Re: opCast cannot implicitly convert a.opCast of type X to Y
On Thursday, 15 February 2018 at 00:34:33 UTC, Meta wrote:
On Thursday, 15 February 2018 at 00:27:40 UTC, Meta wrote:
On Wednesday, 14 February 2018 at 23:46:30 UTC, aliak wrote:
On Wednesday, 14 February 2018 at 15:14:24 UTC, Meta wrote:
Ooh yes, of course! Thank you :)
Even better:
import std.conv;
auto b = a.map!(to!float);
Actually, that won't quite work without redefining map a little:
Optional!U map(alias f, U = typeof(f(t.init)))()
{
etc...
}
Ah yes, true, also auto return would work. But then you'd still
need to do the typeof(f(T.init)) evaluation in the body... plus
you lose being able to see an explicit return type i guess...
hmm. So nevermind :)
Though a free function would be good me thinks. Then you could
use it seamlessly with std.algorithm.map.
Optional!U map(alias f, T, U = typeof(f(T.init)))(Optional!T opt)
{
return Optional!U(f(opt.t));
}
Cheers,
- Ali
Re: opCast cannot implicitly convert a.opCast of type X to Y
On Thursday, 15 February 2018 at 00:27:40 UTC, Meta wrote:
On Wednesday, 14 February 2018 at 23:46:30 UTC, aliak wrote:
On Wednesday, 14 February 2018 at 15:14:24 UTC, Meta wrote:
Ooh yes, of course! Thank you :)
Even better:
import std.conv;
auto b = a.map!(to!float);
Actually, that won't quite work without redefining map a little:
Optional!U map(alias f, U = typeof(f(t.init)))()
{
etc...
}
Re: opCast cannot implicitly convert a.opCast of type X to Y
On Wednesday, 14 February 2018 at 23:46:30 UTC, aliak wrote:
On Wednesday, 14 February 2018 at 15:14:24 UTC, Meta wrote:
I think the best way to do this is to implement `map` for your
optional type.
Optional!U map(U, alias f)()
{
return empty? no!U : some!U(f(t));
}
Optional!int a = 3;
auto b = a.map!(v => cast(float)v);
assert(is(typeof(b) == Optional!float));
Ooh yes, of course! Thank you :)
Even better:
import std.conv;
auto b = a.map!(to!float);
Re: opCast cannot implicitly convert a.opCast of type X to Y
On Wednesday, 14 February 2018 at 15:14:24 UTC, Meta wrote:
I think the best way to do this is to implement `map` for your
optional type.
Optional!U map(U, alias f)()
{
return empty? no!U : some!U(f(t));
}
Optional!int a = 3;
auto b = a.map!(v => cast(float)v);
assert(is(typeof(b) == Optional!float));
Ooh yes, of course! Thank you :)
Re: opCast cannot implicitly convert a.opCast of type X to Y
On Monday, 12 February 2018 at 02:05:16 UTC, aliak wrote:
From spec: Cast expression: "cast ( Type ) UnaryExpression"
converts UnaryExpresssion to Type.
And https://dlang.org/spec/operatoroverloading.html#cast makes
no mention of the return type of opCast. One could think that
the return type of opCast would be the return type. But it
seems it must be the same as the template parameter of opCast
else you get a compile error that seems like it can be much
better.
---
import std.stdio;
struct B(T) {
T t;
}
struct A(T) {
T t;
auto opCast(U)() {
return B!U(cast(U)t);
}
}
void main() {
auto a = A!int(3);
auto b = cast(float)a; // error
}
Error: cannot implicitly convert expression a.opCast() of type
B!float to float
Is this deliberate?
The use case I have is making an optional type that you can
cast to a different type:
auto opCast(U)() const {
static if (isOptional!U)
{
alias V = OptionalTarget!U;
return empty ? no!V : some!V(cast(V)front); // it's a
range so "front" is the raw value
}
else
{
return empty ? no!U : some!U(cast(U)front);
}
}
It would allow for scenarios like:
Optional!int a = 3;
auto b = cast(float)a;
// b == some!float
Cheers
- Ali
I think the best way to do this is to implement `map` for your
optional type.
Optional!U map(U, alias f)()
{
return empty? no!U : some!U(f(t));
}
Optional!int a = 3;
auto b = a.map!(v => cast(float)v);
assert(is(typeof(b) == Optional!float));
Re: opCast cannot implicitly convert a.opCast of type X to Y
On Tuesday, 13 February 2018 at 12:12:30 UTC, Nathan S. wrote:
On Monday, 12 February 2018 at 02:05:16 UTC, aliak wrote:
struct B(T) {
T t;
}
struct A(T) {
T t;
auto opCast(U)() {
return B!U(cast(U)t);
}
}
void main() {
auto a = A!int(3);
auto b = cast(float)a; // error
}
Having the result of "cast(float) a" not be a float would be
evil.
Ya :p dunno what I was thinking.
Re: opCast cannot implicitly convert a.opCast of type X to Y
On Monday, 12 February 2018 at 02:05:16 UTC, aliak wrote:
struct B(T) {
T t;
}
struct A(T) {
T t;
auto opCast(U)() {
return B!U(cast(U)t);
}
}
void main() {
auto a = A!int(3);
auto b = cast(float)a; // error
}
Having the result of "cast(float) a" not be a float would be evil.
Re: opCast cannot implicitly convert a.opCast of type X to Y
On Monday, 12 February 2018 at 02:05:16 UTC, aliak wrote: From spec: Cast expression: "cast ( Type ) UnaryExpression" converts UnaryExpresssion to Type. And https://dlang.org/spec/operatoroverloading.html#cast makes no mention of the return type of opCast. One could think that the return type of opCast would be the return type. But it seems it must be the same as the template parameter of opCast else you get a compile error that seems like it can be much better. - Ali In my opinion, you should fill a documentation enhancement request on issues.dlang.org.
opCast cannot implicitly convert a.opCast of type X to Y
From spec: Cast expression: "cast ( Type ) UnaryExpression"
converts UnaryExpresssion to Type.
And https://dlang.org/spec/operatoroverloading.html#cast makes no
mention of the return type of opCast. One could think that the
return type of opCast would be the return type. But it seems it
must be the same as the template parameter of opCast else you get
a compile error that seems like it can be much better.
---
import std.stdio;
struct B(T) {
T t;
}
struct A(T) {
T t;
auto opCast(U)() {
return B!U(cast(U)t);
}
}
void main() {
auto a = A!int(3);
auto b = cast(float)a; // error
}
Error: cannot implicitly convert expression a.opCast() of type
B!float to float
Is this deliberate?
The use case I have is making an optional type that you can cast
to a different type:
auto opCast(U)() const {
static if (isOptional!U)
{
alias V = OptionalTarget!U;
return empty ? no!V : some!V(cast(V)front); // it's a
range so "front" is the raw value
}
else
{
return empty ? no!U : some!U(cast(U)front);
}
}
It would allow for scenarios like:
Optional!int a = 3;
auto b = cast(float)a;
// b == some!float
Cheers
- Ali
