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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);
}

0 1

The 0b prefix indicates a binary constant. Sure, 1, 01, or 0x01 would also work, 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);
}

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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 a 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);
}

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🧙 Even though C++ wizards understand bool (*p)(int) = odd; just fine, we realize that 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, always having names:

cout << 355/113.0;              // easy as

3.14159

int numerator=355;
float denominator=113.0;
cout << numerator/denominator;  // tedious

3.14159

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

Trailing return type

There are two ways to declare a function return type:

• Standard declaration, with the return type first, before the arguments are declared.
• Trailing return type, with an auto placeholder, then the arguments, so we can use the arguments to figure out the return type in a template.

bool odd_and(int n) { return n & 0b1; }
auto odd_mod(int n) -> bool { return n % 2 != 0; }

int main() {
    cout << odd_and(12) << ' ' << odd_mod(13);
}

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Patience—neither of these are lambda functions, because they have names.

Deduced Return Type

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

bool odd_and(int n) { return n & 0b1; }
auto odd_mod(int n) { return n % 2 != 0; }

int main() {
    cout << odd_and(12) << ' ' << odd_mod(13);
}

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This use of auto does not indicate any sort of run-time flexibility. Instead, it instructs the compiler to figure it out at compile time, the same as auto answer=42; figures out the type of answer from the type of 42. In neither case is there any run-time flexibility.

The return type of odd_mod is not auto, it’s the actual type that is returned.

Deduced Return Type

However, the return type must be unambiguous:

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

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

c.cc:4: 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);

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• [](int n) -> bool { return n & 0b1; } is a λ-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
    • optional, if empty
  • -> bool specifies the trailing 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);

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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 radically different than using auto as the return value.
• Instead, this is more like a templated function, in that it defines a family of functions, taking different argument types.
• + works very differently for the different argument types: int, double, and char do addition, whereas std::string concatenates strings.

No arguments

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

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

3108632

Of course, this example could be accomplished with a simple pointer:

auto unique_id = getpid;     // no parens
cout << unique_id();

3108633

Use

• Well, this is all very impressive, but what good is it?
• Surely, it’s easier just to use regular functions.
• It’s more general than bind(), which is the older way of doing this.
• Functors and λ-expressions are useful with algorithms.