I have a continuation chain using lambda expressions where one task assigns to a variable and the next task reads from that variable. Microsoft suggests using a shared_ptr to wrap the variable even when the variable is a reference-counted handle (^). Wouldn't a reference-counted handle increment its reference count when captured by value by the lambda expression? Why then is it necessary to wrap a reference-counted handle with a shared_ptr?
The documentation makes it clear that the cases they are concerned about are ones where
one task in a continuation chain assigns to a variable, and another task reads that variable
(Emphasis is mine.) This is not a question of object lifetime, but rather a question of object identity.
Take this example from the Hilo project, paying close attention to the decoder variable (which is a shared_ptr<BitmapDecoder^>):
task<InMemoryRandomAccessStream^> ThumbnailGenerator::CreateThumbnailFromPictureFileAsync(
StorageFile^ sourceFile,
unsigned int thumbSize)
{
(void)thumbSize; // Unused parameter
auto decoder = make_shared<BitmapDecoder^>(nullptr);
auto pixelProvider = make_shared<PixelDataProvider^>(nullptr);
auto resizedImageStream = ref new InMemoryRandomAccessStream();
auto createThumbnail = create_task(
sourceFile->GetThumbnailAsync(
ThumbnailMode::PicturesView,
ThumbnailSize));
return createThumbnail.then([](StorageItemThumbnail^ thumbnail)
{
IRandomAccessStream^ imageFileStream =
static_cast<IRandomAccessStream^>(thumbnail);
return BitmapDecoder::CreateAsync(imageFileStream);
}).then([decoder](BitmapDecoder^ createdDecoder)
{
(*decoder) = createdDecoder;
return createdDecoder->GetPixelDataAsync(
BitmapPixelFormat::Rgba8,
BitmapAlphaMode::Straight,
ref new BitmapTransform(),
ExifOrientationMode::IgnoreExifOrientation,
ColorManagementMode::ColorManageToSRgb);
}).then([pixelProvider, resizedImageStream](PixelDataProvider^ provider)
{
(*pixelProvider) = provider;
return BitmapEncoder::CreateAsync(
BitmapEncoder::JpegEncoderId,
resizedImageStream);
}).then([pixelProvider, decoder](BitmapEncoder^ createdEncoder)
{
createdEncoder->SetPixelData(BitmapPixelFormat::Rgba8,
BitmapAlphaMode::Straight,
(*decoder)->PixelWidth,
(*decoder)->PixelHeight,
(*decoder)->DpiX,
(*decoder)->DpiY,
(*pixelProvider)->DetachPixelData());
return createdEncoder->FlushAsync();
}).then([resizedImageStream]
{
resizedImageStream->Seek(0);
return resizedImageStream;
});
}
The decoder variable is first defined outside of the continuations, since it is needed in multiple continuations. At that point, its value is null. It is obtained and set within the second continuation, and properties of that object (PixelWidth etc) are used within the fourth continuation.
Were you to instead have decoder be defined as a BitmapDecoder^, set it to nullptr, and then assign it a value within the second continuation, that change would not propagate to subsequent continuations because the change cannot reflected back to the initial handle (the lambda has made a copy of the handle, essentially copying the memory address 0x00000000).
In order to update the original version (and subsequent references), you would need an additional indirection (e.g. a BitmapDecoder^*). A shared_ptr<BitmapDecoder^> is one such indirection, and a useful one in that you don't need to manage the lifetime of the pointer unlike with a raw pointer, which is why it is recommended in documentation.
There are other cases where capturing an Object^ would be sufficient, for example if I created a TextBlock^ outside of my continuation and set some properties of it in the first continuation and read some other properties in a subsequent continuation. In this case, all of the handles are referring to the same underlying object and no continuation is attempting to overwrite the identity of the object itself. (However, as initially mentioned, this is not the use case the documentation is referring to.)
Related
In a SystemVerilog function, is it legal to do multiple assignments to the implicitly-declared return variable? See the following function for an example:
localparam int Q=1,I=0;
function logic [1:0][Q:I][15:0] Cast24to16(input logic [1:0][Q:I][23:0] din);
foreach (Cast24to16[n,iq])
Cast24to16[n][iq] = din[n][iq][23 -: 8];
endfunction
The language reference manual, IEEE Std 1800-2017, sec 13.4.1 states:
Function return values can be specified in two ways, either by using a return statement or by assigning a value to the internal variable with the same name as the function.
This seems a little unclear as to whether you can assign multiple times, like in my example. Furthermore, the example from the LRM directly after this statement and also all other examples I can find online all show the implicit return value only being assigned once. This makes me feel a bit unsettled.
The LRM also says just before the section you quoted
The function definition shall implicitly declare a variable, internal to the function, with the same name as the function.
I think you can safely assume that if there is no explicit return statement, it effectively inserts an implicit return (var_function_name);
Also, if you declare your function with a static lifetime (which is the implicit default lifetime in a module), that implicit return variable has a static lifetime as well. That means it retains its value from the last time you called the function regardless of whether you assign it or not.
module top;
function int countme;
countme++;
endfunction
initial repeat (10) $display(countme());
endmodule
I have been looking here about CSimpleStringT::GetBufferSetLength.
It says:
If you use the pointer returned by GetBufferSetLength to change the
string contents, call ReleaseBuffer to update the internal state of
CSimpleStringT before you use any other CSimpleStringT methods.
The address returned by GetBufferSetLength may not be valid after the
call to ReleaseBuffer because additional CSimpleStringT operations can
cause the CSimpleStringT buffer to be reallocated. The buffer is not
reassigned if you do not change the length of the CSimpleStringT.
The buffer memory is automatically freed when the CSimpleStringT
object is destroyed.
I am actually using CString but given this code:
CString strHostLabel = dlgLabels.GetTrimmedHostLabel();
CString strCoHostLabel = dlgLabels.GetTrimmedCoHostLabel();
MENUITEMINFO sInfo{};
sInfo.cbSize = sizeof(MENUITEMINFO);
sInfo.fMask = MIIM_STRING;
sInfo.cch = strHostLabel.GetLength();
sInfo.dwTypeData = strHostLabel.GetBufferSetLength(sInfo.cch);
SetMenuItemInfo(pMnuSwap->GetSafeHmenu(), SwapAssignment::Host, TRUE, &sInfo);
sInfo.cch = strCoHostLabel.GetLength();
sInfo.dwTypeData = strCoHostLabel.GetBufferSetLength(sInfo.cch);
SetMenuItemInfo(pMnuSwap->GetSafeHmenu(), SwapAssignment::Cohost, TRUE, &sInfo);
strHostLabel.ReleaseBuffer();
strCoHostLabel.ReleaseBuffer();
Since I do not modify the underlying data it seems I do not need to call ReleaseBuffer. Correct?
Short answer: No.
The longer answer is: Don't call GetBuffer/GetBufferSetLength either. Call GetString instead. It will return a pointer to the immutable sequence of characters. This requires a const_cast as well, i.e.
sInfo.dwTypeData = const_cast<TCHAR*>(strHostLabel.GetString());
This is due to the fact that MENUITEMINFO is used both for setting menu items, as well as reading information back. Since C doesn't allow you to specify transitive constness, the structure is forced to use non-const data members, even when the pointed-to data is never changed. The const_cast is both required as well as safe.
Also note that setting the cch member is not required. As the documentation explains:
[...] cch is ignored when the content of a menu item is set by calling SetMenuItemInfo.
How can I know if I actually need to return an l-value when using FALLBACK?
I'm using return-rw but I'd like to only use return where possible. I want to track if I've actually modified %!attrs or have only just read the value when FALLBACK was called.
Or (alternate plan B) can I attach a callback or something similar to my %!attrs to monitor for changes?
class Foo {
has %.attrs;
submethod BUILD { %!attrs{'bar'} = 'bar' }
# multi method FALLBACK(Str:D $name, *#rest) {
# say 'read-only';
# return %!attrs{$name} if %!attrs«$name»:exists;
# }
multi method FALLBACK(Str:D $name, *#rest) {
say 'read-write';
return-rw %!attrs{$name} if %!attrs«$name»:exists;
}
}
my $foo = Foo.new;
say $foo.bar;
$foo.bar = 'baz';
say $foo.bar;
This feels a bit like a X-Y question, so let's simplify the example, and see if that answers helps in your decisions.
First of all: if you return the "value" of a non-existing key in a hash, you are in fact returning a container that will auto-vivify the key in the hash when assigned to:
my %hash;
sub get($key) { return-rw %hash{$key} }
get("foo") = 42;
dd %hash; # Hash %hash = {:foo(42)}
Please note that you need to use return-rw here to ensure the actual container is returned, rather than just the value in the container. Alternately, you can use the is raw trait, which allows you to just set the last value:
my %hash;
sub get($key) is raw { %hash{$key} }
get("foo") = 42;
dd %hash; # Hash %hash = {:foo(42)}
Note that you should not use return in that case, as that will still de-containerize again.
To get back to your question:
I want to track if I've actually modified %!attrs or have only just read the value when FALLBACK was called.
class Foo {
has %!attrs;
has %!unexpected;
method TWEAK() { %!attrs<bar> = 'bar' }
method FALLBACK(Str:D $name, *#rest) is raw {
if %!attrs{$name}:exists {
%!attrs{$name}
}
else {
%!unexpected{$name}++;
Any
}
}
}
This would either return the container found in the hash, or record the access to the unknown key and return an immutable Any.
Regarding plan B, recording changes: for that you could use a Proxy object for that.
Hope this helps in your quest.
Liz's answer is full of useful info and you've accepted it but I thought the following might still be of interest.
How to know if returning an l-value ... ?
Let's start by ignoring the FALLBACK clause.
You would have to test the value. To deal with Scalars, you must test the .VAR of the value. (For non-Scalar values the .VAR acts like a "no op".) I think (but don't quote me) that Scalar|Array|Hash covers all the l-value super-types:
my \value = 42; # Int is an l-value is False
my \l-value-one = $; # Scalar is an l-value is True
my \l-value-too = #; # Array is an l-value is True
say "{.VAR.^name} is an l-value is {.VAR ~~ Scalar|Array|Hash}"
for value, l-value-one, l-value-too
How to know if returning an l-value when using FALLBACK?
Adding "when using FALLBACK" makes no difference to the answer.
How can I know if I actually need to return an l-value ... ?
Again, let's start by ignoring the FALLBACK clause.
This is a completely different question than "How to know if returning an l-value ... ?". I think it's the core of your question.
Afaik, the answer is, you need to anticipate how the returned value will be used. If there's any chance it'll be used as an l-value, and you want that usage to work, then you need to return an l-value. The language/compiler can't (or at least doesn't) help you make that decision.
Consider some related scenarios:
my $baz := foo.bar;
... (100s of lines of code) ...
$baz = 42;
Unless the first line returns an l-value, the second line will fail.
But the situation is actually much more immediate than that:
routine-foo = 42;
routine-foo is evaluated first, in its entirety, before the lhs = rhs expression is evaluated.
Unless the compiler's resolution of the routine-foo call somehow incorporated the fact that the very next thing to happen would be that the lhs will be assigned to, then there would be no way for a singly or multiply dispatched routine-foo to know whether it can safely return an r-value or must return an l-value.
And the compiler's resolution does not incorporate that. Thus, for example:
multi term:<bar> is rw { ... }
multi term:<bar> { ... }
bar = 99; # Ambiguous call to 'term:<bar>(...)'
I can imagine this one day (N years from now) being solved by a combination of allowing = to be an overloadable operator, robust macros that allow overloading of = being available, and routine resolution being modified so the above ambiguous call could do something equivalent to resolving to the is rw multi. But I doubt it will actually come to pass even with N=10. Perhaps there is another way but I can't think of one at the moment.
How can I know if I actually need to return an l-value when using FALLBACK?
Again, adding "when using FALLBACK" makes no difference to the answer.
I want to track if I've actually modified %!attrs or have only just read the value when FALLBACK was called.
When FALLBACK is called it doesn't know what context it's being called in -- r-value or l-value. Any modification comes after it has already returned.
In other words, whatever solution you come up with will being nothing to do per se with FALLBACK (even if you have to use it to implement some other aspect of whatever it is you're trying to do).
(Even if it were, I suspect trying to solve it via FALLBACK itself would just make matters worse. One can imagine writing two FALLBACK multis, one with an is rw trait, but, as explained above, my imagination doesn't stretch to that making any difference any time soon, if ever, and could only happen if the above imaginary things happened (the macros etc.) and the compiler was also modified to pay attention to the two FALLBACK multi variants, and I'm not at all meaning to suggest that that even makes sense.)
Plan B
Or (alternate plan B) can I attach a callback or something similar to my %!attrs to monitor for changes?
As Lizmat notes, that's the realm of Proxys. And thus your next SO question... :)
Coming from a C# background, I would say that the ref keyword is very useful in certain situations where changes to a method parameter are desired to directly influence the passed value for value types of for setting a parameter to null.
Also, the out keyword can come in handy when returning a multitude of various logically unconnected values.
My question is: is it possible to pass a parameter to a function by reference in Haskell? If not, what is the direct alternative (if any)?
There is no difference between "pass-by-value" and "pass-by-reference" in languages like Haskell and ML, because it's not possible to assign to a variable in these languages. It's not possible to have "changes to a method parameter" in the first place in influence any passed variable.
It depends on context. Without any context, no, you can't (at least not in the way you mean). With context, you may very well be able to do this if you want. In particular, if you're working in IO or ST, you can use IORef or STRef respectively, as well as mutable arrays, vectors, hash tables, weak hash tables (IO only, I believe), etc. A function can take one or more of these and produce an action that (when executed) will modify the contents of those references.
Another sort of context, StateT, gives the illusion of a mutable "state" value implemented purely. You can use a compound state and pass around lenses into it, simulating references for certain purposes.
My question is: is it possible to pass a parameter to a function by reference in Haskell? If not, what is the direct alternative (if any)?
No, values in Haskell are immutable (well, the do notation can create some illusion of mutability, but it all happens inside a function and is an entirely different topic). If you want to change the value, you will have to return the changed value and let the caller deal with it. For instance, see the random number generating function next that returns the value and the updated RNG.
Also, the out keyword can come in handy when returning a multitude of various logically unconnected values.
Consequently, you can't have out either. If you want to return several entirely disconnected values (at which point you should probably think why are disconnected values being returned from a single function), return a tuple.
No, it's not possible, because Haskell variables are immutable, therefore, the creators of Haskell must have reasoned there's no point of passing a reference that cannot be changed.
consider a Haskell variable:
let x = 37
In order to change this, we need to make a temporary variable, and then set the first variable to the temporary variable (with modifications).
let tripleX = x * 3
let x = tripleX
If Haskell had pass by reference, could we do this?
The answer is no.
Suppose we tried:
tripleVar :: Int -> IO()
tripleVar var = do
let times_3 = var * 3
let var = times_3
The problem with this code is the last line; Although we can imagine the variable being passed by reference, the new variable isn't.
In other words, we're introducing a new local variable with the same name;
Take a look again at the last line:
let var = times_3
Haskell doesn't know that we want to "change" a global variable; since we can't reassign it, we are creating a new variable with the same name on the local scope, thus not changing the reference. :-(
tripleVar :: Int -> IO()
tripleVar var = do
let tripleVar = var
let var = tripleVar * 3
return()
main = do
let x = 4
tripleVar x
print x -- 4 :(
Could you please explain differences between and definition of call by value, call by reference, call by name and call by need?
Call by value
Call-by-value evaluation is the most common evaluation strategy, used in languages as different as C and Scheme. In call-by-value, the argument expression is evaluated, and the resulting value is bound to the corresponding variable in the function (frequently by copying the value into a new memory region). If the function or procedure is able to assign values to its parameters, only its local copy is assigned — that is, anything passed into a function call is unchanged in the caller's scope when the function returns.
Call by reference
In call-by-reference evaluation (also referred to as pass-by-reference), a function receives an implicit reference to a variable used as argument, rather than a copy of its value. This typically means that the function can modify (i.e. assign to) the variable used as argument—something that will be seen by its caller. Call-by-reference can therefore be used to provide an additional channel of communication between the called function and the calling function. A call-by-reference language makes it more difficult for a programmer to track the effects of a function call, and may introduce subtle bugs.
differences
call by value example
If data is passed by value, the data is copied from the variable used in for example main() to a variable used by the function. So if the data passed (that is stored in the function variable) is modified inside the function, the value is only changed in the variable used inside the function. Let’s take a look at a call by value example:
#include <stdio.h>
void call_by_value(int x) {
printf("Inside call_by_value x = %d before adding 10.\n", x);
x += 10;
printf("Inside call_by_value x = %d after adding 10.\n", x);
}
int main() {
int a=10;
printf("a = %d before function call_by_value.\n", a);
call_by_value(a);
printf("a = %d after function call_by_value.\n", a);
return 0;
}
The output of this call by value code example will look like this:
a = 10 before function call_by_value.
Inside call_by_value x = 10 before adding 10.
Inside call_by_value x = 20 after adding 10.
a = 10 after function call_by_value.
call by reference example
If data is passed by reference, a pointer to the data is copied instead of the actual variable as is done in a call by value. Because a pointer is copied, if the value at that pointers address is changed in the function, the value is also changed in main(). Let’s take a look at a code example:
#include <stdio.h>
void call_by_reference(int *y) {
printf("Inside call_by_reference y = %d before adding 10.\n", *y);
(*y) += 10;
printf("Inside call_by_reference y = %d after adding 10.\n", *y);
}
int main() {
int b=10;
printf("b = %d before function call_by_reference.\n", b);
call_by_reference(&b);
printf("b = %d after function call_by_reference.\n", b);
return 0;
}
The output of this call by reference source code example will look like this:
b = 10 before function call_by_reference.
Inside call_by_reference y = 10 before adding 10.
Inside call_by_reference y = 20 after adding 10.
b = 20 after function call_by_reference.
when to use which
One advantage of the call by reference method is that it is using pointers, so there is no doubling of the memory used by the variables (as with the copy of the call by value method). This is of course great, lowering the memory footprint is always a good thing. So why don’t we just make all the parameters call by reference?
There are two reasons why this is not a good idea and that you (the programmer) need to choose between call by value and call by reference. The reason are: side effects and privacy. Unwanted side effects are usually caused by inadvertently changes that are made to a call by reference parameter. Also in most cases you want the data to be private and that someone calling a function only be able to change if you want it. So it is better to use a call by value by default and only use call by reference if data changes are expected.
call by name
In call-by-name evaluation, the arguments to a function are not evaluated before the function is called — rather, they are substituted directly into the function body (using capture-avoiding substitution) and then left to be evaluated whenever they appear in the function.
call by need
Lazy evaluation, or call-by-need is an evaluation strategy which delays the evaluation of an expression until its value is needed (non-strict evaluation) and which also avoids repeated evaluations