CS253 Functors

Please pronounce carefully.

A Method

once: 💉

twice: 💉💉 thrice: 💉💉💉

Consider this class:

class Foo {
  public:
    int thrice (int n) const { return n*3; }
};

Foo f;
cout << f.thrice(42) << '\n';

int thrice (int n) const
- return type: int
- name of the method: thrice
- argument list: (int n)
- const: it doesn’t change any of the no data in class Foo

An operator function

Let’s use the unary ~ operator, defined as bit-wise complement for integers.

class Foo {
  public:
    const char * operator~ () const { return "hello"; }
};

Foo f;
cout << ~f << '\n';

hello

const char * operator~ () const
- return type: const char *
- name of the method: operator~
- argument list: ()
- const: it doesn’t change any object data
Not to be confused with the destructor, which is named ~Foo.

Another operator function

Let’s use the unary postfix () operator, alias the “function call” operator:

class Foo {
  public:
    double operator() (int i, float f) const { return i+f; }
};

Foo f;
cout << f(3, 1.2) << '\n';

4.2

double operator() (int i, float f) const
- return type: double
- name of the method: operator()
- argument list: (int i, float f)
- const: it doesn’t change any object data
f is not a function! f is a Foo object! It sure behaves as a function.

Sounds useless

😒

Hooray!
We’ve created an alternate way to write a function.
It has the advantage of being more confusing.
However, it can also have state.

Another operator function

Let’s write a functor that returns its argument+1:

class BiggerBy1 {
  public:
    int operator()(int i) const { return i+1; }
};

BiggerBy1 b;
cout << b(17) << '\n';

int operator()(int i) const
- return type: int
- name of the method: operator()
- argument list: (int i)
- const: it doesn’t change any object data

Another operator function

It’s easy to change that to increment by two, instead:

class BiggerBy2 {
  public:
    int operator()(int i) const { return i+2; }
};

BiggerBy2 b;
cout << b(17) << '\n';

However, it would be tedious to have to write many such functors for all the different increment values we might want to use.

A stateful functor

Let’s give the functor a ctor argument:

class Bigger {
    const int increment;
  public:
    Bigger(int inc) : increment(inc) { }
    int operator()(int i) const { return i+increment; }
};

Bigger b(4), c(12);
cout << b(17) << ' ' << b(100) << '\n'
     << b(1000) << ' ' << c(2000) << '\n';

21 104
1004 2012

The parens in the ctor, b(4), are very different than the parens in the expression b(17).
- b(4) constructs an object of type Bigger, named b.
- b(17) is b.operator()(17).
How do I tell if it’s calling the ctor or operator()?
- Well, are you constructing the object? Is this its creation ?

ID numbers

The Local Static lecture generated ID numbers using a local static variable. Now, with a functor:

class ID {
    long id;
    const short inc;
  public:
    ID(long id=800'00'0000, short inc=13) : id(id), inc(inc) { }
    auto operator()() { return id += inc; } // NOT const
};

int main() {
    ID student_id;  // NOT student_id()
    ID employee_id(10000,1);
    cout << student_id()  << ' ' << student_id()  << '\n'
         << employee_id() << ' ' << employee_id() << '\n';
}

800000013 800000026
10001 10002

Another stateful functor

Another use of memory. You recall that NAN means Not a Number. No const, because operator() changes previous.

class Queue {
    double previous = NAN;
  public:
    double operator()(double d) {
        const auto save = previous;
        previous = d;
        return save;
    }
};

Queue q;
cout << q(1.2) << '\n';
cout << q(9.9) << '\n';
cout << q(3.4) << '\n';

nan
1.2
9.9

Or …

class Total {
    int sum=0;
  public:
    int operator()(int n) {
        return sum += n;
    }
};

Total t;
for (int i=0; i<=10; i++)
    cout << "Add " << i << " sum " << t(i) << '\n';

Add 0 sum 0
Add 1 sum 1
Add 2 sum 3
Add 3 sum 6
Add 4 sum 10
Add 5 sum 15
Add 6 sum 21
Add 7 sum 28
Add 8 sum 36
Add 9 sum 45
Add 10 sum 55

Sorting with Functors

How does a container like set know the sorting order?
Sure, it’s easy to answer for int and string, but what about class Student?
- Sort by major?
- Sort by NetId?
- Sort by last name, then first name, then CSUID?
Besides, even for int, you might want to sort in a different order.
You might even want multi-key sorting, like a phone book:
- primary key: last name
- secondary key: first name
- tertiary key: middle name
- quaternary key: street name
- quinary key: house number

The 🞬🞬🞬🞬🞬 🞬🞬🞬🞬🞬🞬

Imagine that you had a semi-brilliant assistant who was very good at sorting.
However, your assistant doesn’t know alphabetical order.
He can do everything that’s needed for sorting except actually determine if, say, “Jay” comes before “David”.
Therefore, he frequently comes to you, asking if this word comes before this other word.
set is like that. It knows the mechanics of sorting, but it doesn’t know what order to use. You need to help it.

Comparison functor

You provide set with a comparison functor.
The set code makes comparisons using the functor.
The functor only compares two items at a time.
- I say again, two items at a time. Only two. The functor compares—it doesn’t sort.
The functor returns true iff the first argument should be before the second argument in the set. That’s all.

Why can’t set use a comparison function, as opposed to a comparison functor ?

Because the set template takes types, not objects or values.

Implicit/explicit

Implicit sorting criterion:

set<int> s = {31, 49, 41, 26, 53, 58, 97, 93, 23, 84};
for (auto v : s)
    cout << v << ' ';

23 26 31 41 49 53 58 84 93 97

Explicit sorting criterion:

set<int, less<int> > s = {31, 49, 41, 26, 53, 58, 97, 93, 23, 84};
for (auto v : s)
    cout << v << ' ';

23 26 31 41 49 53 58 84 93 97

What’s this > > business?

An old problem with the >> token. Compilers used to have problems regarding >> as the shift-right operator, as opposed to two closing angle brackets. It’s no longer needed, but you still see it in old code, or new code written by old coders 🧓.

Reversed

Use greater<int> to sort from biggest to smallest:

set<int, greater<int>> s = {31, 49, 41, 26, 53, 58, 97, 93, 23, 84};
for (auto v : s)
    cout << v << ' ';

97 93 84 58 53 49 41 31 26 23

Explicitly explicit

Use a comparison functor that behaves the same as less<int>:

struct compare {                            // 😱 😱 struct‽ 😱 😱
    bool operator()(int a, int b) const {
        return a < b;
    }
};

set<int, compare> s = {31, 49, 41, 26, 53, 58, 97, 93, 23, 84};
for (auto v : s)
    cout << v << ' ';

23 26 31 41 49 53 58 84 93 97

Make sure that the comparison functor is const. Compilers are getting fussier about this for comparison functors used with sort() or STL containers.

Two-level

Use a comparison functor that does a two-level comparison; primarily by proximity to 45, secondarily by value.

struct compare {
    bool operator()(int a, int b) const {
        int da = abs(a-45), db = abs(b-45); // distance from 45
        if (da != db)
            return da < db; // Primary sort: distance from 45
        return a < b;       // Secondary sort: value of number
    }
};

set<int, compare> s = {31, 49, 41, 26, 53, 58, 97, 93, 23, 84};
for (auto v : s)
    cout << v << ' ';

41 49 53 58 31 26 23 84 93 97

Ternary operator

Another way to write a two-level comparison functor:

struct compare {
    bool operator()(int a, int b) const {
        int da = abs(a-45), db = abs(b-45); // distance from 45
        return (da != db) ? da < db : a < b;
    }
};

set<int, compare> s = {31, 49, 41, 26, 53, 58, 97, 93, 23, 84};
for (auto v : s)
    cout << v << ' ';

41 49 53 58 31 26 23 84 93 97

Let `pair::operator<` do the work.

A cool technique: offload the multi-level comparison to pair, which naturally compares that way:

struct compare {
    bool operator()(int a, int b) const {
        pair pa = {abs(a-45), a},          // Love that CTAD!
             pb = {abs(b-45), b};
        return pa < pb;
    }
};

set<int, compare> s = {31, 49, 41, 26, 53, 58, 97, 93, 23, 84};
for (auto v : s)
    cout << v << ' ';

41 49 53 58 31 26 23 84 93 97

“But that’s less efficient.” “Did you measure it, or are you guessing?”

Let `pair::operator<` do the work.

Or, eliminating the pair variables that are only used once:

struct compare {
    bool operator()(int a, int b) const {
        return pair{abs(a-45), a} < pair{abs(b-45), b};
    }
};

set<int, compare> s = {31, 49, 41, 26, 53, 58, 97, 93, 23, 84};
for (auto v : s)
    cout << v << ' ';

41 49 53 58 31 26 23 84 93 97

Curious criteria

This sorts first by final digit, then by value:

struct compare {
    bool operator()(int a, int b) const {
        int last_a = a%10, last_b = b%10;  // Final digits
        if (last_a != last_b)
            return last_a < last_b; // Primary sort: by last digit
        return a < b;               // Secondary sort: value of number
    }
};

set<int, compare> s = {31, 49, 41, 26, 53, 58, 97, 93, 23, 84};
for (auto v : s)
    cout << v << ' ';

31 41 23 53 93 84 26 97 58 49

Multi-level

struct student {
    int id; float gpa; string name;
};
struct compare {
    bool operator()(const student &a, const student &b) const {
        return (a.gpa != b.gpa) ? a.gpa > b.gpa : a.id < b.id;
    }
};
set<student,compare> s = {
    {81234567, 3.9, "Jack Applin"},
    {82828282, 0.5, "Craig Partridge"},
    {84444444, 2.2, "Homecoming Queen"},
    {83333333, 2.2, "Joe Kolledge"},
};
for (const auto &v : s)
    cout << v.id << ' ' << v.gpa << ' ' << v.name << '\n';

81234567 3.9 Jack Applin
83333333 2.2 Joe Kolledge
84444444 2.2 Homecoming Queen
82828282 0.5 Craig Partridge

Not all

Of course, not all collections require a comparison functor.
vector, array, and list just go in the order you specify, and so have no comparison functor.
unordered_set needs only an equality comparison (and a hash functor), so its comparison functor defaults to equal_to, not less.

Example #1

We’re going to build up to using functors and λ-expressions.

char embiggen(char c) {
    if ('a' <= c && c <= 'z')  // 'a' <= c <= 'z' will NOT work
        return c - 'a' + 'A';
    else
        return c;
}

int main() {
    string name = "Beverly Hills Chihuahua";
    for (char &c : name)    // & for reference
        c = embiggen(c);
    cout << name << '\n';
}

BEVERLY HILLS CHIHUAHUA

Example #2

Use a ternary expression, instead:

char embiggen(char c) {
    return ('a'<=c && c<='z') ? c-'a'+'A' : c;
}

int main() {
    string name = "Beverly Hills Chihuahua";
    for (char &c : name)
        c = embiggen(c);
    cout << name << '\n';
}

BEVERLY HILLS CHIHUAHUA

Example #3

Use the transform() algorithm, rather than an explicit loop:

char embiggen(char c) {
    return ('a'<=c && c<='z') ? c-'a'+'A' : c;
}

int main() {
    string name   = "Beverly Hills Chihuahua";
    string result = name;
    transform(name.begin(), name.end(),
              result.begin(), embiggen);
    cout << result << '\n';
}

BEVERLY HILLS CHIHUAHUA

Why assign name to result?

Can’t write to a string with no space allocated.

Example #4

This example uses the same buffer for input & output of transform().

char embiggen(char c) {
    return ('a'<=c && c<='z') ? c-'a'+'A' : c;
}

int main() {
    string name = "Beverly Hills Chihuahua";
    transform(name.begin(), name.end(),
              name.begin(), embiggen);
    cout << name << '\n';
}

BEVERLY HILLS CHIHUAHUA

Example #5

This code uses an actual functor, as opposed to a function.

class embiggen {
  public:
    char operator()(char c) const {
        return ('a'<=c && c<='z') ? c-'a'+'A' : c;
    }
};

string name = "Beverly Hills Chihuahua";
embiggen biggifier;
transform(name.begin(), name.end(),
          name.begin(), biggifier);
cout << name << '\n';

BEVERLY HILLS CHIHUAHUA

Example #6

Create a temporary functor object:

class embiggen {
  public:
    char operator()(char c) const {
        return ('a'<=c && c<='z') ? c-'a'+'A' : c;
    }
};

string name = "Beverly Hills Chihuahua";
transform(name.begin(), name.end(),
          name.begin(), embiggen());
cout << name << '\n';

BEVERLY HILLS CHIHUAHUA

Example #7

Instead of a functor, a lambda-expression can also be used.

string name = "Beverly Hills Chihuahua";

transform(
    name.begin(), name.end(),
    name.begin(),
    [](char c){ return 'a'<=c && c<='z' ? c-'a'+'A' : c; }
);
cout << name << '\n';

BEVERLY HILLS CHIHUAHUA

Example #8

Don’t re-invent the wheel:

string name = "Beverly Hills Chihuahua";
for (char &c : name)
    c = toupper(c);
cout << name << '\n';

BEVERLY HILLS CHIHUAHUA

Sprechen Sie Deutsch?

Warning notice on the traffic light system of the Dresden Roads and Civil Engineering Office

The previous eight examples assume:

Each lowercase letter fits in a char.
Each uppercase letter is also one byte.
Each lowercase letter has an uppercase equivalent.

However:

Eszett, ß, U+00DF, an sz ligature, is two bytes in UTF-8.
The capital letter, ẞ, U+1E9E, is three bytes in UTF-8 (because it came later?).
Pre-2017, ẞ was unofficial in Germany, so “Straße” became “STRASSE” in uppercase. Post-2017, “STRAẞE” is also acceptable.

CS253: Software Development with C++

Fall 2022

Functors

CS253 Functors

A Method

An operator function

Another operator function

Sounds useless

Another operator function

Another operator function

A stateful functor

ID numbers

Another stateful functor

Or …

Sorting with Functors

The 🞬🞬🞬🞬🞬 🞬🞬🞬🞬🞬🞬

Comparison functor

Implicit/explicit

Reversed

Explicitly explicit

Two-level

Ternary operator

Let `pair::operator<` do the work.

Let `pair::operator<` do the work.

Curious criteria

Multi-level

Not all

Example #1

Example #2

Example #3

Example #4

Example #5

Example #6

Example #7

Example #8

Sprechen Sie Deutsch?

CS253 Functors

A Method

An operator function

Another operator function

Sounds useless

Another operator function

Another operator function

A stateful functor

ID numbers

Another stateful functor

Or …

Sorting with Functors

The 🞬🞬🞬🞬🞬 🞬🞬🞬🞬🞬🞬

Comparison functor

Implicit/explicit

Reversed

Explicitly explicit

Two-level

Ternary operator

Let pair::operator< do the work.

Let pair::operator< do the work.

Curious criteria

Multi-level

Not all

Example #1

Example #2

Example #3

Example #4

Example #5

Example #6

Example #7

Example #8

Sprechen Sie Deutsch?

Let `pair::operator<` do the work.

Let `pair::operator<` do the work.