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Why is there a StringExpression in the following code?

f["a b c"] //. x_String :> StringReplace[x, (StartOfString ~~ x1__ ~~ "b" ~~ x2__ ~~ EndOfString) :> f["BBB"]] (* f[StringExpression[f[BBB]]] *)

Though I can remove it, I'd rather it not appear in the first place, because I'm not sure this method is safe:

f["a b c"] //. x_String :> StringReplace[x, (StartOfString ~~ x1__ ~~ "b" ~~ x2__ ~~ EndOfString) :> f["BBB"]] /. StringExpression -> Sequence (* f[f[BBB]] *)
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    $\begingroup$ This is clearly covered in the docs for StringReplace (point 3 under details): "If the Subscript[sp, i] in the replacements Subscript[s, i]->Subscript[sp, i] do not evaluate to strings, StringReplace will yield a StringExpression rather than an ordinary string." $\endgroup$ Commented Jul 5, 2013 at 5:39
  • $\begingroup$ @rm-rf oh, I see , thanks, f["a b c b "]//.x_String:>StringReplace[x,Shortest[x1__~~"b"~~x2__]:>f["BBB"]] this maybe more apparently why need StringExpression. $\endgroup$ Commented Jul 5, 2013 at 5:46
  • $\begingroup$ Related: mathematica.stackexchange.com/q/7008/121 $\endgroup$ Commented Apr 14, 2014 at 10:13

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As R.M. described this behavior is documented:

If the spi in the replacements si -> spi do not evaluate to strings, StringReplace will yield a StringExpression rather than an ordinary string.

I recommend stripping StringExpression as part of the right-hand side of the primary replacement, so that the main expression never has StringExpression substituted into it. A shorter replacement form, e.g. _[x_] :> x may be used as the return from StringReplace should be either a String or a StringExpression object, or you could simply Apply Sequence:

f["a b c"] //. x_String :> Sequence @@ StringReplace[x, (StartOfString ~~ x1__ ~~ "b" ~~ x2__ ~~ EndOfString) :> f["BBB"]]

This works because Sequence @@ "string" returns "string".

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