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digitalmars.D - The type of an element-wise arithmetic expression

reply Eduardo Cavazos <wayo.cavazos gmail.com> writes:
Hello,

This program:

----------------------------------------------------------------------
import std.stdio ;

void f0 ( double [2] a ) { writeln ( a ) ; }

void main ()
{
   double [2] a = [ 1.0 , 2.0 ] ;
   double [2] b = [ 3.0 , 4.0 ] ;

   f0 ( a[] - b[] ) ;
}
----------------------------------------------------------------------

produces an error:

$ rdmd array_wise_a.d
array_wise_a.d(11): Error: function array_wise_a.f0 (double[2u] a) is 
not callable using argument types (double[])
array_wise_a.d(11): Error: cannot implicitly convert expression (a[] - 
b[]) of type double[] to double[2u]

Since the sub-expressions a[] and b[] are each of type double[2], I 
would expect the the type of a[]-b[] to also be of double[2], and thus 
compatible with f0.

Ed
Aug 21 2010
parent bearophile <bearophileHUGS lycos.com> writes:
Eduardo Cavazos:

 import std.stdio ;
 
 void f0 ( double [2] a ) { writeln ( a ) ; }
 
 void main ()
 {
    double [2] a = [ 1.0 , 2.0 ] ;
    double [2] b = [ 3.0 , 4.0 ] ;
 
    f0 ( a[] - b[] ) ;
 }


I am not sure this code is supported. I think that currently the right way to
use array ops is to give a wide enough as lvalue. But note this syntax can't be
used (I have a bug report on this too) (the usage of [] has to become
obligatory):

double[2] c[] = a[] - b[];

So you need to use something like:
double[2] c = void;
c[] = a[] - b[];

Bye,
bearophile
Aug 21 2010