If you want all the paths, use recursion.
Using an adjacency list, preferably, create a function f() that attempts to fill in a current list of visited vertices. Like so:
void allPaths(vector<int> previous, int current, int destination)
{
previous.push_back(current);
if (current == destination)
//output all elements of previous, and return
for (int i = 0; i < neighbors[current].size(); i++)
allPaths(previous, neighbors[current][i], destination);
}
int main()
{
//...input
allPaths(vector<int>(), start, end);
}
Due to the fact that the vector is passed by value (and thus any changes made further down in the recursive procedure aren't permanent), all possible combinations are enumerated.
You can gain a bit of efficiency by passing the previous vector by reference (and thus not needing to copy the vector over and over again) but you'll have to make sure that things get popped_back() manually.
One more thing: if the graph has cycles, this won't work. (I assume in this case you'll want to find all simple paths, then) Before adding something into the previous vector, first check if it's already in there.
If you want all shortest paths, use Konrad's suggestion with this algorithm.