Why can't I create a vector of lambdas (of the same type) in C++11?

I was trying to create a vector of lambda, but failed:

auto ignore = [&]() { return 10; };  //1
std::vector<decltype(ignore)> v;     //2
v.push_back([&]() { return 100; });  //3

Up to line #2, it compiles fine. But the line#3 gives compilation error:

error: no matching function for call to 'std::vector<main()::<lambda()>>::push_back(main()::<lambda()>)'

I don't want a vector of function pointers or vector of function objects. However, vector of function objects which encapsulate real lambda expressions, would work for me. Is this possible?

Every lambda has a different type—even if they have the same signature. You must use a run-time encapsulating container such as std::function if you want to do something like that.

e.g.:

std::vector<std::function<int()>> functors;
functors.push_back([&] { return 100; });
functors.push_back([&] { return  10; });

All lambda expressions have a different type, even if they are identical character-by-character. You're pushing a lambda of a different type (because it's another expression) into the vector, and that obviously won't work.

One solution is to make a vector of std::function<int()> instead.

auto ignore = [&]() { return 10; };
std::vector<std::function<int()>> v;
v.push_back(ignore);
v.push_back([&]() { return 100; });

On another note, it's not a good idea to use [&] when you're not capturing anything.

While what others have said is relevant, it is still possible to declare and use a vector of lambda, although it's not very useful:

auto lambda = [] { return 10; };
std::vector<decltype(lambda)> vec;
vec.push_back(lambda);

So, you can store any number of lambdas in there, so long as it's a copy/move of lambda!

If your lambda is stateless, i.e., [](...){...}, C++11 allows it to degrade into a function pointer. In theory, a C++11 compliant compiler would be able to compile this:

auto ignore = []() { return 10; };  //1 note misssing & in []!
std::vector<int (*)()> v;     //2
v.push_back([]() { return 100; });  //3