Using std::for_each on call a function with multiple arguments

Question

I did search, but the closest link , among those found, even doesn't match my problem. I have a std::vector<double> mydata array. I like to use for_each for that mydata array in calling a member function. That member function accepts two arguments. One is each element of mydata array and another one is a int* of another array. I do like

::std::for_each (mydata.begin(), mydata.end(), train(net));

That gives me a compilation error of train function does not take one argument. I know how to use for_each if there isn't int*.

My train function is

void train(double const & data, int* d){}

How can I make it work? Thanks

Can you not supply a default for int* d?

Simon Bosley
– Simon Bosley

2014-02-20 09:49:51 +00:00
Commented Feb 20, 2014 at 9:49 — Simon Bosley
– Simon Bosley, Commented Feb 20, 2014 at 9:49

Some programmer dude · Accepted Answer · 2014-02-20 09:49:20Z

7

If you have C++11 you can use std::bind:

using namespace std::placeholders;  // For `_1`

std::for_each (mydata.begin(), mydata.end(),
               std::bind(&MyClass::train, this, _1, net));

Here the member-function will be called with three arguments: The first is the this pointer, which is always a hidden first argument in all member functions. The second will be the first argument passed to the callable object created by std::bind, and the third argument is the net variable.

answered Feb 20, 2014 at 9:49

Some programmer dude

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3 Comments

James Kanze Over a year ago

Why use std::bind, when lambda is a lot simpler. (In pre-C++11, this would be a good use of boost::bind.)

Some programmer dude Over a year ago

@JamesKanze Didn't think of it until I'd already written my answer. And that's why I upvoted the answer from Alexander. :)

Martin J. Over a year ago

The semantics of bind are clearer than that of a lambda function, in my opinion.

Alexander · Accepted Answer · 2014-02-20 09:50:56Z

6

try to use a lambda-function:

std::for_each (mydata.begin(), mydata.end(), [&](double d)
{
  train(d, net);
});

answered Feb 20, 2014 at 9:50

Alexander

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1 Comment

MSalters Over a year ago

Why take net by reference? I'd do [net](double d){ train(d, net); } - just copy the pointer.

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