In C++11, is there a way to template a lambda function? Or is it inherently too specific to be templated?
I understand that I can define a classic templated class/functor instead, but the question is more like: does the language allow templating lambda functions?
UPDATE 2018: C++20 will come with templated and conceptualized lambdas. The feature has already been integrated into the standard draft.
UPDATE 2014: C++14 has been released this year and now provides Polymorphic lambdas with the same syntax as in this example. Some major compilers already implement it.
At it stands (in C++11), sadly no. Polymorphic lambdas would be excellent in terms of flexibility and power.
The original reason they ended up being monomorphic was because of concepts. Concepts made this code situation difficult:
template <Constraint T>
void foo(T x)
{
auto bar = [](auto x){}; // imaginary syntax
}
In a constrained template you can only call other constrained templates. (Otherwise the constraints couldn't be checked.) Can foo invoke bar(x)? What constraints does the lambda have (the parameter for it is just a template, after all)?
Concepts weren't ready to tackle this sort of thing; it'd require more stuff like late_check (where the concept wasn't checked until invoked) and stuff. Simpler was just to drop it all and stick to monomorphic lambdas.
However, with the removal of concepts from C++0x, polymorphic lambdas become a simple proposition again. However, I can't find any proposals for it. :(
In C++20 this is possible using the following syntax:
auto lambda = []<typename T>(T t){
// do something
};
[]<>(){}?
auto lambda = []<>(){}; and it said "lambda template parameter list cannot be empty". So, at least auto lambda = []<typename>(){};.
T t parameter didn’t exist in the example above), then you can do so using this syntax: lambda.template operator()<int>(). See stackoverflow.com/questions/49392738 for more info.
lambda<int>(). Is there a technical limitation to this?
lambda is a function object. To use the template angle brackets you need a type (or a constexpr template). I assume that per the specification the compiler will have to treat the < in your example as a less than operator on the lambda object which will obviously fail. A future C++ version might allow for a templated operator() to be called in this way.
lambda<a>(b) could either be lambda.operator()<a>(b) or lambda.operator<(a).operator>(b)As others have said, you can have template arguments in C++20 lambdas.
auto doSomething = []<typename T>(T x) {
// blabla
};
However, other answers forget to tell you how to call those lambdas.
In the previous example it's easy because all template types can be deduced from the provided parameters.
doSomething(2);
But the problem comes when template types can NOT be deduced from the parameters.
auto readValue = [&buffer]<typename T>() -> T{
return *(const T*)buffer;
};
You would expect to call it just like a regular templated function:
readValue<int>(); // WRONG!
But this does NOT compile. You have to write it like this:
readValue.template operator()<int>(); // VALID!
readValue<int>(); This is what any sane person would try first and expect to work, right ? And i mean oversight as in by the C++ designers or committee.operator(). As such, this problem is carried over there: godbolt.org/z/Evah16z5TreadValue.template operator()<5>();C++11 lambdas can't be templated as stated in other answers but decltype() seems to help when using a lambda within a templated class or function.
#include <iostream>
#include <string>
using namespace std;
template<typename T>
void boring_template_fn(T t){
auto identity = [](decltype(t) t){ return t;};
std::cout << identity(t) << std::endl;
}
int main(int argc, char *argv[]) {
std::string s("My string");
boring_template_fn(s);
boring_template_fn(1024);
boring_template_fn(true);
}
Prints:
My string
1024
1
I've found this technique is helps when working with templated code but realize it still means lambdas themselves can't be templated.
T would work fine in place of decltype(t) in this example.In C++11, lambda functions can not be templated, but in the next version of the ISO C++ Standard (often called C++14), this feature will be introduced. [Source]
Usage example:
auto get_container_size = [] (auto container) { return container.size(); };
Note that though the syntax uses the keyword auto, the type deduction will not use the rules of auto type deduction, but instead use the rules of template argument deduction. Also see the proposal for generic lambda expressions(and the update to this).
auto type deduction are specifically defined to be the same as those of template function argument deduction.I am aware that this question is about C++11. However, for those who googled and landed on this page, templated lambdas are now supported in C++14 and go by the name Generic Lambdas.
[info] Most of the popular compilers support this feature now. Microsoft Visual Studio 2015 supports. Clang supports. GCC supports.
I wonder what about this:
template <class something>
inline std::function<void()> templateLamda() {
return [](){ std::cout << something.memberfunc() };
}
I used similar code like this, to generate a template and wonder if the compiler will optimize the "wrapping" function out.
There is a gcc extension which allows lambda templates:
// create the widgets and set the label
base::for_each(_widgets, [] <typename Key_T, typename Widget_T>
(boost::fusion::pair<Key_T, Widget_T*>& pair) -> void {
pair.second = new Widget_T();
pair.second->set_label_str(Key_T::label);
}
);
where _widgets is a std::tuple< fusion::pair<Key_T, Widget_T>... >
I've been playing with the latest clang version 5.0.1 compiling with the -std=c++17 flag and there is now some nice support for auto type parameters for lambdas:
#include <iostream>
#include <vector>
#include <stdexcept>
int main() {
auto slice = [](auto input, int beg, int end) {
using T = decltype(input);
const auto size = input.size();
if (beg > size || end > size || beg < 0 || end < 0) {
throw std::out_of_range("beg/end must be between [0, input.size())");
}
if (beg > end) {
throw std::invalid_argument("beg must be less than end");
}
return T(input.begin() + beg, input.begin() + end);
};
auto v = std::vector<int> { 1,2,3,4,5 };
for (auto e : slice(v, 1, 4)) {
std::cout << e << " ";
}
std::cout << std::endl;
}
Have a look at Boost.Phoenix for polymorphic lambdas: http://www.boost.org/doc/libs/1_44_0/libs/spirit/phoenix/doc/html/index.html Does not require C++0x, by the way :)
I'm not sure why nobody else has suggested this, but you can write a templated function that returns lambda functions. The following solved my problem, the reason I came to this page:
template <typename DATUM>
std::function<double(DATUM)> makeUnweighted() {
return [](DATUM datum){return 1.0;};
}
Now whenever I want a function that takes a given type of argument (e.g. std::string), I just say
auto f = makeUnweighted<std::string>()
and now f("any string") returns 1.0.
That's an example of what I mean by "templated lambda function." (This particular case is used to automatically provide an inert weighting function when somebody doesn't want to weight their data, whatever their data might be.)
Another workaround for C++11 is by defining a template function and wrapping it within a lambda expression. However; this needs to define a new function for different templated lambdas:
struct ST{ int x; };
template<class T>
T templateFunc(T variable)
{
return variable;
}
void func()
{
ST st{10};
auto lambda = [&](){return templateFunc<ST>(st);};
auto res = lambda();
}
Here is one solution that involves wrapping the lamba in a structure:
template <typename T>
struct LamT
{
static void Go()
{
auto lam = []()
{
T var;
std::cout << "lam, type = " << typeid(var).name() << std::endl;
};
lam();
}
};
To use do:
LamT<int>::Go();
LamT<char>::Go();
#This prints
lam, type = i
lam, type = c
The main issue with this (besides the extra typing) you cannot embed this structure definition inside another method or you get (gcc 4.9)
error: a template declaration cannot appear at block scope
I also tried doing this:
template <typename T> using LamdaT = decltype(
[](void)
{
std::cout << "LambT type = " << typeid(T).name() << std::endl;
});
With the hope that I could use it like this:
LamdaT<int>();
LamdaT<char>();
But I get the compiler error:
error: lambda-expression in unevaluated context
So this doesn't work ... but even if it did compile it would be of limited use because we would still have to put the "using LamdaT" at file scope (because it is a template) which sort of defeats the purpose of lambdas.
It's a bit verbose, but you can "unroll" a lambda into a callable object and use templates on the class definition. You can even capture variables, store the object into a variable as you would a lambda, or instantiate a new object for a single call. The possibilities are endless.
template <typename T>
struct Callable
{
Callable(int& i) : capturedInt(i) {}
T operator()(T a, T b)
{
T total = a;
for (int i = 0; i < capturedInt; i++)
total *= b;
return total;
}
private:
int& capturedInt;
};
int main() {
int i = 5;
Callable<float> callable(i);
cout << "Result of callable(5)(10, 1.5f): " << callable(10, 1.5f) << '\n';
i = 6;
cout << "Result of callable(6)(1.5f, 10): " << callable(1.5f, 10) << '\n';
i = 10;
cout << "Result of Callable<int>(10)(2, 3): " << Callable<int>(i)(2, 3) << '\n';
return 0;
}
Output:
Result of callable(5)(10, 1.5f): 75.9375
Result of callable(6)(1.5f, 10): 1.5e+06
Result of Callable<int>(10)(2, 3): 118098
