Keep arguments of a function unevaluated after replacement rules

Question

Consider an expression of type

f[a,b]+a+b

and a set of replacement rules

sub={a->1,b->2}

If I evaluate now the expression

f[a,b]+a+b/.sub

the output will be

3+f[1,2]

However, I need that the arguments of f are left unevaluated after the replacement rules are applied. In particular, I need my output to be

3+f[a,b]

I tried with HoldAllComplete and similars as attributes to f, but this seems not to work

Answer 1

You can change your rule slightly to reflect the fact that you want to leave the f expression unchanged:

f[a, b] + a + b /. {expr : f[__] :> expr, a -> 1, b -> 2}

3 + f[a, b]