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CS253 Lambda Functions

Lambda

λ

This is the Greek letter lambda.

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

Functions

• 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.

A function

Here’s a boring ordinary function:

bool odd(int n) { return n & 0b1; }

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

0 1

0b indicates a binary constant. Sure, I could have written 0b1 as 1, 01, or 0x01, but 0b1 stresses that it’s a bitmask.

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

Pointer to a function

bool odd(int n) { return n & 0b1; }

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

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• 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, * is optional when calling the function.

auto is your friend

bool odd(int n) { return n & 0b1; }

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

0 1

auto sure made that declaration easier!

There must be a better way.

• It seems a shame to have to name the function odd.
• Imagine if numbers worked that way:

cout << 355/113.0 << '\n';  // easy   

3.14159

int a=355;
float b=113.0;
cout << a/b << '\n';        // tedious

3.14159

• It’s convenient to have unnamed numeric constants.
• If only we could have unnamed functions …

Alternate Function Syntax

There are two ways to declare a function return type:

int alpha() { return 42; }
auto beta() -> int { return 66; }

int main() {
    cout << alpha() << ' ' << beta();
}

42 66

The alternate syntax exists to make template programming simpler; it allows us to deduce the return type using the arguments.

Patience—neither of these are lambda functions.

Alternate Function Syntax

If the return type is not given, it is deduced:

int alpha() { return 42; }
auto gamma() { return 88; }

int main() {
    cout << alpha() << ' ' << gamma();
}

42 88

Alternate Function Syntax

However, the return type must be unambiguous:

auto delta(bool flag) {
    if (flag)
        return 5;
    else
        return 6.7;
}

int main() {
    cout << delta(true);
}

c.cc:5: error: inconsistent deduction for auto return type: 'int' and then 
   'double'

λ-expressions

auto p = [](int n) -> bool { return n & 0b1; };
cout << p(42) << ' ' << p(43) << '\n';

0 1

• [](int n) -> int { return n & 0b1; } is a lambda-expression
  • [] is the capture-specification
    • Stuff can go inside [], but we’ll ignore that.
    • Think of it as “here comes a λ-function”.
  • (int n) is the argument list
  • -> bool specifies the return type
    • optional, if the compiler can figure it out
  • { return n & 0b1; } is the body
• I often forget the ; at the end of the first line. It’s an assignment statement, and so ends with a semicolon.

λ-expressions

The return-type may be deduced, if omitted:

auto p = [](int n) { return n & 0b1; };
cout << p(42) << ' ' << p(43) << '\n';

0 1

What is the return-type of the lambda-expression pointed to by p?

Generic λ-expressions

Even the arguments can be auto:

auto twice = [](auto v) { return v + v; };

cout << twice(3) << '\n'
     << twice(2.34) << '\n'
     << twice("Jack"s) << '\n'
     << twice('!') << '\n';

6
4.68
JackJack
66

This is more like a templated function, in that it defines a family of functions.

No arguments

If no arguments are needed, the () can be omitted:

auto unique_id = [] { return getpid(); };
cout << unique_id() << '\n';

1021000

Use

• Well, this is all very impressive, but what good is it?
• Surely, it’s easier just to use regular functions.
• Functors and λ-expressions are useful with algorithms.