Question

This works …

auto x = 4;
typedef decltype(x) x_t;
x_t y = 5;

… so why doesn’t this?

int j = 4;  
auto func = [&] (int i) { cout << "Hello: i=" << i << "  j=" << j << endl;};
typedef decltype(func) lambda_t;
lambda_t func2 = [&] (int i) { cout << "Bye: i=" << i << "  j=" << j << endl;};

… and how would I declare lambda_t manually using std::function?

Answer 1

… so why doesn’t this [work]?

Because each lexical instance of a lambda has a different type. It does not matter if the same characters are used.

.. and how would I declare lambda_t manually using std::function?

The lambda takes an int argument and does not return anything… Therefore:

typedef std::function<void(int)> lambda_t;