CS253 Iterators

Foundation

Before we talk about iterators, let’s review just what a class can do.
We all know a class can provide public methods & data.
It can also provide types.

class Foo {
  public:
    string bar() { return "Yow!\n"; }
    int x = 42;
    using cash = float;
};

Foo::cash salary = 2.43;
Foo f;
cout << salary << ' ' << f.x << ' ' << f.bar();

2.43 42 Yow!

Array Traversal

You have an array, and you want to traverse (walk through) it.

int a[] = {2, 3, 5, 7, 11, 13, 17, 19};
for (int i=0; i!=8; ++i)
    cout << a[i] << ' ';

2 3 5 7 11 13 17 19

a[i] is the same as *(a+i).
- It requires addition, scaling, and indirection.
  - Though they’re often combined in a single machine instruction.
I usually say i<8, rather than i!=8.
Similarly, I usually say i++ rather than ++i.
Patience.

Problems with indexing

We’re really used to traversing an array like that.
We’re so much used to it, that it seems the right & natural way.
After all, mathematicians do that with a_i, a_i+1, a_i+2, …
Trouble is, it doesn’t extend naturally. We can speak of the nth element of a linked list or a binary tree, but we don’t compute it that way.
Instead, we think in terms of pointers.
We naturally use pointers to walk a linked list, or to traverse a binary tree.

Array Traversal via Pointers

int a[] = {2, 3, 5, 7, 11, 13, 17, 19};
for (int *p = &a[0]; p != &a[8]; ++p)
    cout << *p << ' ';

2 3 5 7 11 13 17 19

This uses pointers.
- Instead of int i, we use int *p. It points to each element, one-by-one.
- p initially points to &a[0], the address of the first element.
- a contains eight elements, a[0]…a[7]. Therefore, when we get to the address &a[8], it’s time to stop.
This is a bit faster: instead of a[i] (addition+scaling+indirection) we have just indirection.
Not really a radical improvement, however.

`vector` traversal

You can traverse a vector in the same way:

vector<int> a = {2, 3, 5, 7, 11, 13, 17, 19};
for (int *p = &a[0]; p != &a[8]; ++p)
    cout << *p << ' ';

2 3 5 7 11 13 17 19

or, avoiding the magic number 8:

vector<int> a = {2, 3, 5, 7, 11, 13, 17, 19};
for (int *p = &a[0]; p != &a[a.size()]; ++p)
    cout << *p << ' ';

2 3 5 7 11 13 17 19

Methods

vector<int> a = {2, 3, 5, 7, 11, 13, 17, 19};
for (vector<int>::iterator it = a.begin(); it != a.end(); ++it)
    cout << *it << ' ';

2 3 5 7 11 13 17 19

iterator is a type provided by the templated vector class.
You can treat it like a pointer.
You don’t care what type it really is. You know that:
- * and ++ work on it
- you can assign .begin() to it
- you can compare it to .end()
What type is it?
- might be int *, might not (it’s unspecified!)

`auto` is your friend

This is prettier:

vector<int> a = {2, 3, 5, 7, 11, 13, 17, 19};
for (auto it = a.begin(); it != a.end(); ++it)
    cout << *it << ' ';

2 3 5 7 11 13 17 19

All that we did is we used auto to tell the compiler to make it the same type as a.begin() returns, namely, vector<int>::iterator.

`for` loop

This is (nearly) exactly the same, and gorgeous:

vector<int> a = {2, 3, 5, 7, 11, 13, 17, 19};
for (auto v : a)
    cout << v << ' ';

2 3 5 7 11 13 17 19

The for loop is defined to use .begin() and .end(), just as the previous code.

It’s simple

Really, the for-loop like that (sometimes called a for…each loop) is just a mechanical source-level transformation. It rewrites this:

set<string> s = {"my", "dog", "has", "fleas"};
for (auto v : s)
    cout << v << '\n';

dog
fleas
has
my

as this:

set<string> s = {"my", "dog", "has", "fleas"};
for (auto it = s.begin(); it != s.end(); ++it) {
    auto v = *it;
    cout << v << '\n';
}

dog
fleas
has
my

Really

See this error message:

double zulu;
for (auto v : zulu)
    cout << v;

c.cc:2: error: 'begin' was not declared in this scope

“begin”? Why is it complaining about “begin”? Because it rewrote that for loop to use .begin() and .end().

Variations

Which is best? Why?

vector<int> a = {2, 3, 5, 7, 11, 13, 17, 19};
for (int v : a)
    cout << v << ' ';

2 3 5 7 11 13 17 19

vector<int> a = {2, 3, 5, 7, 11, 13, 17, 19};
for (const auto v : a)
    cout << v << ' ';

2 3 5 7 11 13 17 19

vector<int> a = {2, 3, 5, 7, 11, 13, 17, 19};
for (auto &v : a)
    cout << v << ' ';

2 3 5 7 11 13 17 19

vector<int> a = {2, 3, 5, 7, 11, 13, 17, 19};
for (const auto &v : a)
    cout << v << ' ';

2 3 5 7 11 13 17 19

Other containers

The same iterator code works for all STL containers.

forward_list<char> l = {'a', 'c', 'k', 'J'};
for (auto it = l.begin(); it != l.end(); ++it)
    cout << *it << ' ';

a c k J

unordered_set<string> u = {"a", "c", "k", "J"};
for (auto it = u.begin(); it != u.end(); ++it)
    cout << *it << ' ';

J k c a

set<char> s = {'a', 'c', 'k', 'J'};
for (auto it = s.begin(); it != s.end(); ++it)
    cout << *it << ' ';

J a c k

Of the top 100 ♀ + top 100 ♂ U.S. names, 1918–2017, the ones in ASCII order: Amy, Ann, Betty, Billy, Gary, Harry, Henry, Jack, Jerry, Kelly, Larry, Mary, Roy, Scott, and Terry.

`iterator` type

What type is set<int>::iterator?
- It’s not an int *, that’s for sure!
- It’s a user-defined type, contining an internal node pointer, with operator* and operator++ defined.
Similarly for list<int>::iterator and unordered_set<int>::iterator.
Most iterator types cannot be a native pointer.
- What happens when you increment a plain pointer?
- It goes to the next memory location.
- Is that what you want for a linked list, a tree, or a hash? No!
- It must be that operator++ is overloaded.
- Can’t do that for a plain pointer.
  - C++ is extensible, not mutable.
- Therefore, most iterators cannot be plain pointers.

iterator vs. pointer

All pointers are iterators.
Not all iterators are pointers.
Some iterators are pointers.
- string, vector, perhaps.
- list, set, no.

Iterator classifications

BidirectionalIterator (list, set)
- ++, --, ==, !=, *, ->, =
RandomAccessIterator (std::array, vector, string)
- ++, --, +=int, -=int, +int, -int, iter-iter, ==, !=, <, <=, >, >=, *, ->, =
ForwardIterator (forward_list, unordered_set)
- ++, ==, !=, *, ->, =

These are not C++ types. These are descriptions, used to discuss the behavior of a given iterator.

`begin()` & `end()`

.begin() & .end() return iterators. Not necessarily pointers.
.begin() is, conceptually, a pointer to the first element of the container.
.end() is, conceptually, a pointer one past the end of the container.
It’s a half-open interval!

string s = "bonehead";
cout << "First char: " << *s.begin() << '\n';
cout << "Badness: " << *s.end() << '\n';

First char: b
Badness: ␀

string s = "genius";
cout << "First char: " << *s.begin() << '\n';
cout << "Last char:  " << *(s.end()-1) << '\n';

First char: g
Last char:  s

Intervals

Let’s get the concept of a half-open interval clear, because it's important.
In mathematics, [1.0, 3.0] means the infinitude of real numbers, starting & including 1.0, and ending & including 3.0.
(4.0, 4.5) means the infinitude of real numbers, starting at but excluding 4.0, and ending at but excluding 4.5.
[6.2, 9.0) means the infinitude of real numbers, starting & including 6.2, and ending at but excluding 9.0. Therefore, 8.99999 is included, but 9.0 is not.
C++ positively loves the last one.

Intervals

It’s all about [square brackets] vs. (round parens).

[a,b] includes a and b.
(a,b) doesn’t include a, and doesn’t include b.
[a,b) includes a, but doesn’t include b.
- This is a half-open interval.

Half-open intervals

string alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
string::iterator c = alpha.begin()+2, f = c+3;
string foo(c, f);
cout << *c << '\n';
cout << *f << '\n';
cout << foo << '\n';

C
F
CDE

-or-

string alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
auto c = &alpha[2], f = c+3;
cout << *c << '\n';
cout << *f << '\n';
cout << string(c,f) << '\n';

C
F
CDE

foo is not CDEF, because it’s a half-open interval.
Are both f variables the same type? The first is string::iterator, the second is char *. Are those the same?

Half-open Intervals

int a[] = {00, 11, 22, 33, 44, 55, 66, 77};
vector<int> v(a+2, a+5);
for (auto n : v)
    cout << n << ' ';

22 33 44

00	11	22	33	44	55	66	77
		↑			↑
		start			end

The end location is not included.

`.front()` & `.back()`

Some containers have .front() and .back(), which return references to the first and last elements.

list<double> c = {2.3, 5.7, 11.13, 17.19, 23.29};
cout << "First: " << c.front() << '\n';
cout << "Last:  " << c.back() << '\n';

First: 2.3
Last:  23.29

These are not iterators; they are references to the actual data.
The return type of list<double>::front() is double &, not list<double>::iterator.
Note the difference between the types double and double &.

string name = "Jack";
name.front() = 'H';
cout << name;

Hack

Comparisons, part one

This won’t work:

list<string> l = {"kappa", "alpha", "gamma"};
for (auto it = l.begin(); it < l.end(); ++it)
    cout << *it << ' ';

c.cc:2: error: no match for 'operator<' in 'it < 
   l.std::__cxx11::list<std::__cxx11::basic_string<char> >::end()' (operand 
   types are 'std::_List_iterator<std::__cxx11::basic_string<char> >' and 
   'std::__cxx11::list<std::__cxx11::basic_string<char> >::iterator' {aka 
   'std::_List_iterator<std::__cxx11::basic_string<char> >'})

Read the message. It says that < isn’t defined for those iterators.

Comparisons, part two

This will work:

list<string> l = {"kappa", "alpha", "gamma"};
for (auto it = l.begin(); it != l.end(); ++it)
    cout << *it << ' ';

kappa alpha gamma

list<>::iterator is a BidirectionalIterator, not a RandomAccessIterator, and so < isn’t defined. What would it compare? The addresses of the linked list nodes? That’s not useful.

Constructors

Nearly all containers accept a pair of iterators as ctor arguments. These do not have to be iterators for the same type of container.

char date[] = __DATE__;    // E.g., Nov 22 2024
string now(date, date+6);
cout << "month & day: " << now << '\n';

string day_of_month(now.begin()+4, now.begin()+6);
cout << "day: " << day_of_month << '\n';

multiset<int> ms(now.begin(), now.end());
for (auto n : ms)
    cout << n << ' ';
cout << '\n';

month & day: Nov 22
day: 22
32 50 50 78 111 118

`begin()` and `end()` functions

Let’s copy from a C array to a C++ string:

int fido[] = {68, 79, 71};
string current(fido, fido+2);
cout << current << '\n';

DO

Oh dear, I counted wrong. Why am I counting!? Am I a computer?

int fido[] = {68, 79, 71};
string current(fido.begin(), fido.end());
cout << current << '\n';

c.cc:2: error: request for member 'begin' in 'fido', which is of non-class type 
   'int [3]'

Alas, fido has no methods.

`begin()` and `end()` functions

There are also free functions begin() and end(), which work on arrays (not pointers) and all standard containers:

int fido[] = {68, 79, 71};
string current(begin(fido), end(fido));
cout << current << '\n';

DOG

Try that again

It was crazy to use integer ASCII values. Let’s just use a C string:

char fido[] = "DOG";
string current(begin(fido), end(fido));
cout << current << '\n';

DOG␀

What fresh hell is this? What is sizeof(fido)?

Invalidation

Consider this poor code:

int *p = new int(42);
cout << "Before: " << *p << '\n';
delete p;
cout << "After:  " << *p << '\n';

Before: 42
After:  3732

Nothing mysterious here. p was a valid pointer, then it became invalid.
Just because a pointer exists, doesn’t mean that it points to something that you may access.
I know the address of the White House, but they won’t let me move in.

More Invalidation

Another way to invalidate a pointer:

string *p;
{
    string name = "John Jacob Jingleheimer Schmidt";
    p = &name;
    cout << p->size() << ' ' << *p << '\n';
}
cout << p->size() << ' ' << *p << '\n';

There’s some undefined behavior!

Yet More Invalidation

Yet another way to invalidate a value:

vector<int> &foo() {
    vector<int> v = {11,22,33,44,55,66,77};
    auto &r = v;
    return r;
}

int main() {
    for (auto val : foo())
        cout << val << '\n';
}

More undefined behavior.

Iterator invalidation

vector<long> v = {253};
vector<long>::iterator it = v.begin();

cout << "Before: " << *it << '\n';

for (long i=1; i<1000; i++)
    v.push_back(i);

cout << "After:  " << *it << '\n';

Before: 253
After:  5561

Sure, it “pointed” to v[0] just fine, at first.
As we added all those elements to v, unavoidably, v’s data got reallocated to make more room.
The poor iterator referring to v[0] continued to point to the old, obsolete, location.

Lipstick on a pig

Using auto makes the code prettier, but no better:

vector<long> v = {253};
auto it = v.begin();

cout << "Before: " << *it << '\n';

for (long i=1; i<1000; i++)
    v.push_back(i);

cout << "After: " << *it << '\n';

Before: 253
After: 5303

Reservation

Using .reserve() pre-allocates memory:

vector<long> v = {253};
v.reserve(1005);
auto it = v.begin();

cout << "Before: " << *it << '\n';

for (long i=1; i<1000; i++)
    v.push_back(i);

cout << "After:  " << *it << '\n';

Before: 253
After:  253

How often?

How often does re-allocation happen? We can find out, for any particular implemention:

vector<int> v;

for (int i=1; i<=1000; i++) {
    auto before = v.capacity();
    v.push_back(i);
    auto after = v.capacity();
    if (before != after)
        cout << i << ' ' << after << '\n';
}

.capacity(): how much memory is allocated
.size(): how much memory is used

How often?

Similarly, because a std::string is not much more than a vector<char>:

string s;

cout << s.size() << ' ' << s.capacity() << '\n';
for (int i=1; i<10000; i++) {
    auto before = s.capacity();
    s += 'x';
    auto after = s.capacity();
    if (before != after)
        cout << s.size() << ' ' << after << '\n';
}

What happened to our nice powers of two?

Curious String Behavior

for (string s; s.size()<60; s+="abcde")
    cout << s.size() << ' ' << (void *) s.data() << '\n';

0 0x7fffa22aae80
5 0x7fffa22aae80
10 0x7fffa22aae80
15 0x7fffa22aae80
20 0x8c52c0
25 0x8c52c0
30 0x8c52c0
35 0x8c52f0
40 0x8c52f0
45 0x8c52f0
50 0x8c52f0
55 0x8c52f0

Casting? Why can’t we use << s.data() or << &s[0] to get the address of the string data?

Small String Optimization

This is the popular small string optimization.
Big strings (>15 bytes, for g++ 8) are stored in the heap, as usual.
Small strings are kept right in the string object, in the same space as the pointer and lengths.
Small strings are much faster: everything’s on the stack, in cache, no pointers!

Small String Optimization

Possible small string implementation:

class string {
  private:
    size_t size;            // actual string length
    union {
      struct {
        size_t alloc_size;  // amount of heap data
        char *heap;         // ptr to heap data
      };
      char local[16];       // or, put it here
    };
  public:
    char &operator[](size_t i) {
        return (i < sizeof(local)) ? return local[i] : return heap[i];
    }
    … and many more methods …
};

Order Calculation

Still, the small string optimization only applies to, well, small strings.
Bigger data still undergoes reallocation and copying.
All that reallocation looks expensive.
However, science is better than first impressions.
What is the big-O time cost for adding n items to a vector?

Big-O

Let’s push 0…77 onto a vector. Reallocation will occur at power of two sizes:
- Push 0: grow to size=1, copy nothing
- Push 1: grow to size=2, copy 1 int
- Push 2: grow to size=4, copy 2 ints
- Push 4: grow to size=8, copy 4 ints
- Push 8: grow to size=16, copy 8 ints
- Push 16: grow to size=32, copy 16 ints
- Push 32: grow to size=64, copy 32 ints
- Push 64: grow to size=128, copy 64 ints
That’s 1+2+4+8+16+32+64 copies for 78 ints, 127÷78≈1.6 copies/int. O(1)!

Does it scale?

vector<char> v;
int copies = 0, iterations = 1000000;

for (int i=0; i<iterations; i++) {
    auto before = v.capacity();
    v.push_back(i);
    if (before != v.capacity())
        copies += before;
}

cout << double(copies)/iterations;

1.04858

Pre-allocation helps

Again, if we know how many items we’re going to add, we can .reserve() the space:

vector<int> v;
v.reserve(900);

cout << v.size() << ' ' << v.capacity() << '\n';
for (int i=1; i<1000; i++) {
    auto before = v.capacity();
    v.push_back(i);
    auto after = v.capacity();
    if (before != after)
        cout << v.size() << ' ' << after << '\n';
}

0 900
901 1800

Half-open interval

.begin() returns an iterator that “points” to the first element
.end() returns an iterator that “points” one past the last element
They form a half-open interval.
Remember, iterators are often not pointers.

`.begin()` and `.end()`

vector<int> v = {1, 1, 2, 3, 5, 8, 13, 21, 34};
for (vector<int>::iterator it = v.begin(); it != v.end(); ++it)
    cout << *it << ' ';

1 1 2 3 5 8 13 21 34

or:

vector<int> v = {1, 1, 2, 3, 5, 8, 13, 21, 34};
for (auto it = v.begin(); it != v.end(); ++it)
    cout << *it << ' ';

1 1 2 3 5 8 13 21 34

`const`

Why does this fail?

class Foo {
    int sum() const {
        int total = 0;
        for (vector<int>::iterator it = data.begin(); it != data.end(); ++it)
            total += *it;
        return total;
    }
    vector<int> data;
};

c.cc:4: error: conversion from '__normal_iterator<const int*,[...]>' to 
   non-scalar type '__normal_iterator<int*,[...]>' requested

`const`

.sum() is a const method.
Therefore, inside .sum(), data is a const vector<int>.
- Let that sink in for a moment.
- How is a const method implemented, anyway?
- It’s not that the compiler checks for changes to member data inside a const method.
- Instead, the compiler simply regards all data members as const when compiling a const method.
Therefore, data.begin() and data.end() return iterators of type const_iterator, not type iterator.
- They’re overloaded methods!
You can’t assign const_iterator to iterator.

Restate the problem

This is the same problem:

const vector<int> v = {1, 1, 2, 3, 5, 8, 13, 21, 34};
for (vector<int>::iterator it = v.begin(); it != v.end(); ++it)
    cout << *it << ' ';

c.cc:2: error: conversion from '__normal_iterator<const int*,[...]>' to 
   non-scalar type '__normal_iterator<int*,[...]>' requested

Solution #1

One solution—use the correct type:

const vector<int> v = {1, 1, 2, 3, 5, 8, 13, 21, 34};
for (vector<int>::const_iterator it = v.begin(); it != v.end(); ++it)
    cout << *it << ' ';

1 1 2 3 5 8 13 21 34

Solution #2

A better solution—let auto figure it out:

const vector<int> v = {1, 1, 2, 3, 5, 8, 13, 21, 34};
for (auto it = v.begin(); it != v.end(); ++it)
    cout << *it << ' ';

1 1 2 3 5 8 13 21 34

Solution #3

The best solution—avoid all of this:

const vector<int> v = {1, 1, 2, 3, 5, 8, 13, 21, 34};
for (auto val : v)
    cout << val << ' ';

1 1 2 3 5 8 13 21 34

`.cbegin()` and `.cend()`

.begin() is an overloaded method. It returns a const_iterator if the object is const, iterator otherwise.
.end() is an overloaded method. It returns a const_iterator if the object is const, iterator otherwise.
.cbegin(): like .begin(), but always returns const_iterator

.cend(): like .end(), but always returns const_iterator

vector<int> v = {1, 1, 2, 3, 5, 8, 13, 21, 34};
for (auto it = v.cbegin(); it != v.cend(); ++it)
    cout << *it << ' ';

1 1 2 3 5 8 13 21 34

`.rbegin()` and `.rend()`

.rbegin(): like .begin(), but goes the other way
.rend(): like .end(), but goes the other way
.crbegin(): like .cbegin(), but goes the other way

.crend(): like .cend(), but goes the other way

array<int, 9> v = {1, 1, 2, 3, 5, 8, 13, 21, 34};
for (auto it = v.rbegin(); it != v.rend(); ++it)
    cout << *it << ' ';

34 21 13 8 5 3 2 1 1

list<int> v = {1, 1, 2, 3, 5, 8, 13, 21, 34};
for (auto it = v.crbegin(); it != v.crend(); ++it)
    cout << *it << ' ';

34 21 13 8 5 3 2 1 1

Don’t get too excited

Not all containers have the reversed versions.
- For example, forward_list, as a singly-linked list, doesn’t have reverse iterators.
For some containers, iterator and const_iterator are the same.
- A set has inherently-constant iterators, because it’s in order. If you could change the values via iterators, then it wouldn’t be in order any more.
- An unordered_set, a hash, has its own special order, and changing the values in place would be bad.

Plain old data types:

How about old C-style data?

int a[] = {1, 1, 2, 3, 5, 8, 13, 21, 34};
for (auto it = a.begin(); it != a.end(); ++it)
    cout << *it << ' ';

c.cc:2: error: request for member 'begin' in 'a', which is of non-class type 
   'int [9]'

That failed miserably. Perhaps it’s because arrays are not objects.

Free functions

Fortunately, there exist begin() and end() functions, which work for C arrays or STL containers.

begin(con) is defined as, conceptually:

    if con is a C-style array
    then
        con
    else
        con.begin()

Similarly, end(con) is:

    if con is a C-style array
    then
        con+sizeof(con)/sizeof(con[0])
    else
        con.end()

Free functions

int a[] = {1, 1, 2, 3, 5, 8, 13, 21, 34};
for (auto it = begin(a); it != end(a); ++it)
    cout << *it << ' ';

1 1 2 3 5 8 13 21 34

These work for objects, as well:

deque<int> s = {1, 1, 2, 3, 5, 8, 13, 21, 34};
for (auto it = begin(s); it != end(s); ++it)
    cout << *it << ' ';

1 1 2 3 5 8 13 21 34

What use is that? Generality for the sake of generality?

for loop

Consider any for-loop:

forward_list<int> fl = {1, 1, 2, 3, 5, 8, 13, 21, 34};
for (double v : fl)
    cout << v << ' ';

1 1 2 3 5 8 13 21 34

The compiler turns this into (approximately):

forward_list<int> fl = {1, 1, 2, 3, 5, 8, 13, 21, 34};
for (auto it = begin(fl); it != end(fl); ++it) {
    double v = *it;
    cout << v << ' ';
}

1 1 2 3 5 8 13 21 34

Which works for any container type, even C-style arrays.

Efficiency

Foo operator++(int) {
    const auto save = *this;
    ++*this;
    return save;
}

Why do our examples use preincrement (++it) instead of postincrement (it++)?
Preincrement’s generally faster. Postincrement copies & preincrements.
Even if efficiency isn’t a consideration, there’s no clear readability difference, so we may as well be fast.
A smart compiler might make them equivalent.
For actual pointers (int *) or ints, do what you like. Some programmers always use preincrement, as a good habit.

CS253 Iterators

Foundation

Array Traversal

Problems with indexing

Array Traversal via Pointers

vector traversal

Methods

auto is your friend

for loop

It’s simple

Really

Variations

Other containers

iterator type

iterator vs. pointer

Iterator classifications

begin() & end()

Intervals

Intervals

Half-open intervals

Half-open Intervals

.front() & .back()

Comparisons, part one

Comparisons, part two

Constructors

begin() and end() functions

begin() and end() functions

Try that again

Invalidation

More Invalidation

Yet More Invalidation

Iterator invalidation

Lipstick on a pig

Reservation

How often?

How often?

Curious String Behavior

Small String Optimization

Small String Optimization

Order Calculation

Big-O

Does it scale?

Pre-allocation helps

Half-open interval

.begin() and .end()

const

const

Restate the problem

Solution #1

Solution #2

Solution #3

.cbegin() and .cend()

.rbegin() and .rend()

Don’t get too excited

Plain old data types:

Free functions

Free functions

for loop

Efficiency

`vector` traversal

`auto` is your friend

`for` loop

`iterator` type

`begin()` & `end()`

`.front()` & `.back()`

`begin()` and `end()` functions

`begin()` and `end()` functions

`.begin()` and `.end()`

`const`

`const`

`.cbegin()` and `.cend()`

`.rbegin()` and `.rend()`