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I don’t know much about Haskell, but I looked up `let`’s documentation, and I think it already does support embedded structures like you ask for. You can just w

I don’t know much about Haskell, but I looked up `let`’s documentation, and I think it already does support embedded structures like you ask for. You can just write `p2 = delay (1/4) $ p1` (and you can also leave out the `$` because `p1` is already atomic). I think the site’s example repeats the definition of `p1` just to make reading the example clearer, not because you are forced to code like that.

cstrahan 12y ago
Some extra background on `$`, for the curious: Function application in Haskell is expressed like so: > f x Where `f` is a function, and `x` is it's argument. Function application is left associative, so > f x y Is equivalent to > (f x) y In this case, `(f x)` yields a new function, to which `y` is then applied (see http://en.wikipedia.org/wiki/Currying http://en.wikipedia.org/wiki/Currying). The function `$` that you mention is just application: > f $ x = f x So, using `$` infix like so > f $ x Is equivalent to > f x And likewise > delay (1/4) $ p1 is equivalent to > (delay (1/4)) p1 with extra parenthesis to make associativity clear. This begs the question: when is `$` ever a useful function? There are two cases that come to mind: 1) Precedence. `$` has the lowest precedence of any function, and can thus be used to omit otherwise necessary parens; this > f (g x) is equivalent to > f $ g x 2) Use in higher order functions. Imagine you have a function `all` that takes a predicate function `f` and a list of values `xs` and returns `True` if `f x` returns `True` for all `x` in `xs`. So `all (> 3) [4,5,6]` returns `True`. But, imagine now that you have a list of predicates and one value: can you still make use of `all` in this case? Yes, you could do something like this: > all (\f -> f 42) [even, (> 3), (< 100)] But that's a little verbose. If we look at `(\f -> f 42)`, all it's doing is taking a function and immediately applying `42` to it - hey, that sounds like `$` if it was partially applied to `42`! Here's a simpler version of the previous expression: > all ($ 42) [even, (> 3), (< 100)] So that's pretty much all there is to `$` and why one may omit it in the original expression.

cstrahan 12y ago

Some extra background on `$`, for the curious: Function application in Haskell is expressed like so: > f x Where `f` is a function, and `x` is it's argument. Function application is left associative, so > f x y Is equivalent to > (f x) y In this case, `(f x)` yields a new function, to which `y` is then applied (see http://en.wikipedia.org/wiki/Currying http://en.wikipedia.org/wiki/Currying). The function `$` that you mention is just application: > f $ x = f x So, using `$` infix like so > f $ x Is equivalent to > f x And likewise > delay (1/4) $ p1 is equivalent to > (delay (1/4)) p1 with extra parenthesis to make associativity clear. This begs the question: when is `$` ever a useful function? There are two cases that come to mind: 1) Precedence. `$` has the lowest precedence of any function, and can thus be used to omit otherwise necessary parens; this > f (g x) is equivalent to > f $ g x 2) Use in higher order functions. Imagine you have a function `all` that takes a predicate function `f` and a list of values `xs` and returns `True` if `f x` returns `True` for all `x` in `xs`. So `all (> 3) [4,5,6]` returns `True`. But, imagine now that you have a list of predicates and one value: can you still make use of `all` in this case? Yes, you could do something like this: > all (\f -> f 42) [even, (> 3), (< 100)] But that's a little verbose. If we look at `(\f -> f 42)`, all it's doing is taking a function and immediately applying `42` to it - hey, that sounds like `$` if it was partially applied to `42`! Here's a simpler version of the previous expression: > all ($ 42) [even, (> 3), (< 100)] So that's pretty much all there is to `$` and why one may omit it in the original expression.