• Benjamin Herr (14/14) Oct 23 2004 Hello, D!
• Tyro (3/19) Oct 23 2004 a = b[foo .. bar];
• Benjamin Herr (14/20) Oct 23 2004 I think there is the subtle difference that a[] = b[..] copies while a =...
• Walter (3/16) Oct 30 2004 Just overload the .. operator with opSlice and have it return an array.

Benjamin Herr <ben 0x539.de> writes:

Hello, D!

I was taught that overloaded operators should emulate their function on 
primitive data types.
With ints (and probably structs), a += b and a = a + b is essentially 
the same.
With objects, a += b is a.opAddAssign(b) or the reverse and a = a + b is 
a = a.opAdd(b), which is clearly different as the former cannot emulate 
the latter's reference assignment.
This leads me to the assumption that we lack a operator for valueish 
assignment that is different from the reference assignment one.
What am I missing?

On a completely unrelated note, how can I implement arraylike a[] = 
b[foo ... bar]?

   -ben (the newbie one)

Tyro <ridimz_at yahoo.dot.com> writes:

Benjamin Herr wrote:
 Hello, D!
 
 I was taught that overloaded operators should emulate their function on 
 primitive data types.
 With ints (and probably structs), a += b and a = a + b is essentially 
 the same.
 With objects, a += b is a.opAddAssign(b) or the reverse and a = a + b is 
 a = a.opAdd(b), which is clearly different as the former cannot emulate 
 the latter's reference assignment.
 This leads me to the assumption that we lack a operator for valueish 
 assignment that is different from the reference assignment one.
 What am I missing?
 
 On a completely unrelated note, how can I implement arraylike a[] = 
 b[foo ... bar]?

a = b[foo .. bar];

Unless I completely misunderstood your question!

   -ben (the newbie one)

Benjamin Herr <ben 0x539.de> writes:

Tyro wrote:
 Benjamin Herr wrote:
 how can I implement arraylike a[] = b[foo ... bar]?

 
 a = b[foo .. bar];
 
 Unless I completely misunderstood your question!

I think there is the subtle difference that a[] = b[..] copies while a = 
b makes a refer the same data as b.

int main() {
         char[] a = "Hello, World".dup;
         char[] b, c;
         b = a[0 .. length];
         c.length = a.length;
         c[] = a[0 .. length];

         a[7 .. 12] = "Tyro!";

         printf("b is: %.*s\n", b);
         printf("c is: %.*s\n", c);
         return 0;
}

"Walter" <newshound digitalmars.com> writes:

"Benjamin Herr" <ben 0x539.de> wrote in message
news:clelpa$oae$1 digitaldaemon.com...
 Hello, D!

 I was taught that overloaded operators should emulate their function on
 primitive data types.
 With ints (and probably structs), a += b and a = a + b is essentially
 the same.
 With objects, a += b is a.opAddAssign(b) or the reverse and a = a + b is
 a = a.opAdd(b), which is clearly different as the former cannot emulate
 the latter's reference assignment.
 This leads me to the assumption that we lack a operator for valueish
 assignment that is different from the reference assignment one.
 What am I missing?

 On a completely unrelated note, how can I implement arraylike a[] =
 b[foo ... bar]?

Just overload the .. operator with opSlice and have it return an array.