Different results using f@expr and expr // f in Mathematica

I was playing around with the Prefix and Postfix operators (@ and // respectively) and I ran into the following issue.

Given the following code, they evaluate in the same exact way:

Hold[MatrixPlot@Sort@data] // FullForm
(* Hold[MatrixPlot[Sort[data]]] *)

Hold[data // Sort // MatrixPlot] // FullForm
(* Hold[MatrixPlot[Sort[data]]] *)

However, given the following expressions, I get different results:

FunctionExpand@Abs'[0]
(* Abs'[0] *)

Abs'[0] // FunctionExpand
(* 0 *)

I'm开发者_运维技巧 not quite sure really why this is. In dozens of other snippets of code I've had, f@expr, expr // f, and f[expr] all evaluate to the same result. Why does this one particular case give this result?

This is a precedence issue. @ has higher precedence than //. To see what is going on, place the cursor on FunctionExpand in both cases, then either cmd+. (on OS X) or ctrl+. on anything else, and you end up selecting things by precedence.

Another way to see it is to use Trace:

FunctionExpand@Abs'[0] // Trace
(*
-> {{{FunctionExpand[Abs],Abs},Abs^\[Prime]},(Abs^\[Prime])[0]}
*)

while

Abs'[0] // FunctionExpand//Trace
(*
-> {FunctionExpand[(Abs^\[Prime])[0]],0}
*)

In particular, notice how in the first case mma first evaluates FunctionExpand[Abs], obtaining Abs, then continuing. This is precisely due to how strongly @ binds as compared to //.

EDIT: Inspired by @Leonid's comment, this is also informative:

Hold[FunctionExpand@Abs'[0]] // FullForm
Hold[Abs'[0] // FunctionExpand] // FullForm
(*
-> Hold[Derivative[1][FunctionExpand[Abs]][0]]
   Hold[FunctionExpand[Derivative[1][Abs][0]]]
*)

which is a much better demonstration of what is going on.