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);
}