• estew (15/15) Jan 30 2013 void main() {
• Jacob Carlborg (6/13) Jan 30 2013 There are dynamic arrays and static arrays. Dynamic arrays are reference...
• estew (1/5) Jan 31 2013 Got it. Thanks very much for the help.
• Timon Gehr (17/22) Jan 31 2013 It fails at compile time?
• FG (7/8) Jan 31 2013 It's because floating point literals are double by default.

"estew" <estewh gmail.com> writes:

void main() {
      float[3] v1 = [1.0, 2.0, 3.0];    // No error
      float[3] v = [1.0, 2.0, 3.0].dup; // Fails at runtime with
error message
}


Why does the array assignment work when "dup" is not used. My
understanding is that arrays, like classes, are references.

So I declare v1 as a float[3] point it at a double[3] literal. Is
v1 now a double[3] or a float[3]? Is there an implicit cast from
double[3] to float[3]?

The dup, gives a compile time error and if cast to float[] it
gives a runtime error. This is all good, I just don't quite
understand how the ref assignment is working...


Thanks,
Stewart

Jacob Carlborg <doob me.com> writes:

On 2013-01-31 05:48, estew wrote:
 void main() {
       float[3] v1 = [1.0, 2.0, 3.0];    // No error
       float[3] v = [1.0, 2.0, 3.0].dup; // Fails at runtime with
 error message
 }


 Why does the array assignment work when "dup" is not used. My
 understanding is that arrays, like classes, are references.

There are dynamic arrays and static arrays. Dynamic arrays are reference 
types, static arrays are value types. You have declared a static array.

http://dlang.org/arrays.html

-- 
/Jacob Carlborg

"estew" <estewh gmail.com> writes:

 There are dynamic arrays and static arrays. Dynamic arrays are 
 reference types, static arrays are value types. You have 
 declared a static array.

 http://dlang.org/arrays.html

Got it. Thanks very much for the help.

Timon Gehr <timon.gehr gmx.ch> writes:

On 01/31/2013 05:48 AM, estew wrote:
 void main() {
       float[3] v1 = [1.0, 2.0, 3.0];    // No error
       float[3] v = [1.0, 2.0, 3.0].dup; // Fails at runtime with error message
 }
 ...

It fails at compile time?

The reason is that array literals have special conversion rules:

Eg:

bool[] x = [0,1,0,1,0,1,1];

An array literal is converted element-wise. This means an array literal 
sometimes behaves differently from other expressions of the same type:

import std.stdio;

void main() {
	int[] a = [0,2,0,1];
	bool[] x = cast(bool[])[0,2,0,1];
	bool[] y = cast(bool[])a;
	writeln(x,"\n",y);
}

[false, true, false, true]
[false, false, false, false, true, false, false, false, false, false, 
false, false, true, false, false, false]

FG <home fgda.pl> writes:

On 2013-01-31 10:47, Timon Gehr wrote:
 The reason is that array literals have special conversion rules

It's because floating point literals are double by default.
In the first assignment float type can be deduced from v1.
To make the second one work, you can explicitly make them float:

     float[3] v1 = [1.0, 2.0, 3.0];
     float[3] v = [1.0f, 2.0f, 3.0f].dup;

or duplicate v1:    float[3] v = v1.dup;