F# function, how it really works
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I got function like this one.
let c = (fun a b ‐> let d = a 10 in d ‐ 4*b) (fun c ‐> c + c) 5
let x = c‐6
That's what I understand so far:
"fun c ‐> c + c" is first argument for "fun a b" and "5" is second.
"let d" is kind a function that takes "a" and "b" and returns the result.
But if somebody could explain what exactly happened in:
let d = a 10 in d ‐ 4*b
f#
add a comment |
I got function like this one.
let c = (fun a b ‐> let d = a 10 in d ‐ 4*b) (fun c ‐> c + c) 5
let x = c‐6
That's what I understand so far:
"fun c ‐> c + c" is first argument for "fun a b" and "5" is second.
"let d" is kind a function that takes "a" and "b" and returns the result.
But if somebody could explain what exactly happened in:
let d = a 10 in d ‐ 4*b
f#
add a comment |
I got function like this one.
let c = (fun a b ‐> let d = a 10 in d ‐ 4*b) (fun c ‐> c + c) 5
let x = c‐6
That's what I understand so far:
"fun c ‐> c + c" is first argument for "fun a b" and "5" is second.
"let d" is kind a function that takes "a" and "b" and returns the result.
But if somebody could explain what exactly happened in:
let d = a 10 in d ‐ 4*b
f#
I got function like this one.
let c = (fun a b ‐> let d = a 10 in d ‐ 4*b) (fun c ‐> c + c) 5
let x = c‐6
That's what I understand so far:
"fun c ‐> c + c" is first argument for "fun a b" and "5" is second.
"let d" is kind a function that takes "a" and "b" and returns the result.
But if somebody could explain what exactly happened in:
let d = a 10 in d ‐ 4*b
f#
f#
asked Dec 31 '18 at 12:07
silvanestisilvanesti
133
133
add a comment |
add a comment |
1 Answer
1
active
oldest
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in
is used here as part of verbose syntax. You can rewrite it in leightweight syntax to make it a bit more readable. It would be something like that:
fun a b =
let d = a 10
d - 4 * b
a
is invoked with 10
as argument and the results is assigned to d
. Later on d - 4 * b
is calculated and returned from the function.
For the example you have a
is fun c -> c + c
so invoking it with 10
returns 20
.
20 - 4 * 5 = 0
so c
is set to 0
and x
will be 0 - 6 = -6
.
From F# interactive:
> let c = (fun a b -> let d = a 10 in d - 4*b) (fun c -> c + c) 5;;
val c : int = 0
I did not consider the possibility that "a" could be a function and takes "10" as an argument. Thanks, that really helped me.
– silvanesti
Dec 31 '18 at 12:57
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
in
is used here as part of verbose syntax. You can rewrite it in leightweight syntax to make it a bit more readable. It would be something like that:
fun a b =
let d = a 10
d - 4 * b
a
is invoked with 10
as argument and the results is assigned to d
. Later on d - 4 * b
is calculated and returned from the function.
For the example you have a
is fun c -> c + c
so invoking it with 10
returns 20
.
20 - 4 * 5 = 0
so c
is set to 0
and x
will be 0 - 6 = -6
.
From F# interactive:
> let c = (fun a b -> let d = a 10 in d - 4*b) (fun c -> c + c) 5;;
val c : int = 0
I did not consider the possibility that "a" could be a function and takes "10" as an argument. Thanks, that really helped me.
– silvanesti
Dec 31 '18 at 12:57
add a comment |
in
is used here as part of verbose syntax. You can rewrite it in leightweight syntax to make it a bit more readable. It would be something like that:
fun a b =
let d = a 10
d - 4 * b
a
is invoked with 10
as argument and the results is assigned to d
. Later on d - 4 * b
is calculated and returned from the function.
For the example you have a
is fun c -> c + c
so invoking it with 10
returns 20
.
20 - 4 * 5 = 0
so c
is set to 0
and x
will be 0 - 6 = -6
.
From F# interactive:
> let c = (fun a b -> let d = a 10 in d - 4*b) (fun c -> c + c) 5;;
val c : int = 0
I did not consider the possibility that "a" could be a function and takes "10" as an argument. Thanks, that really helped me.
– silvanesti
Dec 31 '18 at 12:57
add a comment |
in
is used here as part of verbose syntax. You can rewrite it in leightweight syntax to make it a bit more readable. It would be something like that:
fun a b =
let d = a 10
d - 4 * b
a
is invoked with 10
as argument and the results is assigned to d
. Later on d - 4 * b
is calculated and returned from the function.
For the example you have a
is fun c -> c + c
so invoking it with 10
returns 20
.
20 - 4 * 5 = 0
so c
is set to 0
and x
will be 0 - 6 = -6
.
From F# interactive:
> let c = (fun a b -> let d = a 10 in d - 4*b) (fun c -> c + c) 5;;
val c : int = 0
in
is used here as part of verbose syntax. You can rewrite it in leightweight syntax to make it a bit more readable. It would be something like that:
fun a b =
let d = a 10
d - 4 * b
a
is invoked with 10
as argument and the results is assigned to d
. Later on d - 4 * b
is calculated and returned from the function.
For the example you have a
is fun c -> c + c
so invoking it with 10
returns 20
.
20 - 4 * 5 = 0
so c
is set to 0
and x
will be 0 - 6 = -6
.
From F# interactive:
> let c = (fun a b -> let d = a 10 in d - 4*b) (fun c -> c + c) 5;;
val c : int = 0
answered Dec 31 '18 at 12:16
MarcinJuraszekMarcinJuraszek
108k10138215
108k10138215
I did not consider the possibility that "a" could be a function and takes "10" as an argument. Thanks, that really helped me.
– silvanesti
Dec 31 '18 at 12:57
add a comment |
I did not consider the possibility that "a" could be a function and takes "10" as an argument. Thanks, that really helped me.
– silvanesti
Dec 31 '18 at 12:57
I did not consider the possibility that "a" could be a function and takes "10" as an argument. Thanks, that really helped me.
– silvanesti
Dec 31 '18 at 12:57
I did not consider the possibility that "a" could be a function and takes "10" as an argument. Thanks, that really helped me.
– silvanesti
Dec 31 '18 at 12:57
add a comment |
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