Re: Why can't templates with default arguments be instantiated without the bang syntax?
On Thu, 15 Sep 2011 17:54:19 +0200, Christophe wrote: "Simen Kjaeraas" , dans le message (digitalmars.D.learn:29539), a écrit : On Thu, 15 Sep 2011 16:46:24 +0200, Andrej Mitrovic wrote: struct Foo(T = int) {} void main() { Foo foo; // fail Foo!() bar; // ok } It would be very convenient to be able to default to one type like this. For example, in CairoD there's a Point structure which takes doubles as its storage type, and then there's PointInt that takes ints. The reason they're not both a template Point() that takes a type argument is because in most cases the user will use the Point structure with doubles, and only in rare cases Point with ints. So to simplify code one doesn't have to write Point!double in all of their code, but simply Point. If the bang syntax wasn't required in presence of default arguments then these workarounds wouldn't be needed. How would you then pass a single-argument template as a template alias parameter? Example: template Foo( ) { template Bar( ) { } } template Baz(alias A) { mixin A!(); } void main( ) { mixin Baz!Foo; } Does this mixin Foo or Bar to main's scope? I don't get the problem. Maybe I am not used to mixin enough. Can you mixin normal templates, and not only mixin templates ? Anyway, why would this mixin Bar ? As I understand the proposition, only "mixin Baz!(Foo.Bar);" and of course "mixin Baz!(Foo!().Bar)" should mixin Bar. Sorry, you're right. I meant: template Foo( ) { template Foo( ) { } } -- Simen
Re: Why can't templates with default arguments be instantiated without the bang syntax?
On 2011-09-15 16:46, Andrej Mitrovic wrote: struct Foo(T = int) {} void main() { Foo foo; // fail Foo!() bar; // ok } It would be very convenient to be able to default to one type like this. For example, in CairoD there's a Point structure which takes doubles as its storage type, and then there's PointInt that takes ints. The reason they're not both a template Point() that takes a type argument is because in most cases the user will use the Point structure with doubles, and only in rare cases Point with ints. So to simplify code one doesn't have to write Point!double in all of their code, but simply Point. If the bang syntax wasn't required in presence of default arguments then these workarounds wouldn't be needed. I've wondered the same thing, why this doesn't work: template Foo (T = int) {} mixin Foo; But this works: template Foo () {} mixin Foo; -- /Jacob Carlborg
Re: Why can't templates with default arguments be instantiated without the bang syntax?
On Thu, 15 Sep 2011 10:46:24 -0400, Andrej Mitrovic wrote: struct Foo(T = int) {} void main() { Foo foo; // fail Foo!() bar; // ok } It would be very convenient to be able to default to one type like this. For example, in CairoD there's a Point structure which takes doubles as its storage type, and then there's PointInt that takes ints. The reason they're not both a template Point() that takes a type argument is because in most cases the user will use the Point structure with doubles, and only in rare cases Point with ints. So to simplify code one doesn't have to write Point!double in all of their code, but simply Point. If the bang syntax wasn't required in presence of default arguments then these workarounds wouldn't be needed. Perhaps a different approach: struct PointT(T) {...} alias PointT!(double) Point; // and if so desired: alias PointT!int PointInt; Just a thought... -Steve
Re: Why can't templates with default arguments be instantiated without the bang syntax?
On Thursday, September 15, 2011 16:46:24 Andrej Mitrovic wrote: > struct Foo(T = int) {} > > void main() > { > Foo foo; // fail > Foo!() bar; // ok > } > > It would be very convenient to be able to default to one type like this. > > For example, in CairoD there's a Point structure which takes doubles > as its storage type, and then there's PointInt that takes ints. The > reason they're not both a template Point() that takes a type argument > is because in most cases the user will use the Point structure with > doubles, and only in rare cases Point with ints. So to simplify code > one doesn't have to write Point!double in all of their code, but > simply Point. > > If the bang syntax wasn't required in presence of default arguments > then these workarounds wouldn't be needed. There is no type inference for templated structs, so you need the bang. Only functions get type inference. They way to fix this is to create a separate factory function to construct the type. std.container does this with redBlackTree for RedBlackTree. - Jonathan M Davis
Re: Why can't templates with default arguments be instantiated without the bang syntax?
"Simen Kjaeraas" , dans le message (digitalmars.D.learn:29539), a écrit : > On Thu, 15 Sep 2011 16:46:24 +0200, Andrej Mitrovic > wrote: > >> struct Foo(T = int) {} >> >> void main() >> { >> Foo foo; // fail >> Foo!() bar; // ok >> } >> >> It would be very convenient to be able to default to one type like this. >> >> For example, in CairoD there's a Point structure which takes doubles >> as its storage type, and then there's PointInt that takes ints. The >> reason they're not both a template Point() that takes a type argument >> is because in most cases the user will use the Point structure with >> doubles, and only in rare cases Point with ints. So to simplify code >> one doesn't have to write Point!double in all of their code, but >> simply Point. >> >> If the bang syntax wasn't required in presence of default arguments >> then these workarounds wouldn't be needed. > > How would you then pass a single-argument template as a template alias > parameter? > > Example: > > template Foo( ) { > template Bar( ) { > } > } > > template Baz(alias A) { > mixin A!(); > } > > void main( ) { > mixin Baz!Foo; > } > > Does this mixin Foo or Bar to main's scope? I don't get the problem. Maybe I am not used to mixin enough. Can you mixin normal templates, and not only mixin templates ? Anyway, why would this mixin Bar ? As I understand the proposition, only "mixin Baz!(Foo.Bar);" and of course "mixin Baz!(Foo!().Bar)" should mixin Bar. -- Christophe
Re: Why can't templates with default arguments be instantiated without the bang syntax?
On Thu, 15 Sep 2011 16:46:24 +0200, Andrej Mitrovic wrote: struct Foo(T = int) {} void main() { Foo foo; // fail Foo!() bar; // ok } It would be very convenient to be able to default to one type like this. For example, in CairoD there's a Point structure which takes doubles as its storage type, and then there's PointInt that takes ints. The reason they're not both a template Point() that takes a type argument is because in most cases the user will use the Point structure with doubles, and only in rare cases Point with ints. So to simplify code one doesn't have to write Point!double in all of their code, but simply Point. If the bang syntax wasn't required in presence of default arguments then these workarounds wouldn't be needed. How would you then pass a single-argument template as a template alias parameter? Example: template Foo( ) { template Bar( ) { } } template Baz(alias A) { mixin A!(); } void main( ) { mixin Baz!Foo; } Does this mixin Foo or Bar to main's scope? -- Simen
Why can't templates with default arguments be instantiated without the bang syntax?
struct Foo(T = int) {} void main() { Foo foo; // fail Foo!() bar; // ok } It would be very convenient to be able to default to one type like this. For example, in CairoD there's a Point structure which takes doubles as its storage type, and then there's PointInt that takes ints. The reason they're not both a template Point() that takes a type argument is because in most cases the user will use the Point structure with doubles, and only in rare cases Point with ints. So to simplify code one doesn't have to write Point!double in all of their code, but simply Point. If the bang syntax wasn't required in presence of default arguments then these workarounds wouldn't be needed.