Re: Working with ranges: mismatched function return type inference
On Tuesday, October 11, 2016 10:42:42 Ali Çehreli via Digitalmars-d-learn
wrote:
> Those interfaces already exist in Phobos: :)
>
>https://dlang.org/phobos/std_range_interfaces.html
>
> auto foo(int[] ints) {
>import std.range;
>if (ints.length > 10) {
>return
> cast(RandomAccessFinite!int)inputRangeObject(chain(ints[0..5], ints[8..$]));
> } else {
>return cast(RandomAccessFinite!int)inputRangeObject(ints);
>}
> }
>
> void main() {
> import std.stdio;
> import std.range;
> import std.algorithm;
> writeln(foo([1, 2, 3]));
> writeln(foo(iota(20).array));
> }
And in this case, if you were considering doing that, you might as well just
concatenate the dynamic arrays rather than chaining them, because using
interfaces means allocating on the heap just like you would with
concatenating.
About the only time that using interfaces is the right solution with ranges
is when you're dealing with virtual functions (which can't be templatized),
and even then, it's not necessarily the best choice. Here, IMHO, it makes no
sense at all.
- Jonathan M Davis
Re: Working with ranges: mismatched function return type inference
On Tuesday, 11 October 2016 at 18:09:26 UTC, ag0aep6g wrote: You've got some options: Wow, thanks everyone, great information! I think I understand my options now.
Re: Working with ranges: mismatched function return type inference
On 10/11/2016 09:55 AM, orip wrote:
auto foo(int[] ints) {
import std.range;
if (ints.length > 10) {
return chain(ints[0..5], ints[8..$]);
} else {
//return ints; // Error: mismatched function return type inference
of int[] and Result
return chain(ints[0..0], ints[0..$]); // This workaround compiles
}
}
Is there a compatible return type that can be used, or some other
workaround?
You've got some options:
1) OOP with std.range.interfaces. Ali already showed how this work.
Comes at the cost of extra allocations and indirections.
2) std.range.choose wraps two different range types and uses forwards to
one of them based on a condition. Should be cheap. But you need
restructure your code a little:
auto foo(int[] ints) {
import std.range: chain, choose;
return choose(ints.length > 10,
chain(ints[0..5], ints[8..$]),
ints);
}
3) The workaround you already discovered: making a seemingly pointless
call to `chain` to get the types to match. Possibly the most efficient
solution. Looks a little odd.
Re: Working with ranges: mismatched function return type inference
On Tuesday, October 11, 2016 07:55:36 orip via Digitalmars-d-learn wrote:
> I get "Error: mismatched function return type inference" errors
> with choosing the return type for functions that work on ranges
> using, e.g, std.algorithm or std.range functions, but have
> different behavior based on runtime values. The return type is
> always a range with the same underlying type.
>
> Here's an example:
>
> auto foo(int[] ints) {
>import std.range;
>if (ints.length > 10) {
> return chain(ints[0..5], ints[8..$]);
>} else {
> //return ints; // Error: mismatched function return type
> inference of int[] and Result
> return chain(ints[0..0], ints[0..$]); // This workaround
> compiles
>}
> }
>
> Is there a compatible return type that can be used, or some other
> workaround?
> I couldn't find one when searching for the error or looking at
> the phobos source code.
>
> Thanks! orip
You're workaround is basically doing what you need to do. A function can
only return one type. The fact that both return statements are returning
ranges over the same kind of elements is irrelevant. They have to be
_exactly_ the same type. So, either you need to convert the range for the
first return statement into int[] so that it matches the second (e.g. by
calling array on the result or just using ~), or you need to call chain on
two int[]s for the second return statement so that it matches the first.
The second option (which your workaround does) is better if you don't intend
to convert the result to an array, since it avoids allocating an array, but
if you're just going to convert the result to int[] anyway, the first option
would be better.
Regardless, you can't have a function returning different types from
different return statements - even with auto. The compiler needs to know
exactly what the return type is whether you type it or not; auto just
infers it for you rather than requiring you to type it out.
- Jonathan M Davis
Re: Working with ranges: mismatched function return type inference
On 10/11/2016 10:28 AM, TheFlyingFiddle wrote:
On Tuesday, 11 October 2016 at 15:46:20 UTC, orip wrote:
On Tuesday, 11 October 2016 at 13:06:37 UTC, pineapple wrote:
Rewrite `return chain(ints[0..5], ints[8..$]);` as `return ints[0..5]
~ ints[8..$];`
The `chain` function doesn't return an array, it returns a
lazily-evaluated sequence of an entirely different type from `int[]`.
Of course it does! I would like the function to return an "input range
of int", no matter which one specifically. Is this possible?
It is, but you will have to use an interface / class to achieve this
behavior (or use some sort of polymorphic struct). Something like this
will do the trick:
import std.range;
import std.stdio;
interface IInputRange(T)
{
bool empty();
T front();
void popFront();
}
final class InputRange(Range) if(isInputRange!Range)
: IInputRange!(ElementType!Range)
{
Range r;
this(Range r)
{
this.r = r;
}
bool empty() { return r.empty; }
ElementType!Range front() { return r.front; }
void popFront() { r.popFront; }
}
auto inputRange(Range)(Range r)
{
return new InputRange!Range(r);
}
IInputRange!int foo(int[] ints)
{
import std.range;
if(ints.length > 10) {
return inputRange(chain(ints[0 .. 5], ints[8 .. $]));
} else {
return inputRange(ints);
}
}
void main()
{
auto ir = foo([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]);
auto ir2 = foo([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]);
writeln(ir);
writeln(ir2);
}
Those interfaces already exist in Phobos: :)
https://dlang.org/phobos/std_range_interfaces.html
auto foo(int[] ints) {
import std.range;
if (ints.length > 10) {
return
cast(RandomAccessFinite!int)inputRangeObject(chain(ints[0..5], ints[8..$]));
} else {
return cast(RandomAccessFinite!int)inputRangeObject(ints);
}
}
void main() {
import std.stdio;
import std.range;
import std.algorithm;
writeln(foo([1, 2, 3]));
writeln(foo(iota(20).array));
}
Ali
Re: Working with ranges: mismatched function return type inference
On Tuesday, 11 October 2016 at 15:46:20 UTC, orip wrote:
On Tuesday, 11 October 2016 at 13:06:37 UTC, pineapple wrote:
Rewrite `return chain(ints[0..5], ints[8..$]);` as `return
ints[0..5] ~ ints[8..$];`
The `chain` function doesn't return an array, it returns a
lazily-evaluated sequence of an entirely different type from
`int[]`.
Of course it does! I would like the function to return an
"input range of int", no matter which one specifically. Is this
possible?
It is, but you will have to use an interface / class to achieve
this behavior (or use some sort of polymorphic struct). Something
like this will do the trick:
import std.range;
import std.stdio;
interface IInputRange(T)
{
bool empty();
T front();
void popFront();
}
final class InputRange(Range) if(isInputRange!Range)
: IInputRange!(ElementType!Range)
{
Range r;
this(Range r)
{
this.r = r;
}
bool empty() { return r.empty; }
ElementType!Range front() { return r.front; }
void popFront() { r.popFront; }
}
auto inputRange(Range)(Range r)
{
return new InputRange!Range(r);
}
IInputRange!int foo(int[] ints)
{
import std.range;
if(ints.length > 10) {
return inputRange(chain(ints[0 .. 5], ints[8 .. $]));
} else {
return inputRange(ints);
}
}
void main()
{
auto ir = foo([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]);
auto ir2 = foo([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]);
writeln(ir);
writeln(ir2);
}
Re: Working with ranges: mismatched function return type inference
11.10.2016 18:46, orip пишет: On Tuesday, 11 October 2016 at 13:06:37 UTC, pineapple wrote: Rewrite `return chain(ints[0..5], ints[8..$]);` as `return ints[0..5] ~ ints[8..$];` The `chain` function doesn't return an array, it returns a lazily-evaluated sequence of an entirely different type from `int[]`. Of course it does! I would like the function to return an "input range of int", no matter which one specifically. Is this possible? it doesn't. Using runtime argument you can't choose compile time parameter - returned type. So it's impossible. Almost - b/c you can use Algebraic. Again you can do the following: ```D return chain(ints[0..5], ints[8..$]).array; // it returns int[] ```
Re: Working with ranges: mismatched function return type inference
On Tuesday, 11 October 2016 at 13:06:37 UTC, pineapple wrote: Rewrite `return chain(ints[0..5], ints[8..$]);` as `return ints[0..5] ~ ints[8..$];` The `chain` function doesn't return an array, it returns a lazily-evaluated sequence of an entirely different type from `int[]`. Of course it does! I would like the function to return an "input range of int", no matter which one specifically. Is this possible?
Re: Working with ranges: mismatched function return type inference
On Tuesday, 11 October 2016 at 07:55:36 UTC, orip wrote:
I get "Error: mismatched function return type inference" errors
with choosing the return type for functions that work on ranges
using, e.g, std.algorithm or std.range functions, but have
different behavior based on runtime values. The return type is
always a range with the same underlying type.
Here's an example:
auto foo(int[] ints) {
import std.range;
if (ints.length > 10) {
return chain(ints[0..5], ints[8..$]);
} else {
//return ints; // Error: mismatched function return type
inference of int[] and Result
return chain(ints[0..0], ints[0..$]); // This workaround
compiles
}
}
Is there a compatible return type that can be used, or some
other workaround?
I couldn't find one when searching for the error or looking at
the phobos source code.
Thanks! orip
Rewrite `return chain(ints[0..5], ints[8..$]);` as `return
ints[0..5] ~ ints[8..$];`
The `chain` function doesn't return an array, it returns a
lazily-evaluated sequence of an entirely different type from
`int[]`.
Working with ranges: mismatched function return type inference
I get "Error: mismatched function return type inference" errors
with choosing the return type for functions that work on ranges
using, e.g, std.algorithm or std.range functions, but have
different behavior based on runtime values. The return type is
always a range with the same underlying type.
Here's an example:
auto foo(int[] ints) {
import std.range;
if (ints.length > 10) {
return chain(ints[0..5], ints[8..$]);
} else {
//return ints; // Error: mismatched function return type
inference of int[] and Result
return chain(ints[0..0], ints[0..$]); // This workaround
compiles
}
}
Is there a compatible return type that can be used, or some other
workaround?
I couldn't find one when searching for the error or looking at
the phobos source code.
Thanks! orip
