CS253 Templates

This presentation was stolen from Prof. Chuck Anderson.

Initial Code

double *copyDoubles(const double *src, int num) {
    double *dest = new double[num];
    for (int i=0; i<num; i++) dest[i] = src[i];
    return dest;
}

char *copyLetters(const char *src, int num) {
    char *dest = new char[num];
    for (int i=0; i<num; i++) dest[i] = src[i];
    return dest;
}

void printDoubles(const double *src, int num) {
    for (int i=0; i<num; i++) cout << src[i] << '\n';
}

void printLetters(const char *src, int num) {
    for (int i=0; i<num; i++) cout << src[i] << '\n';
}

int main() {
    double doubles[] = {0.1, 0.2, 0.3, 0.4};
    double *doubles2 = copyDoubles(doubles, 4);
    char *letters2 = copyLetters("bonehead", 8);
    printDoubles(doubles2, 4);
    printLetters(letters2, 8);
}

0.1
0.2
0.3
0.4
b
o
n
e
h
e
a
d

Simplify

Can this code be simplified?
Yep. We don’t need both printDoubles and printLetters.
Both can be named print, because the compiler can disambiguate the calls to print by the types of the arguments.

Simplified

double *copyDoubles(const double *src, int num) {
    double *dest = new double[num];
    for (int i=0; i<num; i++) dest[i] = src[i];
    return dest;
}

char *copyLetters(const char *src, int num) {
    char *dest = new char[num];
    for (int i=0; i<num; i++) dest[i] = src[i];
    return dest;
}

void print(const double *src, int num) {
    for (int i=0; i<num; i++) cout << src[i] << '\n';
}

void print(const char *src, int num) {
    for (int i=0; i<num; i++) cout << src[i] << '\n';
}

int main() {
    double doubles[] = {0.1, 0.2, 0.3, 0.4};
    double *doubles2 = copyDoubles(doubles, 4);
    char *letters2 = copyLetters("bonehead", 8);
    print(doubles2, 4);
    print(letters2, 8);
}

0.1
0.2
0.3
0.4
b
o
n
e
h
e
a
d

Simplify Again

Can we do the same for the copy functions?
Sure, why not?

Simplified Again

double *copy(const double *src, int num) {
    double *dest = new double[num];
    for (int i=0; i<num; i++) dest[i] = src[i];
    return dest;
}

char *copy(const char *src, int num) {
    char *dest = new char[num];
    for (int i=0; i<num; i++) dest[i] = src[i];
    return dest;
}

void print(const double *src, int num) {
    for (int i=0; i<num; i++) cout << src[i] << '\n';
}

void print(const char *src, int num) {
    for (int i=0; i<num; i++) cout << src[i] << '\n';
}

int main() {
    double doubles[] = {0.1, 0.2, 0.3, 0.4};
    double *doubles2 = copy(doubles, 4);
    char *letters2 = copy("bonehead", 8);
    print(doubles2, 4);
    print(letters2, 8);
}

0.1
0.2
0.3
0.4
b
o
n
e
h
e
a
d

Simplified more?

Do you see further simplifications?
No, unless there is some way to parameterize the type of the arguments when you define the functions.

Templates to the rescue!

template <typename T>
T *copy(const T *src, int num) {
    T *dest = new T[num];     // could use auto
    for (int i=0; i<num; i++) dest[i] = src[i];
    return dest;
}

template <class T>            // old-fashioned syntax
void print(T *src, int num) {
    for (int i=0; i<num; i++) cout << src[i] << '\n';
}

int main() {
    double doubles[] = {0.1, 0.2, 0.3, 0.4};
    double *doubles2 = copy(doubles, 4);
    char *letters2 = copy("bonehead", 8);
    print(doubles2, 4);
    print(letters2, 8);
}

0.1
0.2
0.3
0.4
b
o
n
e
h
e
a
d

T is a placeholder, a compile-time argument representing a type.

Templates to the rescue!

template <typename T>
T *copy(const T *src, int num) {
    T *dest = new T[num];     // could use auto
    for (int i=0; i<num; i++) dest[i] = src[i];
    return dest;
}

template <class T>            // old-fashioned syntax
void print(T *src, int num) {
    for (int i=0; i<num; i++) cout << src[i] << '\n';
}

int main() {
    double doubles[] = {0.1, 0.2, 0.3, 0.4};
    double *doubles2 = copy(doubles, 4);
    char *letters2 = copy("bonehead", 8);
    print(doubles2, 4);
    print(letters2, 8);
}

0.1
0.2
0.3
0.4
b
o
n
e
h
e
a
d

The first template parameter is usually named T, for type.
The second template parameter is usually named U, because U comes after T.
T1 & T2 are also common.
Meaningful names are nice, for larger templates.

Templates to the rescue!

template <typename T>
T *copy(const T *src, int num) {
    T *dest = new T[num];     // could use auto
    for (int i=0; i<num; i++) dest[i] = src[i];
    return dest;
}

template <class T>            // old-fashioned syntax
void print(T *src, int num) {
    for (int i=0; i<num; i++) cout << src[i] << '\n';
}

int main() {
    double doubles[] = {0.1, 0.2, 0.3, 0.4};
    double *doubles2 = copy(doubles, 4);
    char *letters2 = copy("bonehead", 8);
    print(doubles2, 4);
    print(letters2, 8);
}

0.1
0.2
0.3
0.4
b
o
n
e
h
e
a
d

template <class T> is old‑fashioned.
template <typename T> is newer.
Either allows non-class types such as int.

Templates to the rescue!

template <typename T>
T *copy(const T *src, int num) {
    T *dest = new T[num];     // could use auto
    for (int i=0; i<num; i++) dest[i] = src[i];
    return dest;
}

template <class T>            // old-fashioned syntax
void print(T *src, int num) {
    for (int i=0; i<num; i++) cout << src[i] << '\n';
}

int main() {
    double doubles[] = {0.1, 0.2, 0.3, 0.4};
    double *doubles2 = copy(doubles, 4);
    char *letters2 = copy("bonehead", 8);
    print(doubles2, 4);
    print(letters2, 8);
}

0.1
0.2
0.3
0.4
b
o
n
e
h
e
a
d

The use of T in copy is completely separate from the use of T in print.

Templates to the rescue!

template <typename T>
T *copy(const T *src, int num) {
    T *dest = new T[num];     // could use auto
    for (int i=0; i<num; i++) dest[i] = src[i];
    return dest;
}

template <class T>            // old-fashioned syntax
void print(T *src, int num) {
    for (int i=0; i<num; i++) cout << src[i] << '\n';
}

int main() {
    double doubles[] = {0.1, 0.2, 0.3, 0.4};
    double *doubles2 = copy(doubles, 4);
    char *letters2 = copy("bonehead", 8);
    print(doubles2, 4);
    print(letters2, 8);
}

0.1
0.2
0.3
0.4
b
o
n
e
h
e
a
d

Note that copy has a const in the arguments.
print does not.
There are four templatized function calls.
What does T represent in each?

Classes

Say we want to keep track of either integer or real 2-D points:

class PointInt {
    int x, y;
  public:
    PointInt(int xarg, int yarg) : x(xarg), y(yarg) { }
    int getx() const { return x; }
    int gety() const { return y; }
};

class PointDouble {
    double x, y;
  public:
    PointDouble(double xarg, double yarg) : x(xarg), y(yarg) { }
    double getx() const { return x; }
    double gety() const { return y; }
};

int main() {
    PointInt pi(1, 2);
    PointDouble pd(3.4, 2.5);
    cout << '(' << pi.getx() << ',' << pi.gety() << ")\n"
         << '(' << pd.getx() << ',' << pd.gety() << ")\n";
}

(1,2)
(3.4,2.5)

You saw this coming

Can we get rid of the duplications between PointInt and PointDouble?
Only if there is a way to parameterize the type of the class.

Template class

template <typename T>
class Point {
    T x, y;
  public:
    Point(T xarg, T yarg) : x(xarg), y(yarg) { }
    T getx() const { return x; }
    T gety() const { return y; }
};

int main() {
    // This statement fails—must specify the type.
    // Point pi(1, 2);

    Point<int> pi(1, 2);
    Point<double> pd(3.4, 2.5);

    cout << '(' << pi.getx() << ',' << pi.gety() << ")\n"
         << '(' << pd.getx() << ',' << pd.gety() << ")\n";
}

(1,2)
(3.4,2.5)

Multiple template parameters

You can have more than one parameter—just be sure you use the right type in the right place.

template <typename Tx, typename Ty>
class Point {
    Tx x;
    Ty y;
  public:
    Point(Tx xarg, Ty yarg) : x(xarg), y(yarg) { }
    Tx getx() const { return x; }
    Ty gety() const { return y; }
};

int main() {
    Point<int, int> pi(1, 2);
    Point<double, double> pd(3.4, 2.5);
    Point<int, double> pid(5, 6.42);
    cout << '(' << pi.getx()  << ',' << pi.gety()  << ")\n"
         << '(' << pd.getx()  << ',' << pd.gety()  << ")\n"
         << '(' << pid.getx() << ',' << pid.gety() << ")\n";
}

(1,2)
(3.4,2.5)
(5,6.42)

Integer template parameters

You can also include a constant integer to specify things like the size of an array, and can have a default value.

Example

template <typename T, int N = 5>
class Handful {                       // Much like std::array
    T data[N];
  public:
    void setData(int index, const T &item) { data[index] = item; }
    friend ostream &operator<< (ostream &out, const Handful<T, N> &h) {
        for (int i=0; i<N; i++)
            out << h.data[i] << '\n';
        return out;
    }
};

int main() {
    Handful<char, 3> h;
    h.setData(0, 'a');
    h.setData(1, 'b');
    h.setData(2, 'c');
    cout << h;
}

a
b
c

Is operator<< a member function? It’s in the class!

A less successful alternative

template <typename T, int N = 5>
class Handful {
    T data[N];
  public:
    void setData(int index, const T &item) { data[index] = item; }
    friend ostream &operator<< (ostream &out, const Handful<T, N> &h) {
        for (int i=0; i<N; i++)
            out << h.data[i] << '\n';
        return out;
    }
};

int main() {
    Handful<int> h;
    h.setData(0, 42);
    h.setData(1, 365);
    cout << h;
}

Range of template parameters

What can a template parameter be?

Types: typename T or class C (how quaint!)
Integerlike: int N, short N, long N, bool B, char C, …
Pointers or references: int *IP, string *SP, …
Another template: template T (not enough time in the semester)

If the template parameter is typename or class, the actual parameter doesn’t have to be a built-in boring type like int or float. It can be a class type, or even filled-out templated type such as set<long>.

Range of template parameters

What can’t a template parameter be?

a floating-point number
a string
an object

Remember, it’s a compile-time parameter: types, integral values, and pointers.

Understand the difference

This is fine:

template<typename T>
class Foo {
  public:
    T data[3] = {1,2,3};
};

int main() {
    Foo<double> f;
    cout << "😀\n";
}

😀

The template parameter is a typename.
The actual argument is the type double.

Understand the difference

This won’t work:

template<double D>
class Foo {
  public:
    double datum = D;
};

int main() {
    Foo<3.16227766> f;
    cout << "😟\n";
}

c.cc:1: error: 'double' is not a valid type for a template non-type parameter

The formal parameter is a double.
The actual argument is 3.16227766.

Standard containers

This is how the standard containers (the STL) are implemented.
There’s nothing magic about them.
<vector> contains, essentially:

    template <typename T>
    class vector {
        …
    };

Templated `using` declarations

A using declaration can be templated to create a new type:

template<class T>
using Counter = map<T, size_t>;

int main() {
    Counter<string> c;
    c["alpha"]++;
    c["beta"]++;
    c["alpha"]++;
    for (auto v : c)
        cout << v.first << ' ' << v.second << '\n';
}

alpha 2
beta 1

Templated variables

God help us, even variables can be templated:

template <typename T>
constexpr T pi = T(3.14159265358979323846264);

template <>
const string pi<string> = "π";

int main() {
    cout << pi<long double> << '\n';
    cout << pi<int> << '\n';
    cout << pi<string> << '\n';
}

3.14159
3
π

What can be templated?

classes
functions
- methods
- ctors
using aliases
variables

Function specialization

You can have both a templated function and non-templated versions:

template <typename T>
void out(T n) {
    cout << "whatever: " << n << '\n';
}

void out(double n) {
    cout << "double: " << n << '\n';
}

int main() {
    out(11);
    out(2.22);
    out(3.33F);
    out("chocolate");
}

whatever: 11
double: 2.22
whatever: 3.33
whatever: chocolate

The ordinary function is a better match than the templated function.

Template specialization

template <typename T>
class Dozen {
    T data[12];
  public:
    size_t size() const { return sizeof(data); }
};

int main() {
    Dozen<int> di;
    Dozen<void> dv;
    cout << di.size() << ' ' << dv.size() << '\n';
}

c.cc: In instantiation of 'class Dozen<void>':
c.cc:10:   required from here
c.cc:3: error: creating array of 'void'

Sometimes, a template doesn't work for all types.

Template specialization

You can have both a templated class and non-templated versions:

template <typename T>
class Dozen {
    T data[12];
  public:
    size_t size() const { return sizeof(data); }
};

template <>
struct Dozen<void> {
    size_t size() const { return 0; }
};

int main() {
    Dozen<int> di;
    Dozen<void> dv;
    cout << di.size() << ' ' << dv.size() << '\n';
}

48 0

CS253: Software Development with C++

Spring 2020

Templates

CS253 Templates

CS253 Templates

Initial Code

Simplify

Simplified

Simplify Again

Simplified Again

Simplified more?

Templates to the rescue!

Templates to the rescue!

Templates to the rescue!

Templates to the rescue!

Templates to the rescue!

Classes

You saw this coming

Template class

Multiple template parameters

Integer template parameters

Example

A less successful alternative

Range of template parameters

Range of template parameters

Understand the difference

Understand the difference

Standard containers

Templated `using` declarations

Templated variables

What can be templated?

Function specialization

Template specialization

Template specialization

CS253 Templates

CS253 Templates

Initial Code

Simplify

Simplified

Simplify Again

Simplified Again

Simplified more?

Templates to the rescue!

Templates to the rescue!

Templates to the rescue!

Templates to the rescue!

Templates to the rescue!

Classes

You saw this coming

Template class

Multiple template parameters

Integer template parameters

Example

A less successful alternative

Range of template parameters

Range of template parameters

Understand the difference

Understand the difference

Standard containers

Templated using declarations

Templated variables

What can be templated?

Function specialization

Template specialization

Template specialization

Templated `using` declarations