Lambda Expressions¶
“A lambda expression is just that: an expression. It is part of the source code” [Meyer].
In other words, we can say that a lambda expression allows to write a function inside of another function.
auto can be used as a lambda expression, as a type deduction, or to
handle STL containers.
auto as a lambda expression¶
If you place the below expression inside of a main() function (for
example), this will work:
const auto lambda_momentum = [] (const double &px, const double &py){
const double pz = px * 0.6 + py * 0.8;
const double &momentum = std::sqrt(px * px + py * py + pz * pz);
return momentum;
};
auto as type deduction¶
auto can be used as a type. This is very useful when objects are very large
and hard to handle. See the auto documentation for more explanation about
type deduction. For example: std::map<std::string,int>::iterator:
for (auto it=StringIntMap.begin(); it != StringIntMap.end(); it++){
std::cout << "Key: " << it->first << " Value " << it->second << std::endl;
}
In the above example, auto does not work as lambda expression, this works
as type deduction.
Use a lambda expression to handle a STL container¶
You should always prefer to use STL algorithms instead of handwritten calculations [Sutter_Alexandrescu]. Below is an example of how to use lambda expressions to manipulate a STL vector:
std::vector<int>::iterator first_even = // this line can be replaced by "auto", used as Type Deduction
std::find_if(IntVect.begin(), IntVect.end(), [](int it) {
return it % 2==0;
});
std::cout << "The first even number is " << * first_even << std::endl;
References¶
For further, more complete examples, please refer to this code from the August 2016 code sprint.
Effective Modern C++ by Scott Meyers (O’Reilly). Copyright 2015 Scott Meyers, 978-1-491-90399-5.
C++ coding standards : 101 rules, guidelines, and best practices by Herb Sutter, Andrei Alexandrescu (Addison-Wesley Professional). Copyright 2005 Pearson Education, Inc, 0-321-11358-6.
