How to make a binary function associative? (Or define an n-ary function?)

Question

Suppose I have a simple function that I assign to an operator

f[a_, b_] := a + b
CirclePlus = f

Then I want to write

1 (+) 2 (+) 3

But it doesn't work, because it's trying to evaluate f[1,2,3].

So how does one instruct Mathematica that it should instead evaluate f[f[1,2],3], or alternatively how does one work with infix operators? I'd be okay if I had to write my as f[a_List] instead, and then took care of things myself ...

-- Edit:

As the answer was deleted, note that making the function Flat is not the answer, here, as far as I can see.

-- Edit:

Here is an exact copy of the MMA file to reproduce this problem

In[13]:= f[a_,b_]:=a+b;
f[a_,b_,c_]:=f[f[a,b],c]
CirclePlus=f;
1\[CirclePlus]2\[CirclePlus]5\[CirclePlus]6
Out[16]= f[1,2,5,6]

Answer 1

You can tell it by making a definition what it should do if you have more then 2 arguments:

ClearAll[f];
f[a_, b_, c__] := f[f[a, b], c];
CirclePlus = f

Then you get

And of course you have to add the definition of f when it is called as binary function. So for instance, and only for the purpose of showing what happens:

f[a_, b_] := Row[{"(", a, "\[CirclePlus]", b, ")"}]

Here is how it looks when you use more than 3 terms. Note that the pattern is recursively applied until there are only 2 arguments in each call:

Answer 2

halirutan demonstrated how to adapt your function f to handle multiple arguments but that changes the game: f is no longer a binary function.

The other aspect of this problem is that the CirclePlus operator itself is not binary; rather it accepts a series of arguments as illustrated in the question. CircleMinus is a binary operator, and with it your code works as written:

f[a_, b_] := a + b
CircleMinus = f;

q ⊖ r ⊖ s ⊖ t

q + r + s + t

This of course would be a confusing choice of operator for this function but it serves as illustration. An additional property that must be considered is left versus right association. CircleMinus is left-associative:

CircleMinus = foo;

q ⊖ r ⊖ s ⊖ t

foo[foo[foo[q, r], s], t]

Therefore is right-associative:

Therefore = foo;

q ∴ r ∴ s ∴ t

foo[q, foo[r, foo[s, t]]]

Care should be exercised when choosing an operator for your function!