Re: How to instantiate a template struct with a template constructor without relying on auto deduction?
On Wednesday, 21 February 2018 at 14:29:31 UTC, Simen Kjærås
wrote:
On Wednesday, 21 February 2018 at 14:11:10 UTC, ParticlePeter
wrote:
struct Foo(T) {
T bar;
this(S)(S s) {
bar = convert(s);
}
}
auto foo = Foo!int(some_float);
this works because S is deduced as typeof(some_float), but how
would I instantiate the struct without relying on auto
deduction?
Suppose we would have this kind of constructor where auto
deduction is not possible:
this(int n)(float f) {
static foreach( i; 0..n) { do_some_ctfe_magic;}
}
How to instantiate Foo then?
No can do. The solution is to use a factory function:
Feared the same, thanks.
struct Foo(T) {
static Foo create(int n)(float f) {
Foo result;
static foreach( i; 0..n) { do_some_ctfe_magic;}
return result;
}
}
--
Simen
I will consider this, actually I use something quite close, but
my create is not static and does not return anything. It simply
initializes the struct after it has been constructed with the
default ctor. The templated user ctor would have been nice,
though.
Re: How to instantiate a template struct with a template constructor without relying on auto deduction?
On Wednesday, 21 February 2018 at 14:42:56 UTC, Simen Kjærås
wrote:
On Wednesday, 21 February 2018 at 14:29:38 UTC, ixid wrote:
I do not understand what is happening here, I tried to wrote
what I thought would be the answer. If someone could explain
that would be great. I wrote this code:
struct Foo2(T, S) {
T bar;
this(S s) {
bar = s.to!T;
}
}
void main() {
float some_float = 0.5f;
int some_int = 1;
auto foo1 = Foo2!(int, float)(some_float);// Compiles, OK!
auto foo2 = Foo2!(int, float)(some_int); // Compiles,
wat?
}
int n = 1;
float f = n; // Basically this.
Foo2!(int, float) expects a float, and ints are implicitly
convertible to float.
--
Simen
Ah yes, that was silly of me to forget. Thanks!
Re: How to instantiate a template struct with a template constructor without relying on auto deduction?
On Wednesday, 21 February 2018 at 14:29:38 UTC, ixid wrote:
I do not understand what is happening here, I tried to wrote
what I thought would be the answer. If someone could explain
that would be great. I wrote this code:
struct Foo2(T, S) {
T bar;
this(S s) {
bar = s.to!T;
}
}
void main() {
float some_float = 0.5f;
int some_int = 1;
auto foo1 = Foo2!(int, float)(some_float);// Compiles, OK!
auto foo2 = Foo2!(int, float)(some_int); // Compiles, wat?
}
int n = 1;
float f = n; // Basically this.
Foo2!(int, float) expects a float, and ints are implicitly
convertible to float.
--
Simen
Re: How to instantiate a template struct with a template constructor without relying on auto deduction?
On Wednesday, 21 February 2018 at 14:11:10 UTC, ParticlePeter
wrote:
struct Foo(T) {
T bar;
this(S)(S s) {
bar = convert(s);
}
}
auto foo = Foo!int(some_float);
this works because S is deduced as typeof(some_float), but how
would I instantiate the struct without relying on auto
deduction?
Suppose we would have this kind of constructor where auto
deduction is not possible:
this(int n)(float f) {
static foreach( i; 0..n) { do_some_ctfe_magic;}
}
How to instantiate Foo then?
I do not understand what is happening here, I tried to wrote what
I thought would be the answer. If someone could explain that
would be great. I wrote this code:
struct Foo2(T, S) {
T bar;
this(S s) {
bar = s.to!T;
}
}
void main() {
float some_float = 0.5f;
int some_int = 1;
auto foo1 = Foo2!(int, float)(some_float);// Compiles, OK!
auto foo2 = Foo2!(int, float)(some_int); // Compiles, wat?
}
Re: How to instantiate a template struct with a template constructor without relying on auto deduction?
On Wednesday, 21 February 2018 at 14:11:10 UTC, ParticlePeter
wrote:
struct Foo(T) {
T bar;
this(S)(S s) {
bar = convert(s);
}
}
auto foo = Foo!int(some_float);
this works because S is deduced as typeof(some_float), but how
would I instantiate the struct without relying on auto
deduction?
Suppose we would have this kind of constructor where auto
deduction is not possible:
this(int n)(float f) {
static foreach( i; 0..n) { do_some_ctfe_magic;}
}
How to instantiate Foo then?
No can do. The solution is to use a factory function:
struct Foo(T) {
static Foo create(int n)(float f) {
Foo result;
static foreach( i; 0..n) { do_some_ctfe_magic;}
return result;
}
}
--
Simen
How to instantiate a template struct with a template constructor without relying on auto deduction?
struct Foo(T) {
T bar;
this(S)(S s) {
bar = convert(s);
}
}
auto foo = Foo!int(some_float);
this works because S is deduced as typeof(some_float), but how
would I instantiate the struct without relying on auto deduction?
Suppose we would have this kind of constructor where auto
deduction is not possible:
this(int n)(float f) {
static foreach( i; 0..n) { do_some_ctfe_magic;}
}
How to instantiate Foo then?
