qznc:
Operator precedence of . is higher than unary minus.
You are right, I didn't know it, so Typedef and constructors are
not to blame:
double foo(in double x) {
assert (x = 0);
return x;
}
void main() {
assert(-1.foo == -1);
}
Is this a good design of the operator
import std.typecons: Typedef;
alias Foo = Typedef!double;
void main() {
auto a1 = Foo(1);
pragma(msg, typeof(a1));
auto a2 = 1.Foo;
pragma(msg, typeof(a2));
auto a3 = Foo(-1);
pragma(msg, typeof(a3));
auto a4 = -1.Foo;
pragma(msg, typeof(a4));
}
It prints:
On Wednesday, 6 November 2013 at 11:04:05 UTC, bearophile wrote:
import std.typecons: Typedef;
alias Foo = Typedef!double;
void main() {
auto a1 = Foo(1);
pragma(msg, typeof(a1));
auto a2 = 1.Foo;
pragma(msg, typeof(a2));
auto a3 = Foo(-1);
pragma(msg, typeof(a3));
qznc:
Operator precedence of . is higher than unary minus.
Is this good?
However, what is Typedef for?
It's to create a differently named type, useful for stronger
static typing, to increase code clarity and avoid some bugs.
If you have a function like this:
double foo(in double x,
On 11/06/2013 03:04 AM, bearophile wrote:
import std.typecons: Typedef;
alias Foo = Typedef!double;
void main() {
auto a1 = Foo(1);
pragma(msg, typeof(a1));
auto a2 = 1.Foo;
pragma(msg, typeof(a2));
auto a3 = Foo(-1);
pragma(msg, typeof(a3));
auto a4 = -1.Foo;
On Wednesday, 6 November 2013 at 17:10:34 UTC, Ali Çehreli wrote:
I would be very surprised if unary - produced a different
type from the operand:
Ali
Operator does not produce type, it produces value of expression,
and type of expression happens not to be the type you expected.
But such
On 11/06/2013 09:46 AM, Maxim Fomin wrote:
On Wednesday, 6 November 2013 at 17:10:34 UTC, Ali Çehreli wrote:
I would be very surprised if unary - produced a different type from
the operand:
Ali
Operator does not produce type, it produces value of expression, and
type of expression
On Wednesday, 6 November 2013 at 18:02:32 UTC, Ali Çehreli wrote:
But such
expectations need not correspond to language rules (try to
think from
from language laywer perspective).
I still argue that the expression -expr must have the same type
as expr.
In bearophile case, I guess
On Wednesday, 6 November 2013 at 18:16:04 UTC, Maxim Fomin wrote:
I think that reason for such behavior is the way used defined
operator overloading functions are implemented, not the
language per se, so programmers confuse themselves.
What about other possible reason - Typedef implementation
Dicebot:
Typedef implementation sucks? ;)
So do you suggest to open some enhancement request/bug report on
Typedef?
Bye,
bearophile
On Wednesday, 6 November 2013 at 21:57:47 UTC, bearophile wrote:
Dicebot:
Typedef implementation sucks? ;)
So do you suggest to open some enhancement request/bug report
on Typedef?
Bye,
bearophile
Sure. get enough such reports and we may even get it back as
language feature :) (I think
11 matches
Mail list logo
