Re: what is wrong with this code?

From:
SG <s.gesemann@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Wed, 22 Jul 2009 12:07:41 CST
Message-ID:
<3178660f-d62e-4a41-9b6a-7ddf6814d94c@c14g2000yqm.googlegroups.com>
Martin Eisenberg wrote:

SG wrote:

...which makes me wonder why nobody proposed a pointer type such
an expression could decay to for C++0x. Such a pointer type
should be class-agnostic and much easier to implement than
member function pointers [1].

Instead, we have to use mem_fun and bind1st. Unfortunately, the
resulting type of bind1st(mem_fun(.),.) depends on the member
function's class type and requires the use of templates and/or
type erasure.


How about employing a lambda expression? Wouldn't that be passable by
function pointer?


No. A lambda expression creates an rvalue of a unique and anonymous
type which has an overloaded function call operator. Whether you use
std::bind or a lambda expression doesn't make a big difference.

  class B {
  public:
    void foo(int);
  };

  template<typename Func>
  void bar1(Func f) {
    f(23);
  }

  void bar2(function<void(int)> f) {
    f(23);
  }

  int main() {
    B b;
    B* pb = &b;
    bar1(bind(&B::foo,pb,_1));
    bar2(bind(&B::foo,pb,_1));
    bar1([pb](int i){pb->foo(i);});
    bar2([pb](int i){pb->foo(i);});
  }

bar1 is a template. Instead of calling B::foo directly it invokes a
proxy which forwards the call to B::foo. For every class you get a
different specialization of bar1. (one "indirection")

bar2 is not a template. It uses a polymorphic function object (type
erasure). bar2 calls function<void(int)>::operator() which calls a
virtual function which calls another operator() which finally calls
B::foo. (three "indirections")

Cheers!
SG

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