what does this line mean in c++? [duplicate]
Solution 1:
The problem
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.
The new solution
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.
Return types
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;
}
});
}
"Capturing" variables
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 reference
The 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.
Solution 2:
What is a lambda function?
The C++ concept of a lambda function 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++ a lambda function is defined like this
[]() { } // barebone lambda
or in all its glory
[]() mutable -> T { } // T is the return type, still lacking throw()
[] is the capture list, () the argument list and {} the function body.
The capture list
The capture list defines what from the outside of the lambda should be available inside the function body and how. It can be either:
- a value: [x]
- a reference [&x]
- any variable currently in scope by reference [&]
- same as 3, but by value [=]
You can mix any of the above in a comma separated list [x, &y].
The argument list
The argument list is the same as in any other C++ function.
The function body
The code that will be executed when the lambda is actually called.
Return type deduction
If a lambda has only one return statement, the return type can be omitted and has the implicit type of decltype(return_statement).
Mutable
If a lambda is marked mutable (e.g. []() mutable { }) it is allowed to mutate the values that have been captured by value.
Use cases
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.
C++14
In C++14 lambdas have been extended by various proposals.
Initialized Lambda Captures
An element of the capture list can now be initialized with =. 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;};
Generic Lambdas
Lambdas can now be generic (auto would be equivalent to T here if
T were a type template argument somewhere in the surrounding scope):
auto lambda = [](auto x, auto y) {return x + y;};
Improved Return Type Deduction
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.
Solution 3:
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.
Solution 4:
Answers
Q: What is a lambda expression in C++11?
A: Under the hood, it is the object of an autogenerated class with overloading operator() const. Such object is called closure and created by compiler. This 'closure' concept is near with the bind concept from C++11. But lambdas typically generate better code. And calls through closures allow full inlining.
Q: When would I use one?
A: To define "simple and small logic" and ask compiler perform generation from previous question. You give a compiler some expressions which you want to be inside operator(). All other stuff compiler will generate to you.
Q: What class of problem do they solve that wasn't possible prior to their introduction?
A: It is some kind of syntax sugar like operators overloading instead of functions for custom add, subrtact operations...But it 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 less chance to make errors in it (I'm also think so)
Example of usage
auto x = [=](int arg1){printf("%i", arg1); };
void(*f)(int) = x;
f(1);
x(1);
Extras about lambdas, not covered by question. Ignore this section if you're not interest
1. Captured values. What you can to 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 for capture values "by value". In such case captured vars will be copied to the function object (closure).
[captureVar1,captureVar2](int arg1){}
1.3. You can capture be reference. & -- in this context mean reference, not pointers.
[&captureVar1,&captureVar2](int arg1){}
1.4. It exists notation 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. It exists notation to capture all non-static vars by value, or by reference and specify smth. more. 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 then 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 can not capture
3.1. You can capture only local vars, not member variable 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 the extra feature which has named as "init capture" have been added. It allow to perform arbitarily 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);};
Solution 5:
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');
});
}