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?
In C++20 this is possible using the following syntax:
auto lambda = []<typename T>(T t){
// do something
};
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).
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();
}