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edited Apr 24, 2013 at 16:41

Why does Haskell have a built-in if/then/else, which is dependent on the Bool type, instead of having a simple library function? Such as

if :: Bool -> a -> a -> a
if True  x _ = x
if False _ y = y

Why does Haskell have a built-in if/then/else, which is dependent on the Bool type, instead of having a simple library function? Such as

if :: Bool -> a -> a
if True  x _ = x
if False _ y = y

Why does Haskell have a built-in if/then/else, which is dependent on the Bool type, instead of having a simple library function? Such as

if :: Bool -> a -> a -> a
if True  x _ = x
if False _ y = y

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occurred Apr 22, 2013 at 14:00

fix grammar

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edited Apr 22, 2013 at 12:36

Why does Haskell hashave a built-in if/then/else, which is dependent on the Bool type, instead of having a simple library function? Such as

if :: Bool -> a -> a
if True  x _ = x
if False _ y = y

Why Haskell has built-in if/then/else, which is dependent on the Bool type, instead of having a simple library function? Such as

if :: Bool -> a -> a
if True  x _ = x
if False _ y = y

Why does Haskell have a built-in if/then/else, which is dependent on the Bool type, instead of having a simple library function? Such as

if :: Bool -> a -> a
if True  x _ = x
if False _ y = y

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asked Apr 22, 2013 at 12:14

Why does Haskell have built-in "if/then/else" instead of defining it as a simple library function?

Why Haskell has built-in if/then/else, which is dependent on the Bool type, instead of having a simple library function? Such as

if :: Bool -> a -> a
if True  x _ = x
if False _ y = y

haskell

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