• Justin (14/14) Dec 04 2008 I recently discovered D's function literals and wrote a small test to ex...
• Ary Borenszweig (7/25) Dec 04 2008 That is:
• Ary Borenszweig (2/39) Dec 04 2008
• Justin (14/14) Dec 04 2008 Ahh, I thought it would some stupid oversight on my part. This works:
• Denis Koroskin (37/51) Dec 04 2008 Shorter way:
• Christopher Wright (4/24) Dec 04 2008 You don't need to make those static, by the way.

Justin <mrjnewt hotmail.com> writes:

I recently discovered D's function literals and wrote a small test to explore
them. The following code prints out a 15, then a 0. It seems to me that the
second should be 64 and not 0. Can anyone explain what I'm doing wrong?

module functionliteral;
import std.stdio;

static void main() {

	int[] values = [1,2,4,8];
	writefln(Reduce(values, function int(int x, int y) { return x + y; }));
	writefln(Reduce(values, function int(int x, int y) { return x * y; }));
}

static int Reduce(int[] values, int function(int x, int y) operation) {
	int total;
	foreach (int v; values)
		total = operation(total,v);
	return total;
}

Ary Borenszweig <ary esperanto.org.ar> writes:

Justin wrote:
 I recently discovered D's function literals and wrote a small test to explore
them. The following code prints out a 15, then a 0. It seems to me that the
second should be 64 and not 0. Can anyone explain what I'm doing wrong?
 
 module functionliteral;
 import std.stdio;
 
 static void main() {
 
 	int[] values = [1,2,4,8];
 	writefln(Reduce(values, function int(int x, int y) { return x + y; }));
 	writefln(Reduce(values, function int(int x, int y) { return x * y; }));
 }
 
 static int Reduce(int[] values, int function(int x, int y) operation) {
 	int total;

That is:

total = 0

That's why in the second case it's like you are doing:

0*1*2*3*4

A reduce function normally takes the first value to use to reduce the 
others. So "total" would be an argument to your reduce function.


 	foreach (int v; values)
 		total = operation(total,v);
 	return total;
 }

Ary Borenszweig <ary esperanto.org.ar> writes:

Ary Borenszweig wrote:
 Justin wrote:
 I recently discovered D's function literals and wrote a small test to 
 explore them. The following code prints out a 15, then a 0. It seems 
 to me that the second should be 64 and not 0. Can anyone explain what 
 I'm doing wrong?

 module functionliteral;
 import std.stdio;

 static void main() {

     int[] values = [1,2,4,8];
     writefln(Reduce(values, function int(int x, int y) { return x + y; 
 }));
     writefln(Reduce(values, function int(int x, int y) { return x * y; 
 }));
 }

 static int Reduce(int[] values, int function(int x, int y) operation) {
     int total;

 
 That is:
 
 total = 0
 
 That's why in the second case it's like you are doing:
 
 0*1*2*3*4

Well... 0*1*2*4*8

 
 A reduce function normally takes the first value to use to reduce the 
 others. So "total" would be an argument to your reduce function.
 
 
     foreach (int v; values)
         total = operation(total,v);
     return total;
 }

Justin <mrjnewt hotmail.com> writes:

Ahh, I thought it would some stupid oversight on my part. This works:

module functionliteral;

import std.stdio;

static void main() {

	int[] values = [1,2,4,8];
	writefln(Reduce(values, function int(int x, int y) { return x + y; }));
	writefln(Reduce(values, function int(int x, int y) { return x * y; }));
}

static int Reduce(int[] values, int function(int x, int y) operation) {
	int total = values[0];
	foreach (int v; values[1..$])
		total = operation(total,v);
	return total;
}

"Denis Koroskin" <2korden gmail.com> writes:

On Thu, 04 Dec 2008 23:08:51 +0300, Justin <mrjnewt hotmail.com> wrote:

 Ahh, I thought it would some stupid oversight on my part. This works:

 module functionliteral;

 import std.stdio;

 static void main() {

 	int[] values = [1,2,4,8];
 	writefln(Reduce(values, function int(int x, int y) { return x + y; }));
 	writefln(Reduce(values, function int(int x, int y) { return x * y; }));
 }

 static int Reduce(int[] values, int function(int x, int y) operation) {
 	int total = values[0];
 	foreach (int v; values[1..$])
 		total = operation(total,v);
 	return total;
 }

Shorter way:

import std.stdio;
import std.algorithm;

void main()
{
     auto values = [1,2,4,8];

     writefln(reduce!("a + b")(0, values));
     writefln(reduce!("a * b")(1, values));
}

or

import std.stdio;
import std.algorithm;

int add(int a, int b)
{
     return a + b;
}

int mul(int a, int b)
{
     return a * b;
}

void main()
{
     auto values = [1, 2, 4, 8];
     writefln(reduce!(add)(0, values));
     writefln(reduce!(mul)(1, values));
}

but the following doesn't work:

import std.stdio;
import std.algorithm;

void main()
{
     auto values = [1, 2, 4, 8];
     writefln(reduce!((int a, int b) { return a * b; })(1, values));
}

It used to work, IIRC, but now it prints '0'. Does anybody know if it is  
supposed to work?

Christopher Wright <dhasenan gmail.com> writes:

Justin wrote:
 Ahh, I thought it would some stupid oversight on my part. This works:
 
 module functionliteral;
 
 import std.stdio;
 
 static void main() {
 
 	int[] values = [1,2,4,8];
 	writefln(Reduce(values, function int(int x, int y) { return x + y; }));
 	writefln(Reduce(values, function int(int x, int y) { return x * y; }));
 }
 
 static int Reduce(int[] values, int function(int x, int y) operation) {
 	int total = values[0];
 	foreach (int v; values[1..$])
 		total = operation(total,v);
 	return total;
 }
 

You don't need to make those static, by the way.

Reduce doesn't work on all inputs:
writefln(Reduce([], (int x, int y) { return x + y; }));