Taking sum of list in Haskell
I am trying to take the sum of a list in Haskell but it gives the error, please see the below code.
binListToDec :: [Int] -> Int
binListToDec (x:xs) = if length binListToDec == 0 then 1
else x + binListToDec xs
It gives the following error
* No instance for (Foldable ((->) [Int]))
arising from a use of `length'
* In the first argument of `(==)', namely `length binListToDec'
In the expression: length binListToDec == 0
In the expression:
if length binListToDec == 0 then 1 else x + binListToDec xs
|
2 | binListToDec (x:xs) = if length binListToDec == 0 then 1
list haskell
asked Nov 13 '18 at 18:10
Muhammad UsmanMuhammad Usman
67118
add a comment |
I am trying to take the sum of a list in Haskell but it gives the error, please see the below code.
binListToDec :: [Int] -> Int
binListToDec (x:xs) = if length binListToDec == 0 then 1
else x + binListToDec xs
It gives the following error
* No instance for (Foldable ((->) [Int]))
arising from a use of `length'
* In the first argument of `(==)', namely `length binListToDec'
In the expression: length binListToDec == 0
In the expression:
if length binListToDec == 0 then 1 else x + binListToDec xs
|
2 | binListToDec (x:xs) = if length binListToDec == 0 then 1
list haskell
asked Nov 13 '18 at 18:10
Muhammad UsmanMuhammad Usman
67118
What islength binListToDecsupposed to do?
– Willem Van Onsem
Nov 13 '18 at 18:14
take the length of the string passed to the function. and if its 0 ( empty) then return 1 as the sum @WillemVanOnsem
– Muhammad Usman
Nov 13 '18 at 18:15
5
but (a) the list you pass is(x:xs)and (b) the length can never be zero, since(x:xs)is the pattern for a non-empty list.
– Willem Van Onsem
Nov 13 '18 at 18:16
add a comment |
I am trying to take the sum of a list in Haskell but it gives the error, please see the below code.
binListToDec :: [Int] -> Int
binListToDec (x:xs) = if length binListToDec == 0 then 1
else x + binListToDec xs
It gives the following error
* No instance for (Foldable ((->) [Int]))
arising from a use of `length'
* In the first argument of `(==)', namely `length binListToDec'
In the expression: length binListToDec == 0
In the expression:
if length binListToDec == 0 then 1 else x + binListToDec xs
|
2 | binListToDec (x:xs) = if length binListToDec == 0 then 1
list haskell
asked Nov 13 '18 at 18:10
Muhammad UsmanMuhammad Usman
67118
I am trying to take the sum of a list in Haskell but it gives the error, please see the below code.
binListToDec :: [Int] -> Int
binListToDec (x:xs) = if length binListToDec == 0 then 1
else x + binListToDec xs
It gives the following error
* No instance for (Foldable ((->) [Int]))
arising from a use of `length'
* In the first argument of `(==)', namely `length binListToDec'
In the expression: length binListToDec == 0
In the expression:
if length binListToDec == 0 then 1 else x + binListToDec xs
|
2 | binListToDec (x:xs) = if length binListToDec == 0 then 1
list haskell
list haskell
asked Nov 13 '18 at 18:10
Muhammad UsmanMuhammad Usman
67118
asked Nov 13 '18 at 18:10
Muhammad UsmanMuhammad Usman
67118
asked Nov 13 '18 at 18:10
Muhammad UsmanMuhammad Usman
67118
asked Nov 13 '18 at 18:10
Muhammad UsmanMuhammad Usman
67118
asked Nov 13 '18 at 18:10
Muhammad UsmanMuhammad Usman
67118
67118
What islength binListToDecsupposed to do?
– Willem Van Onsem
Nov 13 '18 at 18:14
take the length of the string passed to the function. and if its 0 ( empty) then return 1 as the sum @WillemVanOnsem
– Muhammad Usman
Nov 13 '18 at 18:15
5
but (a) the list you pass is(x:xs)and (b) the length can never be zero, since(x:xs)is the pattern for a non-empty list.
– Willem Van Onsem
Nov 13 '18 at 18:16
add a comment |
What islength binListToDecsupposed to do?
– Willem Van Onsem
Nov 13 '18 at 18:14
take the length of the string passed to the function. and if its 0 ( empty) then return 1 as the sum @WillemVanOnsem
– Muhammad Usman
Nov 13 '18 at 18:15
5
but (a) the list you pass is(x:xs)and (b) the length can never be zero, since(x:xs)is the pattern for a non-empty list.
– Willem Van Onsem
Nov 13 '18 at 18:16
What is
length binListToDec supposed to do?– Willem Van Onsem
Nov 13 '18 at 18:14
What is
length binListToDec supposed to do?– Willem Van Onsem
Nov 13 '18 at 18:14
take the length of the string passed to the function. and if its 0 ( empty) then return 1 as the sum @WillemVanOnsem
– Muhammad Usman
Nov 13 '18 at 18:15
take the length of the string passed to the function. and if its 0 ( empty) then return 1 as the sum @WillemVanOnsem
– Muhammad Usman
Nov 13 '18 at 18:15
5
5
but (a) the list you pass is
(x:xs) and (b) the length can never be zero, since (x:xs) is the pattern for a non-empty list.– Willem Van Onsem
Nov 13 '18 at 18:16
but (a) the list you pass is
(x:xs) and (b) the length can never be zero, since (x:xs) is the pattern for a non-empty list.– Willem Van Onsem
Nov 13 '18 at 18:16
add a comment |
1 Answer
1
active
oldest
votes
Among the myriad ways you could write this, two possibilities would be
binListToDec xs = if length xs == 0 then 0 -- see below
else (head xs) + binListToDec (tail xs)
and
binListToDec = 0
binListToDec (x:xs) = x + binListToDec xs
You appear to be trying to combine bits of each. There's no way to write a single pattern that matches simultaneously 1) an empty list and 2) a non-empty list with 3) its head and tail matched separately.
xsmatches 1) and 2).
all@(x:xs)matches 2) and 3)
1) and 3) cannot be matched, because the pairing is nonsensical: an empty list doesn't have a separate head and tail.and(x:xs)match lists from two non-overlapping sets of possible list values.
Update: there is a lazy pattern match
all@(~(x:xs)). The tilde prevents the match(x:xs)from being attempted until there is a need to
evaluatexorxs. We think of
binListToDec all@(~(x:xs)) = if length all == 0 then 0 else x + binListToDec
as equivalent to
binListToDec all = if length all == 0
then 0
else let (x:xs) = all
in x + binListToDec
A lazy pattern match can still fail, but here we defer using
xandxsuntil we know it won't.
length binListToDec attempts to compute the length of the function itself, not the length of its argument, in your attempt. The correct argument for length is used above. Also, the generally accepted sum of an empty list is 0, not 1.
answered Nov 13 '18 at 18:22
chepnerchepner
248k33234329
all@(~(x:xs))meets all three criteria. ...but that's not to say I recommend it. Just that it's possible. e.g.binListToDec all@(~(x:xs)) = if length all == 0 then 0 else x + binListToDec xswould work, though it would be slow for all the same reasons your first solution is.
– Daniel Wagner
Nov 13 '18 at 21:37
@DanielWagner Thanks; I figured there was some bit of syntax I was forgetting about.
– chepner
Nov 13 '18 at 22:43
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
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oldest
votes
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oldest
votes
Among the myriad ways you could write this, two possibilities would be
binListToDec xs = if length xs == 0 then 0 -- see below
else (head xs) + binListToDec (tail xs)
and
binListToDec = 0
binListToDec (x:xs) = x + binListToDec xs
You appear to be trying to combine bits of each. There's no way to write a single pattern that matches simultaneously 1) an empty list and 2) a non-empty list with 3) its head and tail matched separately.
xsmatches 1) and 2).
all@(x:xs)matches 2) and 3)
1) and 3) cannot be matched, because the pairing is nonsensical: an empty list doesn't have a separate head and tail.and(x:xs)match lists from two non-overlapping sets of possible list values.
Update: there is a lazy pattern match
all@(~(x:xs)). The tilde prevents the match(x:xs)from being attempted until there is a need to
evaluatexorxs. We think of
binListToDec all@(~(x:xs)) = if length all == 0 then 0 else x + binListToDec
as equivalent to
binListToDec all = if length all == 0
then 0
else let (x:xs) = all
in x + binListToDec
A lazy pattern match can still fail, but here we defer using
xandxsuntil we know it won't.
length binListToDec attempts to compute the length of the function itself, not the length of its argument, in your attempt. The correct argument for length is used above. Also, the generally accepted sum of an empty list is 0, not 1.
answered Nov 13 '18 at 18:22
chepnerchepner
248k33234329
all@(~(x:xs))meets all three criteria. ...but that's not to say I recommend it. Just that it's possible. e.g.binListToDec all@(~(x:xs)) = if length all == 0 then 0 else x + binListToDec xswould work, though it would be slow for all the same reasons your first solution is.
– Daniel Wagner
Nov 13 '18 at 21:37
@DanielWagner Thanks; I figured there was some bit of syntax I was forgetting about.
– chepner
Nov 13 '18 at 22:43
add a comment |
Among the myriad ways you could write this, two possibilities would be
binListToDec xs = if length xs == 0 then 0 -- see below
else (head xs) + binListToDec (tail xs)
and
binListToDec = 0
binListToDec (x:xs) = x + binListToDec xs
You appear to be trying to combine bits of each. There's no way to write a single pattern that matches simultaneously 1) an empty list and 2) a non-empty list with 3) its head and tail matched separately.
xsmatches 1) and 2).
all@(x:xs)matches 2) and 3)
1) and 3) cannot be matched, because the pairing is nonsensical: an empty list doesn't have a separate head and tail.and(x:xs)match lists from two non-overlapping sets of possible list values.
Update: there is a lazy pattern match
all@(~(x:xs)). The tilde prevents the match(x:xs)from being attempted until there is a need to
evaluatexorxs. We think of
binListToDec all@(~(x:xs)) = if length all == 0 then 0 else x + binListToDec
as equivalent to
binListToDec all = if length all == 0
then 0
else let (x:xs) = all
in x + binListToDec
A lazy pattern match can still fail, but here we defer using
xandxsuntil we know it won't.
length binListToDec attempts to compute the length of the function itself, not the length of its argument, in your attempt. The correct argument for length is used above. Also, the generally accepted sum of an empty list is 0, not 1.
answered Nov 13 '18 at 18:22
chepnerchepner
248k33234329
all@(~(x:xs))meets all three criteria. ...but that's not to say I recommend it. Just that it's possible. e.g.binListToDec all@(~(x:xs)) = if length all == 0 then 0 else x + binListToDec xswould work, though it would be slow for all the same reasons your first solution is.
– Daniel Wagner
Nov 13 '18 at 21:37
@DanielWagner Thanks; I figured there was some bit of syntax I was forgetting about.
– chepner
Nov 13 '18 at 22:43
add a comment |
Among the myriad ways you could write this, two possibilities would be
binListToDec xs = if length xs == 0 then 0 -- see below
else (head xs) + binListToDec (tail xs)
and
binListToDec = 0
binListToDec (x:xs) = x + binListToDec xs
You appear to be trying to combine bits of each. There's no way to write a single pattern that matches simultaneously 1) an empty list and 2) a non-empty list with 3) its head and tail matched separately.
xsmatches 1) and 2).
all@(x:xs)matches 2) and 3)
1) and 3) cannot be matched, because the pairing is nonsensical: an empty list doesn't have a separate head and tail.and(x:xs)match lists from two non-overlapping sets of possible list values.
Update: there is a lazy pattern match
all@(~(x:xs)). The tilde prevents the match(x:xs)from being attempted until there is a need to
evaluatexorxs. We think of
binListToDec all@(~(x:xs)) = if length all == 0 then 0 else x + binListToDec
as equivalent to
binListToDec all = if length all == 0
then 0
else let (x:xs) = all
in x + binListToDec
A lazy pattern match can still fail, but here we defer using
xandxsuntil we know it won't.
length binListToDec attempts to compute the length of the function itself, not the length of its argument, in your attempt. The correct argument for length is used above. Also, the generally accepted sum of an empty list is 0, not 1.
answered Nov 13 '18 at 18:22
chepnerchepner
248k33234329
Among the myriad ways you could write this, two possibilities would be
binListToDec xs = if length xs == 0 then 0 -- see below
else (head xs) + binListToDec (tail xs)
and
binListToDec = 0
binListToDec (x:xs) = x + binListToDec xs
You appear to be trying to combine bits of each. There's no way to write a single pattern that matches simultaneously 1) an empty list and 2) a non-empty list with 3) its head and tail matched separately.
xsmatches 1) and 2).
all@(x:xs)matches 2) and 3)
1) and 3) cannot be matched, because the pairing is nonsensical: an empty list doesn't have a separate head and tail.and(x:xs)match lists from two non-overlapping sets of possible list values.
Update: there is a lazy pattern match
all@(~(x:xs)). The tilde prevents the match(x:xs)from being attempted until there is a need to
evaluatexorxs. We think of
binListToDec all@(~(x:xs)) = if length all == 0 then 0 else x + binListToDec
as equivalent to
binListToDec all = if length all == 0
then 0
else let (x:xs) = all
in x + binListToDec
A lazy pattern match can still fail, but here we defer using
xandxsuntil we know it won't.
length binListToDec attempts to compute the length of the function itself, not the length of its argument, in your attempt. The correct argument for length is used above. Also, the generally accepted sum of an empty list is 0, not 1.
answered Nov 13 '18 at 18:22
chepnerchepner
248k33234329
edited Nov 13 '18 at 22:43
answered Nov 13 '18 at 18:22
chepnerchepner
248k33234329
answered Nov 13 '18 at 18:22
chepnerchepner
248k33234329
answered Nov 13 '18 at 18:22
chepnerchepner
248k33234329
248k33234329
all@(~(x:xs))meets all three criteria. ...but that's not to say I recommend it. Just that it's possible. e.g.binListToDec all@(~(x:xs)) = if length all == 0 then 0 else x + binListToDec xswould work, though it would be slow for all the same reasons your first solution is.
– Daniel Wagner
Nov 13 '18 at 21:37
@DanielWagner Thanks; I figured there was some bit of syntax I was forgetting about.
– chepner
Nov 13 '18 at 22:43
add a comment |
all@(~(x:xs))meets all three criteria. ...but that's not to say I recommend it. Just that it's possible. e.g.binListToDec all@(~(x:xs)) = if length all == 0 then 0 else x + binListToDec xswould work, though it would be slow for all the same reasons your first solution is.
– Daniel Wagner
Nov 13 '18 at 21:37
@DanielWagner Thanks; I figured there was some bit of syntax I was forgetting about.
– chepner
Nov 13 '18 at 22:43
all@(~(x:xs)) meets all three criteria. ...but that's not to say I recommend it. Just that it's possible. e.g. binListToDec all@(~(x:xs)) = if length all == 0 then 0 else x + binListToDec xs would work, though it would be slow for all the same reasons your first solution is.– Daniel Wagner
Nov 13 '18 at 21:37
all@(~(x:xs)) meets all three criteria. ...but that's not to say I recommend it. Just that it's possible. e.g. binListToDec all@(~(x:xs)) = if length all == 0 then 0 else x + binListToDec xs would work, though it would be slow for all the same reasons your first solution is.– Daniel Wagner
Nov 13 '18 at 21:37
@DanielWagner Thanks; I figured there was some bit of syntax I was forgetting about.
– chepner
Nov 13 '18 at 22:43
@DanielWagner Thanks; I figured there was some bit of syntax I was forgetting about.
– chepner
Nov 13 '18 at 22:43
add a comment |
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What is
length binListToDecsupposed to do?– Willem Van Onsem
Nov 13 '18 at 18:14
take the length of the string passed to the function. and if its 0 ( empty) then return 1 as the sum @WillemVanOnsem
– Muhammad Usman
Nov 13 '18 at 18:15
5
but (a) the list you pass is
(x:xs)and (b) the length can never be zero, since(x:xs)is the pattern for a non-empty list.– Willem Van Onsem
Nov 13 '18 at 18:16