Mixing
How does Haskell evaluate this signature?
ggt_euklid :: Nat1 -> (Nat1 -> Nat1)
I am trying to learn partial application, I know that in this case, if the parentheses would be left out, I would get the same result, but I do not know how this signature should be evaluated.
As far as I have understood, parentheses signify that it is a function? Would that not imply that ggt_euklid takes a value Nat1 and returns a function?
Below is the complete function:
ggt_euklid x y
| x == y = x
|x>y =ggt_euklid(x-y) y
|x<y =ggt_euklid x (y-x)
haskell signature currying partial-application
asked Nov 21 '18 at 23:17
MinimaxMinimax
53312
add a comment |
ggt_euklid :: Nat1 -> (Nat1 -> Nat1)
I am trying to learn partial application, I know that in this case, if the parentheses would be left out, I would get the same result, but I do not know how this signature should be evaluated.
As far as I have understood, parentheses signify that it is a function? Would that not imply that ggt_euklid takes a value Nat1 and returns a function?
Below is the complete function:
ggt_euklid x y
| x == y = x
|x>y =ggt_euklid(x-y) y
|x<y =ggt_euklid x (y-x)
haskell signature currying partial-application
asked Nov 21 '18 at 23:17
MinimaxMinimax
53312
3
"parentheses signify that it is a function?" No, the->signifies that it's a function.
– sepp2k
Nov 21 '18 at 23:36
Possible duplicate of Example of deep understanding of currying
– jberryman
Nov 21 '18 at 23:54
add a comment |
ggt_euklid :: Nat1 -> (Nat1 -> Nat1)
I am trying to learn partial application, I know that in this case, if the parentheses would be left out, I would get the same result, but I do not know how this signature should be evaluated.
As far as I have understood, parentheses signify that it is a function? Would that not imply that ggt_euklid takes a value Nat1 and returns a function?
Below is the complete function:
ggt_euklid x y
| x == y = x
|x>y =ggt_euklid(x-y) y
|x<y =ggt_euklid x (y-x)
haskell signature currying partial-application
asked Nov 21 '18 at 23:17
MinimaxMinimax
53312
ggt_euklid :: Nat1 -> (Nat1 -> Nat1)
I am trying to learn partial application, I know that in this case, if the parentheses would be left out, I would get the same result, but I do not know how this signature should be evaluated.
As far as I have understood, parentheses signify that it is a function? Would that not imply that ggt_euklid takes a value Nat1 and returns a function?
Below is the complete function:
ggt_euklid x y
| x == y = x
|x>y =ggt_euklid(x-y) y
|x<y =ggt_euklid x (y-x)
haskell signature currying partial-application
haskell signature currying partial-application
asked Nov 21 '18 at 23:17
MinimaxMinimax
53312
asked Nov 21 '18 at 23:17
MinimaxMinimax
53312
asked Nov 21 '18 at 23:17
MinimaxMinimax
53312
asked Nov 21 '18 at 23:17
MinimaxMinimax
53312
asked Nov 21 '18 at 23:17
MinimaxMinimax
53312
53312
3
"parentheses signify that it is a function?" No, the->signifies that it's a function.
– sepp2k
Nov 21 '18 at 23:36
Possible duplicate of Example of deep understanding of currying
– jberryman
Nov 21 '18 at 23:54
add a comment |
3
"parentheses signify that it is a function?" No, the->signifies that it's a function.
– sepp2k
Nov 21 '18 at 23:36
Possible duplicate of Example of deep understanding of currying
– jberryman
Nov 21 '18 at 23:54
3
3
"parentheses signify that it is a function?" No, the
-> signifies that it's a function.– sepp2k
Nov 21 '18 at 23:36
"parentheses signify that it is a function?" No, the
-> signifies that it's a function.– sepp2k
Nov 21 '18 at 23:36
Possible duplicate of Example of deep understanding of currying
– jberryman
Nov 21 '18 at 23:54
Possible duplicate of Example of deep understanding of currying
– jberryman
Nov 21 '18 at 23:54
add a comment |
2 Answers
2
active
oldest
votes
Would that not imply that ggt_euklid takes a value Nat1 and returns a
function?
No, it still imply that ggt_euklid takes one argument of type Nat1 and return a function of type Nat1->Nat1, even though, the parentheses be left out.
The Arrow -> always be right-associativity (when no parentheses), i.e.:
Nat1 -> Nat1 -> Nat1
is equivalent to
Nat1 -> (Nat1 -> Nat1)
which is corresponding to the function application always be left-associativity. (when no parentheses) for example:
ggt_euklid 1 2
is equivalent to
(ggt_euklid 1) 2
Here
(ggt_euklid 1) ~ Nat1 -> Nat1
and
(ggt_euklid 1) 2 ~ Nat1
So, no matter whether one or two arguments apply to ggt_euklid, it always return a function of type Nat1 -> Nat1 firstly, if second argument is provided, it applies second argument to the returned function.
answered Nov 22 '18 at 10:26
assembly.jcassembly.jc
2,0891214
add a comment |
You have understood the type signature correctly: it takes one argument and returns a function. That's how "multi-argument" functions in Haskell work: through currying. You can see this in action by trying this equivalent implementation:
ggt_euklid :: Nat1 -> (Nat1 -> Nat1)
ggt_euklid x = y -> result
where result | x == y = x
| x > y = ggt_euklid (x-y) y
| x < y = ggt_euklid x (y-x)
Here I've introduced this fairly pointless result variable as a thing to use your pattern guards on, but the idea is the same.
answered Nov 21 '18 at 23:24
amalloyamalloy
59.7k5105156
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Would that not imply that ggt_euklid takes a value Nat1 and returns a
function?
No, it still imply that ggt_euklid takes one argument of type Nat1 and return a function of type Nat1->Nat1, even though, the parentheses be left out.
The Arrow -> always be right-associativity (when no parentheses), i.e.:
Nat1 -> Nat1 -> Nat1
is equivalent to
Nat1 -> (Nat1 -> Nat1)
which is corresponding to the function application always be left-associativity. (when no parentheses) for example:
ggt_euklid 1 2
is equivalent to
(ggt_euklid 1) 2
Here
(ggt_euklid 1) ~ Nat1 -> Nat1
and
(ggt_euklid 1) 2 ~ Nat1
So, no matter whether one or two arguments apply to ggt_euklid, it always return a function of type Nat1 -> Nat1 firstly, if second argument is provided, it applies second argument to the returned function.
answered Nov 22 '18 at 10:26
assembly.jcassembly.jc
2,0891214
add a comment |
Would that not imply that ggt_euklid takes a value Nat1 and returns a
function?
No, it still imply that ggt_euklid takes one argument of type Nat1 and return a function of type Nat1->Nat1, even though, the parentheses be left out.
The Arrow -> always be right-associativity (when no parentheses), i.e.:
Nat1 -> Nat1 -> Nat1
is equivalent to
Nat1 -> (Nat1 -> Nat1)
which is corresponding to the function application always be left-associativity. (when no parentheses) for example:
ggt_euklid 1 2
is equivalent to
(ggt_euklid 1) 2
Here
(ggt_euklid 1) ~ Nat1 -> Nat1
and
(ggt_euklid 1) 2 ~ Nat1
So, no matter whether one or two arguments apply to ggt_euklid, it always return a function of type Nat1 -> Nat1 firstly, if second argument is provided, it applies second argument to the returned function.
answered Nov 22 '18 at 10:26
assembly.jcassembly.jc
2,0891214
add a comment |
Would that not imply that ggt_euklid takes a value Nat1 and returns a
function?
No, it still imply that ggt_euklid takes one argument of type Nat1 and return a function of type Nat1->Nat1, even though, the parentheses be left out.
The Arrow -> always be right-associativity (when no parentheses), i.e.:
Nat1 -> Nat1 -> Nat1
is equivalent to
Nat1 -> (Nat1 -> Nat1)
which is corresponding to the function application always be left-associativity. (when no parentheses) for example:
ggt_euklid 1 2
is equivalent to
(ggt_euklid 1) 2
Here
(ggt_euklid 1) ~ Nat1 -> Nat1
and
(ggt_euklid 1) 2 ~ Nat1
So, no matter whether one or two arguments apply to ggt_euklid, it always return a function of type Nat1 -> Nat1 firstly, if second argument is provided, it applies second argument to the returned function.
answered Nov 22 '18 at 10:26
assembly.jcassembly.jc
2,0891214
Would that not imply that ggt_euklid takes a value Nat1 and returns a
function?
No, it still imply that ggt_euklid takes one argument of type Nat1 and return a function of type Nat1->Nat1, even though, the parentheses be left out.
The Arrow -> always be right-associativity (when no parentheses), i.e.:
Nat1 -> Nat1 -> Nat1
is equivalent to
Nat1 -> (Nat1 -> Nat1)
which is corresponding to the function application always be left-associativity. (when no parentheses) for example:
ggt_euklid 1 2
is equivalent to
(ggt_euklid 1) 2
Here
(ggt_euklid 1) ~ Nat1 -> Nat1
and
(ggt_euklid 1) 2 ~ Nat1
So, no matter whether one or two arguments apply to ggt_euklid, it always return a function of type Nat1 -> Nat1 firstly, if second argument is provided, it applies second argument to the returned function.
answered Nov 22 '18 at 10:26
assembly.jcassembly.jc
2,0891214
edited Nov 24 '18 at 15:29
answered Nov 22 '18 at 10:26
assembly.jcassembly.jc
2,0891214
answered Nov 22 '18 at 10:26
assembly.jcassembly.jc
2,0891214
answered Nov 22 '18 at 10:26
assembly.jcassembly.jc
2,0891214
2,0891214
add a comment |
add a comment |
You have understood the type signature correctly: it takes one argument and returns a function. That's how "multi-argument" functions in Haskell work: through currying. You can see this in action by trying this equivalent implementation:
ggt_euklid :: Nat1 -> (Nat1 -> Nat1)
ggt_euklid x = y -> result
where result | x == y = x
| x > y = ggt_euklid (x-y) y
| x < y = ggt_euklid x (y-x)
Here I've introduced this fairly pointless result variable as a thing to use your pattern guards on, but the idea is the same.
answered Nov 21 '18 at 23:24
amalloyamalloy
59.7k5105156
add a comment |
You have understood the type signature correctly: it takes one argument and returns a function. That's how "multi-argument" functions in Haskell work: through currying. You can see this in action by trying this equivalent implementation:
ggt_euklid :: Nat1 -> (Nat1 -> Nat1)
ggt_euklid x = y -> result
where result | x == y = x
| x > y = ggt_euklid (x-y) y
| x < y = ggt_euklid x (y-x)
Here I've introduced this fairly pointless result variable as a thing to use your pattern guards on, but the idea is the same.
answered Nov 21 '18 at 23:24
amalloyamalloy
59.7k5105156
add a comment |
You have understood the type signature correctly: it takes one argument and returns a function. That's how "multi-argument" functions in Haskell work: through currying. You can see this in action by trying this equivalent implementation:
ggt_euklid :: Nat1 -> (Nat1 -> Nat1)
ggt_euklid x = y -> result
where result | x == y = x
| x > y = ggt_euklid (x-y) y
| x < y = ggt_euklid x (y-x)
Here I've introduced this fairly pointless result variable as a thing to use your pattern guards on, but the idea is the same.
answered Nov 21 '18 at 23:24
amalloyamalloy
59.7k5105156
You have understood the type signature correctly: it takes one argument and returns a function. That's how "multi-argument" functions in Haskell work: through currying. You can see this in action by trying this equivalent implementation:
ggt_euklid :: Nat1 -> (Nat1 -> Nat1)
ggt_euklid x = y -> result
where result | x == y = x
| x > y = ggt_euklid (x-y) y
| x < y = ggt_euklid x (y-x)
Here I've introduced this fairly pointless result variable as a thing to use your pattern guards on, but the idea is the same.
answered Nov 21 '18 at 23:24
amalloyamalloy
59.7k5105156
answered Nov 21 '18 at 23:24
amalloyamalloy
59.7k5105156
answered Nov 21 '18 at 23:24
amalloyamalloy
59.7k5105156
answered Nov 21 '18 at 23:24
amalloyamalloy
59.7k5105156
59.7k5105156
add a comment |
add a comment |
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3
"parentheses signify that it is a function?" No, the
->signifies that it's a function.– sepp2k
Nov 21 '18 at 23:36
Possible duplicate of Example of deep understanding of currying
– jberryman
Nov 21 '18 at 23:54