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Meaning of a semicolon in lambda expression
What does “static” mean in C?Retrieving Property name from lambda expressionWhy would you use Expression<Func<T>> rather than Func<T>?Why are Python lambdas useful?Distinct() with lambda?What does “use strict” do in JavaScript, and what is the reasoning behind it?What does the star operator mean?Why does ++[[]][+[]]+[+[]] return the string “10”?How can a time function exist in functional programming?What is a lambda expression in C++11?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
Type:
data Command a = Command String (a -> IO a)
Function:
iofunc_ :: String -> (a -> IO ()) -> Command a
iofunc_ s f = Command s (x -> do f x ; return x)
What does the semicolon do in the lambda expression (x -> do f x ; return x)?
haskell lambda syntax
edited Mar 27 at 5:21
dfeuer
35.2k3 gold badges52 silver badges137 bronze badges
asked Dec 25 '14 at 11:54
StankoStanko
1,4402 gold badges11 silver badges37 bronze badges
add a comment |
Type:
data Command a = Command String (a -> IO a)
Function:
iofunc_ :: String -> (a -> IO ()) -> Command a
iofunc_ s f = Command s (x -> do f x ; return x)
What does the semicolon do in the lambda expression (x -> do f x ; return x)?
haskell lambda syntax
edited Mar 27 at 5:21
dfeuer
35.2k3 gold badges52 silver badges137 bronze badges
asked Dec 25 '14 at 11:54
StankoStanko
1,4402 gold badges11 silver badges37 bronze badges
3
It's part of thedo-notation, not the lambda.
– augustss
Dec 25 '14 at 12:08
add a comment |
Type:
data Command a = Command String (a -> IO a)
Function:
iofunc_ :: String -> (a -> IO ()) -> Command a
iofunc_ s f = Command s (x -> do f x ; return x)
What does the semicolon do in the lambda expression (x -> do f x ; return x)?
haskell lambda syntax
edited Mar 27 at 5:21
dfeuer
35.2k3 gold badges52 silver badges137 bronze badges
asked Dec 25 '14 at 11:54
StankoStanko
1,4402 gold badges11 silver badges37 bronze badges
Type:
data Command a = Command String (a -> IO a)
Function:
iofunc_ :: String -> (a -> IO ()) -> Command a
iofunc_ s f = Command s (x -> do f x ; return x)
What does the semicolon do in the lambda expression (x -> do f x ; return x)?
haskell lambda syntax
haskell lambda syntax
edited Mar 27 at 5:21
dfeuer
35.2k3 gold badges52 silver badges137 bronze badges
asked Dec 25 '14 at 11:54
StankoStanko
1,4402 gold badges11 silver badges37 bronze badges
edited Mar 27 at 5:21
dfeuer
35.2k3 gold badges52 silver badges137 bronze badges
asked Dec 25 '14 at 11:54
StankoStanko
1,4402 gold badges11 silver badges37 bronze badges
edited Mar 27 at 5:21
dfeuer
35.2k3 gold badges52 silver badges137 bronze badges
edited Mar 27 at 5:21
dfeuer
35.2k3 gold badges52 silver badges137 bronze badges
edited Mar 27 at 5:21
dfeuer
35.2k3 gold badges52 silver badges137 bronze badges
35.2k3 gold badges52 silver badges137 bronze badges
asked Dec 25 '14 at 11:54
StankoStanko
1,4402 gold badges11 silver badges37 bronze badges
asked Dec 25 '14 at 11:54
StankoStanko
1,4402 gold badges11 silver badges37 bronze badges
asked Dec 25 '14 at 11:54
StankoStanko
1,4402 gold badges11 silver badges37 bronze badges
1,4402 gold badges11 silver badges37 bronze badges
3
It's part of thedo-notation, not the lambda.
– augustss
Dec 25 '14 at 12:08
add a comment |
3
It's part of thedo-notation, not the lambda.
– augustss
Dec 25 '14 at 12:08
3
3
It's part of the
do-notation, not the lambda.– augustss
Dec 25 '14 at 12:08
It's part of the
do-notation, not the lambda.– augustss
Dec 25 '14 at 12:08
add a comment |
2 Answers
2
active
oldest
votes
They just separate the two expression f x and return x in do notation. In fact these all are equivalent in your case:
iofunc_ s f = Command s (x -> do f x ; return x)
iofunc_ s f = Command s (x -> do f x ; return x)
iofunc_ s f = Command s (x -> do f x
return x)
iofunc_ s f = Command s (x -> f x >> return x)
answered Dec 25 '14 at 12:08
SibiSibi
31.2k9 gold badges65 silver badges129 bronze badges
1
When can you omit the curly braces? I always thought they were mandatory if you made the semicolon explicit.
– kqr
Dec 25 '14 at 14:27
@kqr Didn't read the Haskell report fully, but you can see that's what they have demonstrated in the example withlet.
– Sibi
Dec 25 '14 at 15:50
@Sibi Thank you.
– Stanko
Dec 25 '14 at 16:29
add a comment |
Semicolon anywhere is equivalent to a change of line indented to the same level as the previous valid expression.
I saw it by going through how indentation works (https://en.wikibooks.org/wiki/Haskell/Indentation).
answered Oct 12 '17 at 9:49
hhefestohhefesto
1158 bronze badges
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
They just separate the two expression f x and return x in do notation. In fact these all are equivalent in your case:
iofunc_ s f = Command s (x -> do f x ; return x)
iofunc_ s f = Command s (x -> do f x ; return x)
iofunc_ s f = Command s (x -> do f x
return x)
iofunc_ s f = Command s (x -> f x >> return x)
answered Dec 25 '14 at 12:08
SibiSibi
31.2k9 gold badges65 silver badges129 bronze badges
1
When can you omit the curly braces? I always thought they were mandatory if you made the semicolon explicit.
– kqr
Dec 25 '14 at 14:27
@kqr Didn't read the Haskell report fully, but you can see that's what they have demonstrated in the example withlet.
– Sibi
Dec 25 '14 at 15:50
@Sibi Thank you.
– Stanko
Dec 25 '14 at 16:29
add a comment |
They just separate the two expression f x and return x in do notation. In fact these all are equivalent in your case:
iofunc_ s f = Command s (x -> do f x ; return x)
iofunc_ s f = Command s (x -> do f x ; return x)
iofunc_ s f = Command s (x -> do f x
return x)
iofunc_ s f = Command s (x -> f x >> return x)
answered Dec 25 '14 at 12:08
SibiSibi
31.2k9 gold badges65 silver badges129 bronze badges
1
When can you omit the curly braces? I always thought they were mandatory if you made the semicolon explicit.
– kqr
Dec 25 '14 at 14:27
@kqr Didn't read the Haskell report fully, but you can see that's what they have demonstrated in the example withlet.
– Sibi
Dec 25 '14 at 15:50
@Sibi Thank you.
– Stanko
Dec 25 '14 at 16:29
add a comment |
They just separate the two expression f x and return x in do notation. In fact these all are equivalent in your case:
iofunc_ s f = Command s (x -> do f x ; return x)
iofunc_ s f = Command s (x -> do f x ; return x)
iofunc_ s f = Command s (x -> do f x
return x)
iofunc_ s f = Command s (x -> f x >> return x)
answered Dec 25 '14 at 12:08
SibiSibi
31.2k9 gold badges65 silver badges129 bronze badges
They just separate the two expression f x and return x in do notation. In fact these all are equivalent in your case:
iofunc_ s f = Command s (x -> do f x ; return x)
iofunc_ s f = Command s (x -> do f x ; return x)
iofunc_ s f = Command s (x -> do f x
return x)
iofunc_ s f = Command s (x -> f x >> return x)
answered Dec 25 '14 at 12:08
SibiSibi
31.2k9 gold badges65 silver badges129 bronze badges
answered Dec 25 '14 at 12:08
SibiSibi
31.2k9 gold badges65 silver badges129 bronze badges
answered Dec 25 '14 at 12:08
SibiSibi
31.2k9 gold badges65 silver badges129 bronze badges
answered Dec 25 '14 at 12:08
SibiSibi
31.2k9 gold badges65 silver badges129 bronze badges
31.2k9 gold badges65 silver badges129 bronze badges
1
When can you omit the curly braces? I always thought they were mandatory if you made the semicolon explicit.
– kqr
Dec 25 '14 at 14:27
@kqr Didn't read the Haskell report fully, but you can see that's what they have demonstrated in the example withlet.
– Sibi
Dec 25 '14 at 15:50
@Sibi Thank you.
– Stanko
Dec 25 '14 at 16:29
add a comment |
1
When can you omit the curly braces? I always thought they were mandatory if you made the semicolon explicit.
– kqr
Dec 25 '14 at 14:27
@kqr Didn't read the Haskell report fully, but you can see that's what they have demonstrated in the example withlet.
– Sibi
Dec 25 '14 at 15:50
@Sibi Thank you.
– Stanko
Dec 25 '14 at 16:29
1
1
When can you omit the curly braces? I always thought they were mandatory if you made the semicolon explicit.
– kqr
Dec 25 '14 at 14:27
When can you omit the curly braces? I always thought they were mandatory if you made the semicolon explicit.
– kqr
Dec 25 '14 at 14:27
@kqr Didn't read the Haskell report fully, but you can see that's what they have demonstrated in the example with
let.– Sibi
Dec 25 '14 at 15:50
@kqr Didn't read the Haskell report fully, but you can see that's what they have demonstrated in the example with
let.– Sibi
Dec 25 '14 at 15:50
@Sibi Thank you.
– Stanko
Dec 25 '14 at 16:29
@Sibi Thank you.
– Stanko
Dec 25 '14 at 16:29
add a comment |
Semicolon anywhere is equivalent to a change of line indented to the same level as the previous valid expression.
I saw it by going through how indentation works (https://en.wikibooks.org/wiki/Haskell/Indentation).
answered Oct 12 '17 at 9:49
hhefestohhefesto
1158 bronze badges
add a comment |
Semicolon anywhere is equivalent to a change of line indented to the same level as the previous valid expression.
I saw it by going through how indentation works (https://en.wikibooks.org/wiki/Haskell/Indentation).
answered Oct 12 '17 at 9:49
hhefestohhefesto
1158 bronze badges
add a comment |
Semicolon anywhere is equivalent to a change of line indented to the same level as the previous valid expression.
I saw it by going through how indentation works (https://en.wikibooks.org/wiki/Haskell/Indentation).
answered Oct 12 '17 at 9:49
hhefestohhefesto
1158 bronze badges
Semicolon anywhere is equivalent to a change of line indented to the same level as the previous valid expression.
I saw it by going through how indentation works (https://en.wikibooks.org/wiki/Haskell/Indentation).
answered Oct 12 '17 at 9:49
hhefestohhefesto
1158 bronze badges
answered Oct 12 '17 at 9:49
hhefestohhefesto
1158 bronze badges
answered Oct 12 '17 at 9:49
hhefestohhefesto
1158 bronze badges
answered Oct 12 '17 at 9:49
hhefestohhefesto
1158 bronze badges
1158 bronze badges
add a comment |
add a comment |
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3
It's part of the
do-notation, not the lambda.– augustss
Dec 25 '14 at 12:08