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In Haskell, when using the XStrict language extension, is if short-circuiting?

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7

1

Standard Haskell is lazily evaluated, so if myCondition then someValue else doSomeLargeComputation x y z will avoid evaluating doSomeLargeComputation x y z if myCondition is true. My question is, if I enable the language extension XStrict then will doSomeLargeComputation x y z now be evaluated even when myCondition is true?

If so, is there a control flow construct other than explicitly marking doSomeLargeComputation x y z as lazy that can be used to avoid calculating it (like a short-circuiting if statement in a strict language)?

haskell

share|improve this question

asked 9 hours ago

LogicChainsLogicChains

1,72422 gold badges1010 silver badges1616 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  if is non-strict in any language I know -- this is even more important in impure languages, where you do not want to observe the side effects of both branches. In an (insane, IMO) higher-order language with a strict if we could still write (if cond then (_-> doX) else (_-> doY)) () to force it to be lazy, but that hack would need to be used to often that it should be the default.
  
  – chi
  6 hours ago
  
  

  
  
  
  

add a comment | 

7

1

Standard Haskell is lazily evaluated, so if myCondition then someValue else doSomeLargeComputation x y z will avoid evaluating doSomeLargeComputation x y z if myCondition is true. My question is, if I enable the language extension XStrict then will doSomeLargeComputation x y z now be evaluated even when myCondition is true?

If so, is there a control flow construct other than explicitly marking doSomeLargeComputation x y z as lazy that can be used to avoid calculating it (like a short-circuiting if statement in a strict language)?

haskell

share|improve this question

asked 9 hours ago

LogicChainsLogicChains

1,72422 gold badges1010 silver badges1616 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  if is non-strict in any language I know -- this is even more important in impure languages, where you do not want to observe the side effects of both branches. In an (insane, IMO) higher-order language with a strict if we could still write (if cond then (_-> doX) else (_-> doY)) () to force it to be lazy, but that hack would need to be used to often that it should be the default.
  
  – chi
  6 hours ago
  
  

  
  
  
  

add a comment | 

Standard Haskell is lazily evaluated, so if myCondition then someValue else doSomeLargeComputation x y z will avoid evaluating doSomeLargeComputation x y z if myCondition is true. My question is, if I enable the language extension XStrict then will doSomeLargeComputation x y z now be evaluated even when myCondition is true?

If so, is there a control flow construct other than explicitly marking doSomeLargeComputation x y z as lazy that can be used to avoid calculating it (like a short-circuiting if statement in a strict language)?

haskell

share|improve this question

asked 9 hours ago

LogicChainsLogicChains

1,72422 gold badges1010 silver badges1616 bronze badges

share|improve this question

asked 9 hours ago

LogicChainsLogicChains

1,72422 gold badges1010 silver badges1616 bronze badges

share|improve this question

asked 9 hours ago

LogicChainsLogicChains

1,72422 gold badges1010 silver badges1616 bronze badges

asked 9 hours ago

LogicChainsLogicChains

1,72422 gold badges1010 silver badges1616 bronze badges

asked 9 hours ago

LogicChainsLogicChains

1,72422 gold badges1010 silver badges1616 bronze badges

1 Answer
1

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9

No, the branch not taken is not evaluated.

For example:

-# LANGUAGE Strict #-
import Debug.Trace

main :: IO ()
main = print (f (fib 3))

f i =
 if i < 5
 then trace "then" 0
 else trace "else" 1

-- Using fib to avoid inlining or optimization from messing up or test.
fib :: Int -> Int
fib 1 = 1
fib n = n + fib (n-1)

Prints:

*Main> main
else
1

However, if we lift the branches to let bindings then yes, those are strictly evaluated:

f :: Int -> Int
f i =
 let f1 = trace "then" 1 in
 let f2 = trace "else" 2 in
 if i < 5
 then f1
 else f2

Yields:

% ./LogicChains
then
else
2

share|improve this answer

answered 9 hours ago

Thomas M. DuBuissonThomas M. DuBuisson

57.7k77 gold badges9191 silver badges155155 bronze badges

add a comment | 

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9

No, the branch not taken is not evaluated.

For example:

-# LANGUAGE Strict #-
import Debug.Trace

main :: IO ()
main = print (f (fib 3))

f i =
 if i < 5
 then trace "then" 0
 else trace "else" 1

-- Using fib to avoid inlining or optimization from messing up or test.
fib :: Int -> Int
fib 1 = 1
fib n = n + fib (n-1)

Prints:

*Main> main
else
1

However, if we lift the branches to let bindings then yes, those are strictly evaluated:

f :: Int -> Int
f i =
 let f1 = trace "then" 1 in
 let f2 = trace "else" 2 in
 if i < 5
 then f1
 else f2

Yields:

% ./LogicChains
then
else
2

share|improve this answer

answered 9 hours ago

Thomas M. DuBuissonThomas M. DuBuisson

57.7k77 gold badges9191 silver badges155155 bronze badges

add a comment | 

9

No, the branch not taken is not evaluated.

For example:

-# LANGUAGE Strict #-
import Debug.Trace

main :: IO ()
main = print (f (fib 3))

f i =
 if i < 5
 then trace "then" 0
 else trace "else" 1

-- Using fib to avoid inlining or optimization from messing up or test.
fib :: Int -> Int
fib 1 = 1
fib n = n + fib (n-1)

Prints:

*Main> main
else
1

However, if we lift the branches to let bindings then yes, those are strictly evaluated:

f :: Int -> Int
f i =
 let f1 = trace "then" 1 in
 let f2 = trace "else" 2 in
 if i < 5
 then f1
 else f2

Yields:

% ./LogicChains
then
else
2

share|improve this answer

answered 9 hours ago

Thomas M. DuBuissonThomas M. DuBuisson

57.7k77 gold badges9191 silver badges155155 bronze badges

add a comment | 

No, the branch not taken is not evaluated.

For example:

-# LANGUAGE Strict #-
import Debug.Trace

main :: IO ()
main = print (f (fib 3))

f i =
 if i < 5
 then trace "then" 0
 else trace "else" 1

-- Using fib to avoid inlining or optimization from messing up or test.
fib :: Int -> Int
fib 1 = 1
fib n = n + fib (n-1)

Prints:

*Main> main
else
1

However, if we lift the branches to let bindings then yes, those are strictly evaluated:

f :: Int -> Int
f i =
 let f1 = trace "then" 1 in
 let f2 = trace "else" 2 in
 if i < 5
 then f1
 else f2

Yields:

% ./LogicChains
then
else
2

share|improve this answer

answered 9 hours ago

Thomas M. DuBuissonThomas M. DuBuisson

57.7k77 gold badges9191 silver badges155155 bronze badges

-# LANGUAGE Strict #-
import Debug.Trace

main :: IO ()
main = print (f (fib 3))

f i =
 if i < 5
 then trace "then" 0
 else trace "else" 1

-- Using fib to avoid inlining or optimization from messing up or test.
fib :: Int -> Int
fib 1 = 1
fib n = n + fib (n-1)

*Main> main
else
1

f :: Int -> Int
f i =
 let f1 = trace "then" 1 in
 let f2 = trace "else" 2 in
 if i < 5
 then f1
 else f2

% ./LogicChains
then
else
2

share|improve this answer

answered 9 hours ago

Thomas M. DuBuissonThomas M. DuBuisson

57.7k77 gold badges9191 silver badges155155 bronze badges

answered 9 hours ago

Thomas M. DuBuissonThomas M. DuBuisson

57.7k77 gold badges9191 silver badges155155 bronze badges

answered 9 hours ago

Thomas M. DuBuissonThomas M. DuBuisson

57.7k77 gold badges9191 silver badges155155 bronze badges