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Output of the inner function becomes the input of the outer function.

Composition requires output-to-input type match.

` g:: double -> double`

` f:: double -> bool `

Function f expects a single value, however g returns an array.

` g:: int -> [int]`

` f:: int -> bool `

This problem can be solved with monads and Kleisli composition.

`f(g(x))`

?`f(x) = x`^{2} - x - 1

`g(x) = x + 1`

Hint:

Substitute

`x`

in `f`

with the definition of g.autocompose= [](auto f, auto g) { return [=](auto x) { return f(g(x)); }; };

auto op =compose(f2, f1);

Unary function composition only.

doublefahrenheitToCelsius(double celsius){ return (celsius - 32) * 5.0 / 9.0; }

boolisRoomTemperature(double t){ return (t >= 20.0 && t < 25.0) ? true : false; }

Notice the order, we compose from right to left.

auto op =compose(isRoomTemperature, fahrenheitToCelsius); auto val1 =op(71.9); // returns 1, i.e. room temperature auto val2 =op(50.3); // returns 0, i.e. not room temperature

We could have created a distinct function for this purpose:

autoopComposed(double fahr){ return isRoomTemperature(fahreheitToCelsius(fahr)); }

int main() { auto val =opComposed(71.9); // room temperature }

auto vecBool =map(vecTemp,compose(isRoomTemperature, fahrenheitToCelsius));

The composed function can be directly fed into higher-order functions, such as map and filter.

`f(g(x))`

, then `g(f(x))`

`f(x) = 0.5x - 5`

`g(x) = 2x + 10`

Hint:

Try a different function pair.

`f • g ≠ g • f`

The result happened to be the same, because the functions were mutual inverses.

doublebmi(double weight, double height){ return weight / height / height; }

boolisOverweight(double bmi){ return (bmi >= 25.0) ? true : false; }

autocompose= [](auto f, auto g) { return [=](auto&& ... x) { return f(g(x...)); }; };

Three dots

auto op =compose(isOverweight, bmi); auto val1 =op(59.8, 1.7); // returns 0, i.e. false auto val2 =op(73.0, 1.52); // returns 1, i.e. true

`(f • g) • h = f • (g • h) `

`compose(compose(f,g),h)=compose(f,compose(g,h))`

`(f • g) • h `

and ` f • (g • h)`

?

`f(x) = x - 2, g(x) = x`^{2}, h(x) = 2x

,
Full application produces a value, partial application produces a function that "remembers" values.

Hint:

Were concrete values supplied?

Function composition

- produces a more complicated function
- requires output-to-input type match
- its result is another function
- is associative but generally not commutative
- is expressed via nested lambda expressions
- multi-argument composition in C++ requires the use of the parameter pack