• Joseph Rushton Wakeling (26/26) Apr 22 2012 Is there a way in which to pass a function as input to another function,...
• =?UTF-8?B?QWxpIMOHZWhyZWxp?= (4/17) Apr 22 2012 You just need to call the delegate with the function call syntax:
• Jakob Ovrum (26/49) Apr 22 2012 import std.random, std.range, std.stdio;

Joseph Rushton Wakeling <joseph.wakeling webdrake.net> writes:

Is there a way in which to pass a function as input to another function, with 
the arguments of the first function already determined?

The case I'm thinking of is one where I have a function which wants to take a 
random number generation scheme, and use it on several occasions, without
having 
any info on that scheme or its parameters.

Here's a little test attempt I made:

////////////////////////////////////////////////////
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) ... ?

=?UTF-8?B?QWxpIMOHZWhyZWxp?= <acehreli yahoo.com> writes:

On 04/22/2012 04:19 PM, Joseph Rushton Wakeling wrote:
 Is there a way in which to pass a function as input to another function,
 with the arguments of the first function already determined?

 The case I'm thinking of is one where I have a function which wants to
 take a random number generation scheme, and use it on several occasions,
 without having any info on that scheme or its parameters.

 Here's a little test attempt I made:

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

 void printRandomNumbers(RandomNumberGenerator)(RandomNumberGenerator
 rng, size_t n)
 {
 foreach(i; 0..n)
 writeln(rng);

You just need to call the delegate with the function call syntax:

           writeln(rng());

Ali

"Jakob Ovrum" <jakobovrum gmail.com> writes:

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);
}