What is the difference between vector.back() and vector.end()?

Question

I am a new C++ learner, and I read a code block about C++ STL accessing the last element from a vector.

Why does the code at line 6, 7, and 8 need to subtract 1 to be equal to that at line 5?

1.    std::vector<int> v;
2.    v.push_back(999);
3.    //fill up the vector
4.    //...

5.    int j = v.back();
6.    int j = v.[size-1]
7.    int j = v.at(v.size()-1)
8.    int j = *(v.end()-1)

Answer 1

Here's an illustration of which is which

v: [ 1 | 2 | 3 | 4 | ... | 999 ]
     🡑                      🡑     🡑
   front()                back() end()
     🡑
   begin()

where front() and back() return a (const) reference to the first and last element respectively, and end() returns an iterator (sort of a pointer) to one beyond the last element of vector. begin() returns an iterator to the first element of a vector.

These are also explained at std::vector

front access the first element
back access the last element
end/cend returns an iterator to the end
begin/cbegin returns an iterator to the beginning

---

Subtracting one from size is because an index in C or C++ starts at zero, and not one as usually. This means, in order to access the first element of an array, or in this case of a vector, you say

v[0]

and not

v[1]

Equally for the last (nth) element, you wouldn't take size or n of an array (a vector), but rather one less, e.g.

v[size() - 1]

or

v[n - 1]

Answer 2

5. int j = v.back();

std::vector::back is defined to return the last element in the vector. This is straight forward.

7. int j = v.[size-1]

Indices are 0 based in c++. If a sequential container has N elements, the valid indices are between 0 and N-1 inclusively. The last element is therefor N-1, or size()-1.

8. int j = *(v.end()-1)

std::vector::end returns an iterators to one-past-the-end of the container. The element just before is then the last element in the vector.