C++11 Tutorial: Lambda Expressions — The Nuts and Bolts of Functional Programming

One highlight of C++11 is lambda expressions: function-like blocks of executable statements that you can insert where normally a function call would appear. Lambdas are more compact, efficient, and secure than function objects. Danny Kalev shows you how to read lambda expressions and use them in C++11 applications.

Originally, functional programming in C consisted of defining a full-blown functions and calling them from other translation units. Pointers to functions and file-scope (i.e., static) functions were additional means of diversifying function usage in C.

C++ improved matters by adding inline functions, member functions and function objects. However, these improvements, particularly function objects, proved to be labor intensive.

At last, the final evolutionary stage of functional programming has arrived in the form of lambda expressions.

Lambda by Example

Note: According to the C++11 standard, implementations #include  implicitly when necessary. Therefore, you’re not supposed to #include this header in your programs. However, certain compilers (GCC 4.7 for example) aren’t fully compliant with the C++11 standard yet. Therefore, if the new initialization notation causes cryptic compilation errors, add the directive #include  to your code manually.

Before discussing the technicalities, let’s look at a concrete example. The last line in the following code listing is a lambda expression that screens the elements of a vector according to a certain computational criterion:

//C++11

vector  emps {{“Josh”, 2100.0}, {“Kate”, 2900.0}, {“Rose”,1700.0}};

const auto min_wage = 1600.0;

const auto upper_limit = 1.5*min_wage;

//report which accountant has a salary that is within a specific range

std::find_if(emps.begin(), emps.end(),

[=](const accountant& a) {return a.salary()>=min_wage && a.salary() < upper_limit;});

As all lambda expressions, ours begins with the lambda introducer []. Even without knowing the syntactic rules, you can guess what this lambda expression does: The executable statements between the braces look like an ordinary function body. That’s the essence of lambdas; they function (pun unintended) as locally-defined functions. In our example, the lambda expression reports whether the current accountant object in the vector emps gets a salary that is both lower than the upper limit and higher than the minimum wage.

Inline computations such as this are ideal candidates for lambda expressions because they consist of only one statement.

Dissecting a Lambda Expression

Now let’s dissect the syntax. A typical lambda expression looks like this:

[capture clause] (parameters) -> return-type {body}

As said earlier, all lambdas begin with a pair of balanced brackets. What’s inside the brackets is the optional capture clause. I get to that shortly.

The lambda’s parameters appear between the parentheses. Although you can omit the parentheses if the lambda takes no parameters, I recommend you leave them, for the sake of clarity.

Lambda expressions can have an explicit return type that’s preceded by a -> sign after the parameter list (find out more about this new functionality in Let Your Compiler Detect the Types of Your Objects Automatically). If the compiler can work out the lambda’s return type (as was the case in the first example above), or if the lambda doesn’t return anything, you can omit the return type.

Finally, the lambda’s body appears inside a pair of braces. It contains zero or more statements, just like an ordinary function.

Back to the find_if() call. It includes a lambda expression that takes const accountant&. Where did this parameter come from? Recall that find_if() calls its predicate function with an argument of type *InputIterator. In our example, *InputIterator is accountant. Hence, the lambda’s parameter is const accountant&. The find_if() algorithm invokes the lambda expression for every accountant in emps.

Since our lambda’s body consists of the following Boolean expression:

{return a.salary()>= min_wage && a.salary() < upper_limit;}

The compiler figures out that the lambda’s return type is bool. However, you may specify the return type explicitly, like this:

[=](const accountant& a)->bool

{return a.salary()>= min_wage && a.salary() < upper_limit;});

Capture Lists

Unlike an ordinary function, which can only access its parameters and local variables, a lambda expression can also access variables from the enclosing scope(s). Such a lambda is said to have external references. In our example, the lambda accesses, or captures, two variables from its enclosing scope: min_wage and upper_limit.

There are two ways to capture variables with external references:

• Capture by copy
• Capture by reference

The capture mechanism is important because it affects how the lambda expression manipulates variables with external references.

At this stage I’m compelled to divulge another behind-the-scenes secret. Conceptually, the compiler transforms every lambda expression you write into a function object, according to the following guidelines:

• The lambda’s parameter list becomes the parameter list of the overloaded operator().
• The lambda’s body morphs into the body of the overloaded operator().
• The captured variables become data members of the said function object.

The compiler-generated function object is called the closure object. The lambda’s capture clause thus defines which data members the closure will have, and what their types will be. A variable captured by copy becomes a data member that is a copy of the corresponding variable from the enclosing scope. Similarly, a variable captured by reference becomes a reference variable that is bound to the corresponding variable from the enclosing scope.

A default capture specifies the mechanism by which all of the variables from the enclosing scope are captured. A default capture by copy looks like this:

[=] //capture all of the variables from the enclosing scope by value

A default capture by reference looks like this:

[&]//capture all of the variables from the enclosing scope by reference

There is also a third capture form that I will not discuss here for the sake of brevity. It’s used in lambdas defined inside a member function:

[this]//capture all of the data members of the enclosing class

You can also specify the capture mechanism for individual variables. In the following example min_wage is captured by copy and upper_limit by reference:

[min_wage, &upper_limit](const accountant& a)->bool

{return a.salary()>= min_wage && a.salary() < upper_limit;});

Finally, a lambda with an empty capture clause is one with no external references. It accesses only variables that are local to the lambda:

[] (int i, int j) {return i*j;}

Let’s look at a few examples of lambdas with various capture clauses:

vector v1={0,12,4}, v2={10,12,14,16}; //read about the new C++11 initialization notation

[&v1](int k) {vi.push_back(k); };  //capture v1 by reference

[&]  (int m) {v1.push_back(m); v2.push_back(m) };//capture vi and vi2 by ref

[v1]() //capture v1 by copy

{for_each(auto y=v1.begin(), y!=v1.end(), y++) {cout<

auto factorial = [](int i, int j) {return i * j;};

int arr{1,2,3,4,5,6,7,8,9,10,11,12};

long res = std::accumulate(arr, arr+12, 1, factorial);

cout<<"12!="<

inline int factorial (int n, int m)

{

return n*m;

}