What is a C++ lambda expression? Why does C++ have them, what problems do they solve that were not solvable prior to their addition? And how can I benifit from using them?
Please provide an example or two, and/or examples of when & where the lambda expression feature should be applied.
C++ includes useful generic functions like std::for_each and std::transform, which can be very handy. Unfortunately they can also be quite cumbersome to use, particularly if the functor you would like to apply is unique to the particular function.
#include <algorithm> #include <vector> namespace { struct f { void operator()(int) { // do something } }; } void func(std::vector<int>& v) { f f; std::for_each(v.begin(), v.end(), f); } If you only use f once and in that specific place it seems overkill to be writing a whole class just to do something trivial and one off.
In C++03 you might be tempted to write something like the following, to keep the functor local:
void func2(std::vector<int>& v) { struct { void operator()(int) { // do something } } f; std::for_each(v.begin(), v.end(), f); } however this is not allowed, f cannot be passed to a template function in C++03.
C++11 introduces lambdas allow you to write an inline, anonymous functor to replace the struct f. For small simple examples this can be cleaner to read (it keeps everything in one place) and potentially simpler to maintain, for example in the simplest form:
void func3(std::vector<int>& v) { std::for_each(v.begin(), v.end(), [](int) { /* do something here*/ }); } Lambda functions are just syntactic sugar for anonymous functors.
In simple cases the return type of the lambda is deduced for you, e.g.:
void func4(std::vector<double>& v) { std::transform(v.begin(), v.end(), v.begin(), [](double d) { return d < 0.00001 ? 0 : d; } ); } however when you start to write more complex lambdas you will quickly encounter cases where the return type cannot be deduced by the compiler, e.g.:
void func4(std::vector<double>& v) { std::transform(v.begin(), v.end(), v.begin(), [](double d) { if (d < 0.0001) { return 0; } else { return d; } }); } To resolve this you are allowed to explicitly specify a return type for a lambda function, using -> T:
void func4(std::vector<double>& v) { std::transform(v.begin(), v.end(), v.begin(), [](double d) -> double { if (d < 0.0001) { return 0; } else { return d; } }); } So far we've not used anything other than what was passed to the lambda within it, but we can also use other variables, within the lambda. If you want to access other variables you can use the capture clause (the [] of the expression), which has so far been unused in these examples, e.g.:
void func5(std::vector<double>& v, const double& epsilon) { std::transform(v.begin(), v.end(), v.begin(), [epsilon](double d) -> double { if (d < epsilon) { return 0; } else { return d; } }); } You can capture by both reference and value, which you can specify using & and = respectively:
[&epsilon, zeta] captures epsilon by reference and zeta by value[&] captures all variables used in the lambda by reference[=] captures all variables used in the lambda by value[&, epsilon] captures all variables used in the lambda by reference but captures epsilon by value[=, &epsilon] captures all variables used in the lambda by value but captures epsilon by referenceThe generated operator() is const by default, with the implication that captures will be const when you access them by default. This has the effect that each call with the same input would produce the same result, however you can mark the lambda as mutable to request that the operator() that is produced is not const.
const always...
std::function<double(int, bool)> f = [](int a, bool b) -> double { ... }; But usually, we let the compiler deduce the type: auto f = [](int a, bool b) -> double { ... }; (and don't forget to #include <functional>)
return d < 0.00001 ? 0 : d; is guaranteed to return double, when one of the operands is an integer constant (it is because of an implicit promotion rule of the ?: operator where the 2nd and 3rd operand are balanced against each other through the usual arithmetic conversions no matter which one that gets picked). Changing to 0.0 : d would perhaps make the example easier to understand.std::array array instead of raw arrays and then it becomes trivial. (Which is good advice in most cases in C++ anyway)The C++ concept of a lambda expression originates in the lambda calculus and functional programming. A lambda is an unnamed function that is useful (in actual programming, not theory) for short snippets of code that are impossible to reuse and are not worth naming.
In C++, the minimal lambda expression looks like:
[]{} // lambda with no parameters that does nothing [] is the capture list and {} the function body.
The full syntax for a lambda-expression, including attributes, noexcept/throw-specifications, requires-clauses, etc. is more complex.
The capture list defines what from the outside of the lambda should be available inside the function body and how. It can be either:
[x][&x][&][=]this and making member functions callable within the lambda [this]You can mix any of the above in a comma separated list [x, &y].
An element of the capture list can now be initialized with =, which is called init-capture. This allows renaming of variables and to capture by moving. An example taken from the standard:
int x = 4; auto y = [&r = x, x = x+1]()->int { r += 2; return x+2; }(); // Updates ::x to 6, and initializes y to 7. and one taken from Wikipedia showing how to capture with std::move:
auto ptr = std::make_unique<int>(10); // See below for std::make_unique auto lambda = [ptr = std::move(ptr)] {return *ptr;}; Since C++20, lambda expressions can have a template-parameter-list:
[]<int N>() {}; Such a generic lambda is like a non-template struct with a call operator template:
struct __lambda { template <int N> void operator()() const {} }; The parameter-declaration-clause is the same as in any other C++ function. It can be omitted completely when there are no parameters, meaning that [](){} is equivalent to []{}.
Lambdas with an auto parameter are generic lambdas. auto would be equivalent to T here if T were a type template argument somewhere in the surrounding scope):
[](auto x, auto y) { return x + y; } This works just like a C++20 abbreviated function template:
struct __lambda { // C++20 equivalent void operator()(auto x, auto y) const { return x + y; } // pre-C++20 equivalent template <typename T, typename U> void operator()(T x, U y) const { return x + y; } }; If a lambda has only one return statement, the return type can be omitted and has the implicit type of decltype(return_statement).
The return type can also be provided explicitly using trailing return type syntax:
[](int x) -> int { return x; } C++14 allows deduced return types for every function and does not restrict it to functions of the form return expression;. This is also extended to lambdas. By default, the return type of a lambda is deduced as if its return type was declared auto.
If a lambda is marked mutable (e.g. []() mutable { }) it is allowed to mutate the values that have been captured by value.
mutable means that the call operator of the lambda's type does not have a const qualifier.
A block-statement will be executed when the lambda is actually called. This becomes the body of the call operator.
The library defined by the ISO standard benefits heavily from lambdas and raises the usability several bars as now users don't have to clutter their code with small functors in some accessible scope.
r = &x; r += 2;, but this happens to the original value of 4.
any variable currently in scope, what does it mean? it means capture all global variables globally and any local variables in this function?
x inside and outside the lambda is mean. Took me a second to get why that was the result :)Lambda expressions are typically used to encapsulate algorithms so that they can be passed to another function. However, it is possible to execute a lambda immediately upon definition:
[&](){ ...your code... }(); // immediately executed lambda expression is functionally equivalent to
{ ...your code... } // simple code block This makes lambda expressions a powerful tool for refactoring complex functions. You start by wrapping a code section in a lambda function as shown above. The process of explicit parameterization can then be performed gradually with intermediate testing after each step. Once you have the code-block fully parameterized (as demonstrated by the removal of the &), you can move the code to an external location and make it a normal function.
Similarly, you can use lambda expressions to initialize variables based on the result of an algorithm...
int a = []( int b ){ int r=1; while (b>0) r*=b--; return r; }(5); // 5! As a way of partitioning your program logic, you might even find it useful to pass a lambda expression as an argument to another lambda expression...
[&]( std::function<void()> algorithm ) // wrapper section { ...your wrapper code... algorithm(); ...your wrapper code... } ([&]() // algorithm section { ...your algorithm code... }); Lambda expressions also let you create named nested functions, which can be a convenient way of avoiding duplicate logic. Using named lambdas also tends to be a little easier on the eyes (compared to anonymous inline lambdas) when passing a non-trivial function as a parameter to another function. Note: don't forget the semicolon after the closing curly brace.
auto algorithm = [&]( double x, double m, double b ) -> double { return m*x+b; }; int a=algorithm(1,2,3), b=algorithm(4,5,6); If subsequent profiling reveals significant initialization overhead for the function object, you might choose to rewrite this as a normal function.
if statements: if ([i]{ for (char j : i) if (!isspace(j)) return false ; return true ; }()) // i is all whitespace, assuming i is an std::string
[](){}();.
main() {{{{((([](){{}}())));}}}}Answers
Q: What is a lambda expression in C++11?
A: Under the hood, it is the object of an autogenerated class with an overloaded operator() const. Such an object is called a closure and is created by the compiler. This 'closure' concept is similar to the 'bind' concept from C++11, but lambdas typically generate more performant code. Also, calls through closures (instead of typical functions) allow full inlining.
Q: When would I use one?
A: When you want to define "simple and small logic" and ask the compiler to perform generation from previous question. You give a compiler some expressions which you want to be inside operator(). All the other stuff, the compiler will generate for you.
Q: What class of problem do lambdas solve that wasn't possible to solve prior to their introduction?
A: Lambdas are more of a syntactical sugar (i.e., code meant for ease-of-use) like "operator overloading" instead of having to make entire functions for custom add, subtract operations... Lambdas save more lines of unneeded code to wrap 1-3 lines of real logic to some classes, and etc.! Some engineers think that if the number of lines is smaller, then there is a lesser chance to make errors in it (I'm think so too).
Example of usage
auto x = [=](int arg1){printf("%i", arg1); }; // Note the ending semicolon after {}. void(*f)(int) = x; // ^Create function pointer f that takes parameter `int`. // ^See point 4.1 below under "Extras about Lambdas". f(1); // Call function f with parameter `int 1` x(1); // Call function x with parameter `int 1` Extras about lambdas, not covered by question. Ignore this section if you're not interested
1. Captured values. What you can capture
1.1. You can reference to a variable with static storage duration in lambdas. They all are captured.
1.2. You can use lambda to capture values "by value". In such case, captured vars will be copied to the function object (closure).
[captureVar1,captureVar2](int arg1){} //Modification of either captured value inside the Lambda //will *not* modify the value outside the Lambda too, //meaning a variable outside the Lambda that is also named //captureVar1 will be unaffected by what happens inside the Lambda. //I.e., Variable Shadowing will occur. 1.3. You can capture by reference. & -- in this context mean reference, not pointers.
[&captureVar1,&captureVar2](int arg1){} //Modification of either capture value inside the Lambda //will modify the value outside the Lambda too! 1.4. Notation exists to capture all non-static vars by value, or by reference
[=](int arg1){} // capture all not-static vars by value [&](int arg1){} // capture all not-static vars by reference 1.5. Notation exists to capture all non-static vars by value, or by reference, while overriding specific variables to be specifically by-value or by-reference Examples: Capture all not-static vars by value, but by reference capture Param2
[=,&Param2](int arg1){} Capture all not-static vars by reference, but by value capture Param2
[&,Param2](int arg1){} 2. Return type deduction
2.1. Lambda return type can be deduced if lambda is one expression. Or you can explicitly specify it.
[=](int arg1)->trailing_return_type{return trailing_return_type();} If lambda has more than one expression, then return type must be specified via trailing return type. Also, similar syntax can be applied to auto functions and member-functions
3. Captured values. What you cannot capture
3.1. You can capture only local vars, not member variables of the object.
4. Сonversions
4.1 !! Lambda is not a function pointer and it is not an anonymous function, but capture-less lambdas can be implicitly converted to a function pointer.
P.s.
More about lambda grammar information can be found in Working draft for Programming Language C++ #337, 2012-01-16, 5.1.2. Lambda Expressions, p.88
In C++14 an extra feature, named as "init capture", has been added. It allows you to arbitrarily perform declaration of closure data members:
auto toFloat = [](int value) { return float(value);}; auto interpolate = [min = toFloat(0), max = toFloat(255)] (int value)->float { return (value - min) / (max - min);}; 
[&,=Param2](int arg1){} doesn't seem to be valid syntax. The correct form would be [&,Param2](int arg1){}A lambda function is an anonymous function that you create in-line. It can capture variables as some have explained, (e.g. http://www.stroustrup.com/C++11FAQ.html#lambda) but there are some limitations. For example, if there's a callback interface like this,
void apply(void (*f)(int)) { f(10); f(20); f(30); } you can write a function on the spot to use it like the one passed to apply below:
int col=0; void output() { apply([](int data) { cout << data << ((++col % 10) ? ' ' : '\n'); }); } But you can't do this:
void output(int n) { int col=0; apply([&col,n](int data) { cout << data << ((++col % 10) ? ' ' : '\n'); }); } because of limitations in the C++11 standard. If you want to use captures, you have to rely on the library and
#include <functional> (or some other STL library like algorithm to get it indirectly) and then work with std::function instead of passing normal functions as parameters like this:
#include <functional> void apply(std::function<void(int)> f) { f(10); f(20); f(30); } void output(int width) { int col; apply([width,&col](int data) { cout << data << ((++col % width) ? ' ' : '\n'); }); } 
apply was a template that accepted a functor, it would work
One of the best explanation of lambda expression is given from author of C++ Bjarne Stroustrup in his book ***The C++ Programming Language*** chapter 11 (ISBN-13: 978-0321563842):
What is a lambda expression?
A lambda expression, sometimes also referred to as a lambda function or (strictly speaking incorrectly, but colloquially) as a lambda, is a simplified notation for defining and using an anonymous function object. Instead of defining a named class with an operator(), later making an object of that class, and finally invoking it, we can use a shorthand.
When would I use one?
This is particularly useful when we want to pass an operation as an argument to an algorithm. In the context of graphical user interfaces (and elsewhere), such operations are often referred to as callbacks.
What class of problem do they solve that wasn't possible prior to their introduction?
Here i guess every action done with lambda expression can be solved without them, but with much more code and much bigger complexity. Lambda expression this is the way of optimization for your code and a way of making it more attractive. As sad by Stroustup :
effective ways of optimizing
Some examples
via lambda expression
void print_modulo(const vector<int>& v, ostream& os, int m) // output v[i] to os if v[i]%m==0 { for_each(begin(v),end(v), [&os,m](int x) { if (x%m==0) os << x << '\n'; }); } or via function
class Modulo_print { ostream& os; // members to hold the capture list int m; public: Modulo_print(ostream& s, int mm) :os(s), m(mm) {} void operator()(int x) const { if (x%m==0) os << x << '\n'; } }; or even
void print_modulo(const vector<int>& v, ostream& os, int m) // output v[i] to os if v[i]%m==0 { class Modulo_print { ostream& os; // members to hold the capture list int m; public: Modulo_print (ostream& s, int mm) :os(s), m(mm) {} void operator()(int x) const { if (x%m==0) os << x << '\n'; } }; for_each(begin(v),end(v),Modulo_print{os,m}); } if u need u can name lambda expression like below:
void print_modulo(const vector<int>& v, ostream& os, int m) // output v[i] to os if v[i]%m==0 { auto Modulo_print = [&os,m] (int x) { if (x%m==0) os << x << '\n'; }; for_each(begin(v),end(v),Modulo_print); } Or assume another simple sample
void TestFunctions::simpleLambda() { bool sensitive = true; std::vector<int> v = std::vector<int>({1,33,3,4,5,6,7}); sort(v.begin(),v.end(), [sensitive](int x, int y) { printf("\n%i\n", x < y); return sensitive ? x < y : abs(x) < abs(y); }); printf("sorted"); for_each(v.begin(), v.end(), [](int x) { printf("x - %i;", x); } ); } will generate next
0
1
0
1
0
1
0
1
0
1
0 sortedx - 1;x - 3;x - 4;x - 5;x - 6;x - 7;x - 33;
[] - this is capture list or lambda introducer: if lambdas require no access to their local environment we can use it.
Quote from book:
The first character of a lambda expression is always [. A lambda introducer can take various forms:
• []: an empty capture list. This implies that no local names from the surrounding context can be used in the lambda body. For such lambda expressions, data is obtained from arguments or from nonlocal variables.
• [&]: implicitly capture by reference. All local names can be used. All local variables are accessed by reference.
• [=]: implicitly capture by value. All local names can be used. All names refer to copies of the local variables taken at the point of call of the lambda expression.
• [capture-list]: explicit capture; the capture-list is the list of names of local variables to be captured (i.e., stored in the object) by reference or by value. Variables with names preceded by & are captured by reference. Other variables are captured by value. A capture list can also contain this and names followed by ... as elements.
• [&, capture-list]: implicitly capture by reference all local variables with names not men- tioned in the list. The capture list can contain this. Listed names cannot be preceded by &. Variables named in the capture list are captured by value.
• [=, capture-list]: implicitly capture by value all local variables with names not mentioned in the list. The capture list cannot contain this. The listed names must be preceded by &. Vari- ables named in the capture list are captured by reference.
Note that a local name preceded by & is always captured by reference and a local name not pre- ceded by & is always captured by value. Only capture by reference allows modification of variables in the calling environment.
Additional
Lambda expression format
Additional references:
for (int x : v) { if (x % m == 0) os << x << '\n';}
The lambda's in c++ are treated as "on the go available function". yes its literally on the go, you define it; use it; and as the parent function scope finishes the lambda function is gone.
c++ introduced it in c++ 11 and everyone started using it like at every possible place. the example and what is lambda can be find here https://en.cppreference.com/w/cpp/language/lambda
i will describe which is not there but essential to know for every c++ programmer
Lambda is not meant to use everywhere and every function cannot be replaced with lambda. It's also not the fastest one compare to normal function. because it has some overhead which need to be handled by lambda.
it will surely help in reducing number of lines in some cases. it can be basically used for the section of code, which is getting called in same function one or more time and that piece of code is not needed anywhere else so that you can create standalone function for it.
Below is the basic example of lambda and what happens in background.
User code:
int main() { // Lambda & auto int member=10; auto endGame = [=](int a, int b){ return a+b+member;}; endGame(4,5); return 0; } How compile expands it:
int main() { int member = 10; class __lambda_6_18 { int member; public: inline /*constexpr */ int operator()(int a, int b) const { return a + b + member; } public: __lambda_6_18(int _member) : member{_member} {} }; __lambda_6_18 endGame = __lambda_6_18{member}; endGame.operator()(4, 5); return 0; } so as you can see, what kind of overhead it adds when you use it. so its not good idea to use them everywhere. it can be used at places where they are applicable.
Well, one practical use I've found out is reducing boiler plate code. For example:
void process_z_vec(vector<int>& vec) { auto print_2d = [](const vector<int>& board, int bsize) { for(int i = 0; i<bsize; i++) { for(int j=0; j<bsize; j++) { cout << board[bsize*i+j] << " "; } cout << "\n"; } }; // Do sth with the vec. print_2d(vec,x_size); // Do sth else with the vec. print_2d(vec,y_size); //... } Without lambda, you may need to do something for different bsize cases. Of course you could create a function but what if you want to limit the usage within the scope of the soul user function? the nature of lambda fulfills this requirement and I use it for that case.
C++ 11 introduced lambda expression to allow us write an inline function which can be used for short snippets of code
[ capture clause ] (parameters) -> return-type { definition of method } Generally return-type in lambda expression are evaluated by compiler itself and we don’t need to specify that explicitly and -> return-type part can be ignored but in some complex case as in conditional statement, compiler can’t make out the return type and we need to specify that.
// C++ program to demonstrate lambda expression in C++ #include <bits/stdc++.h> using namespace std; // Function to print vector void printVector(vector<int> v) { // lambda expression to print vector for_each(v.begin(), v.end(), [](int i) { std::cout << i << " "; }); cout << endl; } int main() { vector<int> v {4, 1, 3, 5, 2, 3, 1, 7}; printVector(v); // below snippet find first number greater than 4 // find_if searches for an element for which // function(third argument) returns true vector<int>:: iterator p = find_if(v.begin(), v.end(), [](int i) { return i > 4; }); cout << "First number greater than 4 is : " << *p << endl; // function to sort vector, lambda expression is for sorting in // non-decreasing order Compiler can make out return type as // bool, but shown here just for explanation sort(v.begin(), v.end(), [](const int& a, const int& b) -> bool { return a > b; }); printVector(v); // function to count numbers greater than or equal to 5 int count_5 = count_if(v.begin(), v.end(), [](int a) { return (a >= 5); }); cout << "The number of elements greater than or equal to 5 is : " << count_5 << endl; // function for removing duplicate element (after sorting all // duplicate comes together) p = unique(v.begin(), v.end(), [](int a, int b) { return a == b; }); // resizing vector to make size equal to total different number v.resize(distance(v.begin(), p)); printVector(v); // accumulate function accumulate the container on the basis of // function provided as third argument int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; int f = accumulate(arr, arr + 10, 1, [](int i, int j) { return i * j; }); cout << "Factorial of 10 is : " << f << endl; // We can also access function by storing this into variable auto square = [](int i) { return i * i; }; cout << "Square of 5 is : " << square(5) << endl; } Output
4 1 3 5 2 3 1 7 First number greater than 4 is : 5 7 5 4 3 3 2 1 1 The number of elements greater than or equal to 5 is : 2 7 5 4 3 2 1 Factorial of 10 is : 3628800 Square of 5 is : 25 A lambda expression can have more power than an ordinary function by having access to variables from the enclosing scope. We can capture external variables from enclosing scope by three ways :
The syntax used for capturing variables :
#include <bits/stdc++.h> using namespace std; int main() { vector<int> v1 = {3, 1, 7, 9}; vector<int> v2 = {10, 2, 7, 16, 9}; // access v1 and v2 by reference auto pushinto = [&] (int m) { v1.push_back(m); v2.push_back(m); }; // it pushes 20 in both v1 and v2 pushinto(20); // access v1 by copy [v1]() { for (auto p = v1.begin(); p != v1.end(); p++) { cout << *p << " "; } }; int N = 5; // below snippet find first number greater than N // [N] denotes, can access only N by value vector<int>:: iterator p = find_if(v1.begin(), v1.end(), [N](int i) { return i > N; }); cout << "First number greater than 5 is : " << *p << endl; // function to count numbers greater than or equal to N // [=] denotes, can access all variable int count_N = count_if(v1.begin(), v1.end(), [=](int a) { return (a >= N); }); cout << "The number of elements greater than or equal to 5 is : " << count_N << endl; } Output:
First number greater than 5 is : 7 The number of elements greater than or equal to 5 is : 3 The answers given before (in time) are still missing some clarity for the beginner. Especially because of the different forms that a lambda can be defined/invoked. Well, perhaps an answer from another beginner (no 'curse of knowledge') may help clarify these different forms.
#include <iostream> int main() { // lambda| arguments | | body ||invoc.| printf("direct lamda call: %d\n", [] (int x, int y) {return x + y;} (4, 5) ); // it is important to be aware that a lambda with invocation arguments // or () if no arguments are defined, can be inserted anywhere where a // value (whatever type) is expected // assignment (instead of "auto", one could also write "int"): auto retval = [] (int x, int y) {return x + y;}(5, 5); printf("the same, via variable: %d\n", retval); // now the same, without supplying actual arguments (definition only): auto funct = [] (int x, int y) {return x + y;}; printf("function address: %x\n", funct); printf("via named lambda invocation: %d\n", funct(6, 5) ); // in this case (definition only), the lambda can be handed over to another // function as well, e.g. } Output:
$ g++ lambda.cpp && ./a.out direct lamda call: 9 the same, via variable: 10 function address: 4878deb0 via named lambda invocation: 11 (inspiration got via https://dotnettutorials.net/lesson/lambda-expressions-in-cpp/)
To address the further questions of the question poster, a demonstration is shown where data is used together with lambda as function arguments in a data processor function à la map():
#include <iostream> #include <vector> using namespace std; template<typename Function> void mymap(vector<int> a, Function fn) { for (auto &i : a) { printf("%d>%d ", i, fn(a[i-1])); } printf("\n"); } int main() { vector<int> data{1,2,3,4,5}; // |what| how | mymap(data, [=](int x){return x*x;}); } Output:
$ g++ lambda.cpp && ./a.out 1>1 2>4 3>9 4>16 5>25 More information about this typical use case to reduce code repetitions can be found here from one of the founders of stack overflow (2006): https://www.joelonsoftware.com/2006/08/01/can-your-programming-language-do-this/.
BTW: you could think of a lambda expression as if we had normal (mathematical) expressions defined in a function, where these can be used (computed) from within another function. The lambda effectively allows this and even more.
Lambdas and other techniques also help reduce git diff 'size' upon doing code changes. To quote Martin Cracauer in his introduction to Compile Time Computing:
Last but certainly not least, there is also Diff Efficiency. What the heck is “diff efficiency”? It means that if you look at a git diff of your code versus a different state of your code, then you only see the actual functional change. Not a lot of boilerplate. Not a lot of copy and pasted slight variants of that functional change.
(I know that the title of the article is not about lambdas, but around the center of the article, some general useful notes can be found)
One problem it solves: Code simpler than lambda for a call in constructor that uses an output parameter function for initializing a const member
You can initialize a const member of your class, with a call to a function that sets its value by giving back its output as an output parameter.
