How to update an existing element of std::set?

I have a std::set<Foo>, and I'd like to update some value of an existing element therein. Note that the value I'm updating does not change the order in the set:

#include <iostream>
#include <set>
#include <utility>

struct Foo {
  Foo(int i, int j) : id(i), val(j) {}
  int id;
  int val;
  bool operator<(const Foo& other) const {
    return id < other.id;
  }
};

typedef std::set<Foo> Set;

void update(Set& s, Foo f) {
  std::pair<Set::iterator, bool> p = s.insert(f);
  bool alreadyThere = p.second;
  if (alreadyThere)
    p.first->val += f.val; // error: assignment of data-member
                           // ‘Foo::val’ in read-only structure
}

int main(int argc, char** argv){
  Set s;
  update(s, Foo(1, 10));
  update(s, Foo(1, 5));
  // Now there should be one Foo object with val==15 in the set.                                                                
  return 0;
}

Is there any concise way to do this? Or do I have to check if the element is already there, and if so, remove it, add the value and re-insert?

Since val is not involved in comparison, it could be declared mutable

struct Foo {
  Foo(int i, int j) : id(i), val(j) {}
  int id;
  mutable int val;
  bool operator<(const Foo& other) const {
    return id < other.id;
  }
};

This implies that the value of val may change in a logically-const Foo, which means that it shouldn't affect other comparison operators etc.

Or you could just remove and insert, that takes O(1) additional time (compared to accessing and modifying) if insertion uses the position just before just after the old one as the hint.

Something like:

bool alreadyThere = !p.second; // you forgot the !
if (alreadyThere)
{
    Set::iterator hint = p.first;
    hint++;
    s.erase(p.first);
    s.insert(hint, f);
}

Don't try to solve this problem by working around the const-ness of items in a set. Instead, why not use map, which already expresses the key-value relationship you are modeling and provides easy ways to update existing elements.