A question regarding stl set

Here is the code,

Code:

---

#include <set>

using namespace std;

class A
{
public:
        void bar()
        {
                cout<<"bar"<<endl;
        }
};

void foo(A& a)
{
        a.bar();
}

int main()
{
        set<A> setA;

        A a;
        for(set<A>::iterator it=setA.begin();it!=setA.end();++it)
        {
                a = *it;

                foo(*it);
        }

        return 0;
}

---

If I call foo(*it), then I get an error " error C2664: 'foo' : cannot convert parameter 1 from 'const A' to 'A &' ". But if I call foo(a), then it would compile just fine. Why? What is the difference between a and *it. I thought they are basically the same. Thanks.

Because foo() could (potentially) modify the parameter that is passed.

So ... foo(*it) , could modify an element in the set (which is not allowed,
since it could mess up the ordering).

there is a difference...
a = *it; with foo(a)
assigns/copies *it into an existing object a. then you call foo on an object that is not const. which is ok.

in the foo(*it) case, you're trying to pass "const" *it to a function that epxects a modifiable object reference.

if you changed foo to foo(const A& a) then you could call foo(*it).
in this particular case, since foo() calls A::bar(), you would also have to change bar to void bar() const;