc =
match (l1, l2) with
| , -> acc
| (,h2::t2) -> scan t2 h2::acc
| (h1::t1, ) -> scan t1 h1::acc
| (h1::t1,h2::t2) -> scan t1 t2 h1::h2::acc
in scan l1 l2
;;
compiler-errors ocaml type-inference operator-precedence
edited Nov 9 at 19:51
glennsl
8,823102545
asked Nov 9 at 19:42
mvxxx
31
I am trying to write a shuffle function in OCaml, but there is a problem with type inference. Merlin tells me that l1 and l2 are of type 'a list list, which is not true because they are only 'a list. Why does it claims that?
let shuffle l1 l2 =
let rec scan l1 l2 acc =
match (l1, l2) with
| , -> acc
| (,h2::t2) -> scan t2 h2::acc
| (h1::t1, ) -> scan t1 h1::acc
| (h1::t1,h2::t2) -> scan t1 t2 h1::h2::acc
in scan l1 l2
;;
compiler-errors ocaml type-inference operator-precedence
compiler-errors ocaml type-inference operator-precedence
edited Nov 9 at 19:51
glennsl
8,823102545
asked Nov 9 at 19:42
mvxxx
31
edited Nov 9 at 19:51
glennsl
8,823102545
asked Nov 9 at 19:42
mvxxx
31
edited Nov 9 at 19:51
glennsl
8,823102545
edited Nov 9 at 19:51
glennsl
8,823102545
edited Nov 9 at 19:51
glennsl
8,823102545
8,823102545
asked Nov 9 at 19:42
mvxxx
31
asked Nov 9 at 19:42
mvxxx
31
asked Nov 9 at 19:42
mvxxx
31
31
add a comment |
add a comment |
1 Answer
1
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1
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The root cause is that operator precedence isn't determined by your grouping by whitespace. That is, scan t2 h2::acc is interpreted as (scan t2 h2)::acc, not scan t2 (h2::acc), because function application has a higher precedence than ::. The fix is simply to add parentheses where appropriate.
See this table for the precedence and associativity of the different operators in OCaml.
answered Nov 9 at 19:48
glennsl
8,823102545
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The root cause is that operator precedence isn't determined by your grouping by whitespace. That is, scan t2 h2::acc is interpreted as (scan t2 h2)::acc, not scan t2 (h2::acc), because function application has a higher precedence than ::. The fix is simply to add parentheses where appropriate.
See this table for the precedence and associativity of the different operators in OCaml.
answered Nov 9 at 19:48
glennsl
8,823102545
add a comment |
up vote
1
down vote
accepted
The root cause is that operator precedence isn't determined by your grouping by whitespace. That is, scan t2 h2::acc is interpreted as (scan t2 h2)::acc, not scan t2 (h2::acc), because function application has a higher precedence than ::. The fix is simply to add parentheses where appropriate.
See this table for the precedence and associativity of the different operators in OCaml.
answered Nov 9 at 19:48
glennsl
8,823102545
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The root cause is that operator precedence isn't determined by your grouping by whitespace. That is, scan t2 h2::acc is interpreted as (scan t2 h2)::acc, not scan t2 (h2::acc), because function application has a higher precedence than ::. The fix is simply to add parentheses where appropriate.
See this table for the precedence and associativity of the different operators in OCaml.
answered Nov 9 at 19:48
glennsl
8,823102545
The root cause is that operator precedence isn't determined by your grouping by whitespace. That is, scan t2 h2::acc is interpreted as (scan t2 h2)::acc, not scan t2 (h2::acc), because function application has a higher precedence than ::. The fix is simply to add parentheses where appropriate.
See this table for the precedence and associativity of the different operators in OCaml.
answered Nov 9 at 19:48
glennsl
8,823102545
edited Nov 9 at 19:57
answered Nov 9 at 19:48
glennsl
8,823102545
answered Nov 9 at 19:48
glennsl
8,823102545
answered Nov 9 at 19:48
glennsl
8,823102545
8,823102545
add a comment |
add a comment |
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