# C++ Lambda

A basic lambda which just adds two integer values is defined:

[] (int a, int b) -> int { return a + b; }


Lambda anatomy:

• [] is the capture clause:
• [name] passes a const clone of variable name which can be accessed in the body of the lambda
• [&name] passes a mutable reference of name which can be accessed in the body of the lambda
• [x, y, z] performs multiple captures
• [=] allows us to capture anything in the current scope
• [&] allows us to capture anything in the current scope by reference
• (int a, int b) are the lambda parameters.
• Can use auto here to make the lambda generic; however for every new type used for the auto parameter, a new lambda will be generated.
• -> int specifies the return type
• this can be omitted:
• [] (int a, int b) { return a + b; })
• but it’s a good idea to be explicit
• { return a + b; } is the body
• unlike python lambdas, we can put whatever we like here (python only allows an expression).

In c++ lambdas are objects (called functors) which do specific things like:

• store captures as data members
• override the operator() so the object is callable

Even though lambdas are objects, it is preferable that they are kept stateless. Stateless lambdas are easier to understand and are simple to copy or pass without unexpected behavior.

Lambda captures can be made mutable, by making the lambda mutable:

[c] (int a, int b) mutable -> int { c++; return a + b + c; }


However, c becomes a data member of the lambda; therefore, mutating it won’t mutate the original variable c. To mutate the original value of c use a reference:

[&c] (int a, int b) -> int { c++; return a + b + c; }


Captures [&] and [=] are usually favored because there is no penalty for not explicitly defining what a lambda should capture. All of the captures are resolved at compile time, so the compiler will construct the object defintion on only the captures that the lambda needs.

Like any other object which stores a reference, lambda should not outlive the data it references.

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