Why is the code printing the last std::cout??

Question

using namespace std;

template<typename T>
int f(vector<T> &v){
    return v.size();
}

template<typename T>
class B{
public:
    int size(){
        return v.size();
    };
private:
    vector<T> v;
};

int main(int argc, char** argv) {

    B<string> b;

    vector<string> v;

    for(int i=0; i<f<string>(v)-1; i++)
        std::cout << "using fn template" << endl;

    for(int i=0; i<b.size()-1; i++)
        std::cout << "using class B" << endl;

    for(int i=0; i<v.size()-1; i++)
        std::cout << "done" << endl; //Why is this printing???
    return (EXIT_SUCCESS);
}

Answer 1

vector's size() function returns a value of type size_t which is unsigned. So, if size() returns 0 and you subtract 1 from it, you're going to get a very large number rather than -1. That very large number will be greater than 0 and the condition i < v.size() - 1 will therefore be be true since i is 0.

EDIT:

I should probably add that normally when iterating over an array or a vector, you iterate as long as your index is less than the size of the array or vector rather than size - 1.

 for(int i = 0; i < v.size(); ++i)
    std::cout << "done" << endl;

would likely be what you really want to do. Even if you used (int)v.size() - 1 to get rid of the signed vs unsigned issue, the loop would stil be wrong because you'd miss the last element in cases where you actually had elements in the vector.