Using std::transform with std::normal_distribution and std::bind

Question

stl c++11 solution :

auto distribution = std::bind(std::normal_distribution<double>{mean, stddev},
                              std::mt19937(std::random_device{}())
                              );
std::transform(data.begin(), data.end(), data.begin(),
               std::bind(std::plus<double>(), std::placeholders::_1, distribution()));

Easy range-based loop :

for (auto& d : data) {
  d += distribution();
}

My STL solution doesn't work as it always take the first number it generated from the distribution. I tried with the placeholder as the 3rd parameter but it doesn't change anything. All of my data is incremented by the same number which is not something I want. I want the same behaviour as the range-based loop.

It this something possible?

`distribution()` will be called only once in your first code snippet, and the result will be bound to `plus`. However, you want to have `distribution` called for every element. You have to use other semantics. Note that a lambda will make this easier, e.g. `[&distribution](double x){ return x += distribution(); }`. — Zeta, Nov 22 '16 at 17:12
@Zeta, that's what I thought. using a lambda seems the only path? — Maxime Roussin-Bélanger, Nov 22 '16 at 17:12
@Lorac a fairly [simple one](http://coliru.stacked-crooked.com/a/c10b741e2ac091b4), though. — jaggedSpire, Nov 22 '16 at 17:18
`std::bind(std::plus(), std::placeholders::_1, std::bind(distribution))`? — cpplearner, Nov 22 '16 at 17:55
@cpplearner or even just `std::bind(std::plus(), std::placeholders::_1, distribution)`, since `distribution` is already a bind expression. — T.C., Nov 22 '16 at 19:17
@T.C. Did you really just tell someone to rely on the passing bind-expression to bind-expressions feature of bind instead of just using a lambda that says what it does? — Yakk - Adam Nevraumont, Nov 23 '16 at 01:30

score 2 · Accepted Answer · answered Nov 23 '16 at 08:35

Let's rewrite your second bind as lambda, to get a feeling how it actually works:

auto func = std::bind(std::plus<double>(), std::placeholders::_1, distribution())

is equivalent too

auto d = distribution();

auto func = [d](double x){ return std::plus<double>()(x,d); };

or, if we use C++14's initialization feature:

auto func = [d=distribution()](double x){ return std::plus<double>()(x,d); };

As you can see, distribution() gets called only once. However, you don't want to use the return value of distribution, you want to call distribution for every call of func. While it's possible to do that with bind, a lambda will make this a lot easier:

std::transform(data.begin(), data.end(), data.begin(),
               [&distribution](double x){ return x + distribution(); });

And in my opinion, that's a lot easier to read than your previous bind. Note that std::bind (or rather boost::bind) predates lambdas. There were some issues with lambdas in C++11 compared to std::bind, but with C++14, lambdas are usually a lot easier to handle, read and understand without sacrificing much.

Using std::transform with std::normal_distribution and std::bind

1 Answers1