Meaning of (a : _ : 1)

Question

This is an example from Leonid Shifrin's tutorial:

testexpr = Expand[(1 + x)^10]
newexpr = 
 Replace[testexpr, {(a : _ : 1)*x^(y_?EvenQ) :> a*f[x^y], 
   a_ /; FreeQ[a, x] :> f[a]}, 1]

Basic question: what does (a : _ : 1) mean and do?

Evaluating FullForm[Hold[a : _ : 1]] might be instructive. — J. M.'s missing motivation, Aug 18 '17 at 14:25
Ah, I wasn't clear on that it is the same as (a_:1)... but why two different syntaxes? — GambitSquared, Aug 18 '17 at 14:35
The core syntax is the colon, which is another way to spell Pattern. The Blank without preceding colon is a convenience that allows omitting parentheses when the low precedence of the colon would otherwise require them. — Alan, Aug 18 '17 at 14:45

score 4 · Accepted Answer · answered Aug 18 '17 at 14:36

If you evaluate HoldForm@FullForm[a : _ : 1] you get Optional[Pattern[a,Blank[]],1]. This is a pattern object that will will be named a for later use and will be replaced by 1 if missing.

f[x] /. f[(a : _ : 1)] -> g[a]  (* g[x] *)
f[] /. f[(a : _ : 1)] -> g[a]  (* g[1] *)

Meaning of (a : _ : 1)

1 Answers1