How to check if a vector of a vector exists?

You probably want to check if edges[i] is empty:

if (edges[i].empty()) 

I think you might have confused checking whether an object "exists" which is not relevant here with bounds checking. If you want to check whether i is a valid index, just check if it's within the size of the vector:

if (i < edges.size()) 

Use vector.empty():

if (edges[i].empty()){ } 

Or use vector.size():

if (edges[i].size() == 0){ } 

vector<vector<int> > edges; 

Now I would like to check if edges[i] existis.

It seems that std::vector is the wrong (outer) data type for you. Perhaps you should use std::map instead:

std::map<int, std::vector<int>> edges; if (edges.find(i) == edges.end()) { // does not exist } 

NULL (or more correctly, since you tagged C++11, nullptr) is used for checking pointers. You've declared a vector of concrete instances, so pointers don't apply, and if the test

if (i < edges.size()) 

then the vector is initialized - that is, constructed. You can check, next, whether it is populated:

if (i < edges.size() && !edges[i].empty()) 

however, this will not distinguish between uninitialized and currently empty, consider:

std::vector<int> v; v.push_back(1); v.pop_back(); // now it's empty again bool uninitialized = v.empty(); // lies 

It is empty, but it has been initialized and used, if that is your definition of initialized.

There is, however, one trick left, but it is tending towards undefined behavior.

if (i < edges.size() && edges[i].capacity() == 0) 

This will tell you if the vector has any memory allocated. In the worst case scenario, if your vector aggressively frees, it will be equivalent to empty(). But in most standard implementations, a vector that has been populated will have non-zero capacity, whilst a vector that has not yet been used will have no capacity.

Again: it's not perfect, there are ways to tell the vector to release any capacity it has.