Re: Passing a function (with arguments) as function input

[Jakob Ovrum]

On Sunday, 22 April 2012 at 23:19:52 UTC, Joseph Rushton Wakeling 
wrote:
////////////////////////////////////////////////////
import std.random, std.range, std.stdio;
void 
printRandomNumbers(RandomNumberGenerator)(RandomNumberGenerator 
rng, size_t n)
{
      foreach(i; 0..n)
            writeln(rng);
}

void main()
{
      foreach(double upper; iota(1.0, 2.0, 0.2) ) {
            double delegate() rng = () {
                  return uniform(0.0, 1.0);
            };

            printRandomNumbers(rng,10);
      }
}
////////////////////////////////////////////////////

... which just prints out: "double delegate()" over many lines.

What am I doing wrong here?  And is there any way to avoid the 
messy business of defining a delegate and just hand over 
uniform(0.0, 1.0) ... ?

import std.random, std.range, std.stdio;

void printRandomNumbers(double delegate() rng, size_t n)
{
        foreach(i; 0..n)
                writeln(rng());
}

void main()
{
        foreach(upper; iota(1.0, 2.0, 0.2) )
                printRandomNumbers(() => uniform(0.0, 1.0), 10);
}


Although I wouldn't recommend it, you can also use a lazy 
parameter to obviate the lambda syntax:

import std.random, std.range, std.stdio;

void printRandomNumbers(lazy double rng, size_t n)
{
        foreach(i; 0..n)
                writeln(rng());
}

void main()
{
        foreach(upper; iota(1.0, 2.0, 0.2) )
                printRandomNumbers(uniform(0.0, 1.0), 10);
}