CS253 Lambda Functions

Lambda

This is the Greek letter lambda.

In C++, it refers to an anonymous function, an unnamed function, a function literal.

Functions are not first-class objects in C++.
You can’t have a variable of type “function”, or even of type “function taking no arguments and returning int”.
You can’t copy a function.
However, you can have a pointer to a function.

Here’s a boring ordinary function:

bool odd(int n) { return n & 01; }

int main() {
    cout << odd(42) << ' ' << odd(43) << '\n';
}

0 1

Boolean values are displayed as 0 and 1 by default, unless you use cout << boolalpha.

bool odd(int n) { return n & 01; }

int main() {
    bool (*p)(int) = odd;
    cout << (*p)(42) << ' ' << p(43) << '\n';
}

0 1

p is a variable of type “pointer to function taking a int and returning a bool”.
Or, p is a pointer to function taking a int and returning a bool.
So, *p is a function that takes a int and returns a bool.
Some people (not me!) find such declarations hard to understand.
For convenience, you can use a pointer to a function to call it.

bool odd(int n) { return n & 01; }

int main() {
    auto p = odd;
    cout << (*p)(42) << ' ' << p(43) << '\n';
}

0 1

auto sure made that declaration easier!

It seems a shame to have to name the function odd.
Imagine if integers worked that way:
- You couldn’t have an integer constant.
- All you could have would be named integers: int foo=7;
It’s nice to have integer constants, integer literals, unnamed integers.
If only we could have unnamed functions …

auto p = [](int n) -> bool { return n & 01; };
cout << p(42) << ' ' << p(43) << '\n';

0 1

Let’s omit that return-type declaration:

auto p = [](int n) { return n & 01; };
cout << p(42) << ' ' << p(43) << '\n';

0 1