Use the auto keyword in C++ STL

It's additional information, and isn't an answer.

In C++11 you can write:

for (auto& it : s) {
    cout << it << endl;
}

instead of

for (auto it = s.begin(); it != s.end(); it++) {
    cout << *it << endl;
}

It has the same meaning.

Update: See the @Alnitak's comment also.

auto keyword is intended to use in such situation, it is absolutely safe. But unfortunately it available only in C++0x so you will have portability issues with it.

The auto keyword gets the type from the expression on the right of =. Therefore it will work with any type, the only requirement is to initialize the auto variable when declaring it so that the compiler can deduce the type.

Examples:

auto a = 0.0f;  // a is float
auto b = std::vector<int>();  // b is std::vector<int>()

MyType foo()  { return MyType(); }

auto c = foo();  // c is MyType

The auto keyword is simply asking the compiler to deduce the type of the variable from the initialization.

Even a pre-C++0x compiler knows what the type of an (initialization) expression is, and more often than not, you can see that type in error messages.

#include <vector>
#include <iostream>
using namespace std;

int main()
{
    vector<int>s;
    s.push_back(11);
    s.push_back(22);
    s.push_back(33);
    s.push_back(55);
    for (int it=s.begin();it!=s.end();it++){
        cout<<*it<<endl;
    }
}

Line 12: error: cannot convert '__gnu_debug::_Safe_iterator<__gnu_cxx::__normal_iterator<int*, __gnu_norm::vector<int, std::allocator<int> > >, __gnu_debug_def::vector<int, std::allocator<int> > >' to 'int' in initialization

The auto keyword simply allows you to take advantage of this knowledge - if you (compiler) know the right type, just choose for me!