在std :: normal_distribution和std :: bind中使用std :: transform

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?

Answer 1

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.