question about using bind2nd to convert a binary predicate to unary predicate

Posted on 2003-10-29
Last Modified: 2012-05-04
In STL, "bind2nd" is used to convert a binary prediate to unary predicate. I try to declare a unary predicate by bind2nd, but it does not work.
Please tell me the reason.


using namespace std;
int main()
      vector<int> v;
      typedef vector<int>::iterator iterator;


      // this part work well
      iterator it1 = find_if(v.begin(), v.end(), bind2nd(equal_to<int>(), 3));

      // however, the following doesn't work
      unary_function<int,bool> p = bind2nd(equal_to<int>(), 3);      
      iterator it2 = find_if(v.begin(), v.end(), p);//when compile: error C2064: term does not evaluate to a function

      return 0;
Question by:zma64
  • 2

Accepted Solution

PhilipNicoletti earned 125 total points
ID: 9643963
    binder2nd<equal_to<int> > p = bind2nd(equal_to<int>(),3);

Not really sure why. I saw this on the web some time ago and just
put it in my notes.

This also works:

     binder1st<equal_to<int> > p = bind1st(equal_to<int>(),3);
LVL 15

Expert Comment

ID: 9646657
Good answer from PhilipNicoletti.  Here's why it works.

find_if wants a function object as its third parameter.  A function object is an object of a class that has an operator().  unary_function is just a structure with a couple of typedefs.  It doesn't have an operator() itself.  It's intended as a base class for function objects so they can inherit the typedefs.  So the compiler complained because it couldn't call p(), because p didn't have an operator ().

What you want as the type of p is something this is derived from unary_function, but has an operator ().  That's exactly what binder2nd is, and conveniently enough, a binder2nd is what bind2nd returns.  The binder2nd has a function object and a second parameter as members, and its operator() calls the function object.


Expert Comment

ID: 9646827
That makes sense,

when I tried the original code on Intel's cpp compiler, it gave
an error message about needing operator () or function
pointer (I'm not sure of the exact wording - I don't have
access to the compiler right now).

Expert Comment

ID: 10546298
No comment has been added lately, so it's time to clean up this question.
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