Re: [boost] [iterator] function_output_iterator constructed from a lambda function is not assignable

On 2016-02-14 22:42, Mikhail Matrosov wrote:

Consider the following code snippet:

auto f = [](int x) { std::cout << x; };auto it = boost::make_function_output_iterator(f);decltype(it) it2 = it; // Ok, copied it2 = it; // Does not compile, cannot assign!

Since the lambda does not capture anything, you might try to convert it to a function pointer, which would make it assignable. typedef void (*lambda_t)(int); auto it = make_function_output_iterator((lambda_t)f); If you do have to capture variables, you might want to use std::bind to store the captures and a pointer to the lambda function without captures. Or just use std::bind with a normal function.