• Ishan Thilina (18/18) Apr 05 2011 I can use the ElementType() template in the std.rane library to find the...
• Philippe Sigaud (16/31) Apr 05 2011 ElementType acts on types. It takes a type and 'returns' (compiles to,
• Ishan Thilina (11/21) Apr 06 2011 Got it!. Thanks.

Ishan Thilina <ishanthilina gmail.com> writes:

I can use the ElementType() template in the std.rane library to find the type
of the elements of a range,isn't it? But I cant compile the following programme.

I get a "untitled.d(32): Error: template instance ElementType!(listR) does not
match template declaration ElementType(R)" error.

What could be the problem ?
--------------------------------------------------------------------------

import std.stdio;
import std.range;
import std.container;

int main(char[][] args)
{
	auto list=SList!(int)(1,2,3);
	auto listR=list.opSlice();
	writefln("%s",ElementType!(listR));


	return 0;
}
--------------------------------------------------------------

Thank you...!

Philippe Sigaud <philippe.sigaud gmail.com> writes:

 I get a "untitled.d(32): Error: template instance ElementType!(listR) doe=

s not
 match template declaration ElementType(R)" error.

 What could be the problem ?
 -------------------------------------------------------------------------=

-
 import std.stdio;
 import std.range;
 import std.container;

 int main(char[][] args)
 {
 =C2=A0 =C2=A0 =C2=A0 =C2=A0auto list=3DSList!(int)(1,2,3);
 =C2=A0 =C2=A0 =C2=A0 =C2=A0auto listR=3Dlist.opSlice();
 =C2=A0 =C2=A0 =C2=A0 =C2=A0writefln("%s",ElementType!(listR));


 =C2=A0 =C2=A0 =C2=A0 =C2=A0return 0;
 }
 --------------------------------------------------------------

ElementType acts on types. It takes a type and 'returns' (compiles to,
actually) another type. You need to give it typeof(listR).
Then, as ElementType!(typeof(listR)) is a type, you cannot pass it to
writeln. Use .stringof to go from the type to a string representation
of its name.

So:

writefln("%s", ElementType!(typeof(listR)).stringof);

ElementType!(typeof(listR)) is a type like any other. You can create a
variable with it:

ElementType!(typeof(listR)) elem;

Btw, .opSlice() is the name of the '[]' operator. You previous line
can be written like this:

auto listR =3D list[];


    Philippe

Ishan Thilina <ishanthilina gmail.com> writes:

--Philippe wrote::

ElementType acts on types. It takes a type and 'returns' (compiles to,
actually) another type. You need to give it typeof(listR).
Then, as ElementType!(typeof(listR)) is a type, you cannot pass it to
writeln. Use .stringof to go from the type to a string representation
of its name.

So:

writefln("%s", ElementType!(typeof(listR)).stringof);

ElementType!(typeof(listR)) is a type like any other. You can create a
variable with it:

ElementType!(typeof(listR)) elem;

Got it!. Thanks.

I faced this confusion because of the description for "ElementType" which is
available in http://d-programming-language.org/phobos/std_range.html. It says,

"
template ElementType(R):

The element type of R. R does not have to be a range. The element type is
determined as the type yielded by r.front for an object r or type R. For
example,
ElementType!(T[]) is T. If R is not a range, ElementType!R is void.

"
So I thought that just passing ElementType!listR was adequate. :s