Why is block designed to be allocated on the stack unless copied? - ios

Why has Apple designed block to be allocated on the stack unless copied? What's the benefit for such behavior? Why not just make it behave like a regular NSObject - alloc-init it and it goes on the heap automatically?

The reason why blocks are placed on the stack by default is speed. In
the common case where the lifetime of the block is less than that of
the stack function that contains it, this is a very good optimisation.
http://www.cocoawithlove.com/2009/10/how-blocks-are-implemented-and.html

Related

Does dart reuse memory for previously used instances?

It is hard to find a good heading for this, but i think my problem comes clear if i post a small code snipped:
SomeObject instance = SomeObject(importantParameter);
// -> "instance" is now a reference to the instance somewhere in RAM
instance = SomeObject(anotherImportantParameter);
// -> "instance" is now a reference to a different instance somewhere in RAM
My question is now, is the used RAM that was allocated at the first construction reused at the second construction? Or is the RAM of the first instance marked as unused for the garbage collector and the second construction is done with a completely new instance with a different portion of RAM?
If the first is true, what with this:
while(true) {
final SomeObject instance = SomeObject(importantParameter);
}
Will then, each time the while is repeated, the RAM be reused?
It's unspecified. The answer is a resounding "maybe".
The language specification never says what happens to unreachable objects, since it's unobservable to the program. (That's what being unreachable means).
In practice, the native Dart implementation uses a generational garbage collector.
The default behavior would be to allocate a new object in "new-space" and overwrite the reference to the previous object. That makes the previous object unreachable (as long as you haven't store other references to it), and it can therefore be garbage collected. If you really go through objects quickly, that will be cheap, since the unreachable object is completely ignored on the next new-space garbage collection.
Allocating a lot of short-lived objects still has an overhead since it causes new-space GC to happen more often, even if the individual objects don't themselves cost anything.
There is also a number of optimization that may change this behavior.
If your object is sufficiently simple and the compiler can see that no reference to it ever escapes, or is used in an identical check, or ... any other number of relevant restrictions, then it might "allocation sink" the object. That means it never actually allocates the object, it just stores the contents somewhere, perhaps even on the stack, and it also inlines the methods so they refer to the data directly instead of going through a this pointer.
In that case, your code may actually reuse the memory of the previous object, because the compiler recognizes that it can.
Do not try to predict whether an optimization like this happens. The requirements can change at any time. Just write code that is correct and not unnecessarily complex, then the compiler will do its best to optimize in all the ways that it can.

Is a block in Objective-C always guaranteed to capture a variable?

Are there any conditions in Objective-C (Objective-C++) where the compiler can detect that a variable capture in a block is never used and thus decide to not capture the variable in the first place?
For example, assume you have an NSArray that contains a large number of items which might take a long time to deallocate. You need to access the NSArray on the main thread, but once you're done with it, you're willing to deallocate it on a background queue. The background block only needs to capture the array and then immediately deallocate. It doesn't actually have to do anything with it. Can the compiler detect this and, "erroneously", skip the block capture altogether?
Example:
// On the main thread...
NSArray *outgoingRecords = self.records;
self.records = incomingRecords;
dispatch_async(background_queue, ^{
(void)outgoingRecords;
// After this do-nothing block exits, then outgoingRecords
// should be deallocated on this background_queue.
});
Am I guaranteed that outgoingRecords will always be captured in that block and that it will always be deallocated on the background_queue?
Edit #1
I'll add a bit more context to better illustrate my issue:
I have an Objective-C++ class that contains a very large std::vector of immutable records. This could easily be 1+ million records. They are basic structs in a vector and accessed on the main thread to populate a table view. On a background thread, a different set of database records might be read into a separate vector, which could also be quite large.
Once the background read has occurred, I jump over to the main thread to swap Objective-C objects and repopulate the table.
At that point, I don't care at all about the contents of the older vector or its parent Objective-C class. There's no fancy destructors or object-graph to teardown, but deallocating hundreds of megabytes, maybe even gigabytes of memory is not instantaneous. So I'm willing to punt it off to a background_queue and have the memory deallocation occur there. In my tests, that appears to work fine and gives me a little bit more time on the main thread to do other stuff before 16ms elapses.
I'm trying to understand if I can get away with simply capturing the object in an "empty" block or if I should do some sort of no-op operation (like call count) so that the compiler cannot optimize it away somehow.
Edit #2
(I originally tried to keep the question as simple as possible, but it seems like it's more nuanced then that. Based on Ken's answer below, I'll add another scenario.)
Here's another scenario that doesn't use dispatch_queues but still uses blocks, which is the part I'm really interested in.
id<MTLCommandBuffer> commandBuffer = ...
// A custom class that manages an MTLTexture that is backed by an IOSurface.
__block MyTextureWrapper *wrapper = ...
// Issue some Metal calls that use the texture inside the wrapper.
// Wait for the buffer to complete, then release the wrapper.
[commandBuffer addCompletedHandler:^(id<MTLCommandBuffer> cb) {
wrapper = nil;
}];
In this scenario, the order of execution is guaranteed by Metal. Unlike the example above, in this scenario performance is not the issue. Rather, the IOSurface that is backing the MTLTexture is being recycled into a CVPixelBufferPool. The IOSurface is being shared between processes and, from what I can tell, MTLTexture does not appear to increase the useCount on the surface. My wrapper class does. When my wrapper class is deallocated, the useCount is decremented and the bufferPool is then free to recycling the IOSurface.
This is all working as expected but I end up with silly code like above just out of uncertainty whether I need to "use" the wrapper instance in the block to ensure it's captured or not. If the wrapper is deallocated before the completion handler runs, then the IOSurface will be recycled and the texture will get overwritten.
Edit to address question edits:
From the Clang Language Specification for Blocks:
Local automatic (stack) variables referenced within the compound
statement of a Block are imported and captured by the Block as const
copies. The capture (binding) is performed at the time of the Block
literal expression evaluation.
The compiler is not required to capture a variable if it can prove
that no references to the variable will actually be evaluated.
Programmers can force a variable to be captured by referencing it in a
statement at the beginning of the Block, like so:
(void) foo;
This matters when capturing the variable has side-effects, as it can
in Objective-C or C++.
(Emphasis added.)
Note that using this technique guarantees that the referenced object lives at least as long as the block, but does not guarantee it will be released with the block, nor by which thread.
There's no guarantee that the block submitted to the background queue will be the last code to hold a strong reference to the array (even ignoring the question of whether the block captures the variable).
First, the block may in fact run before the context which submitted it returns and releases its strong reference. That is, the code which called dispatch_async() could be swapped off the CPU and the block could run first.
But even if the block runs somewhat later than that, a reference to the array may be in an autorelease pool somewhere and not released for some time. Or there may be a strong reference someplace else that will eventually be cleared but not under you explicit control.

Alternative for ReAllocMem that zeros the new allocation

I need to ReAllocMem a block, but the extension should be zero'ed.
For this reason ReAllocMem is no good:
From the helpfile:
ReallocMem reallocates a memory block.
[...]
The content of the newly allocated memory is not set to zero.
I've looked at ReAllocMemory, but the help does not state anything about zeroing the new allocation, it only states:
Note: ReallocMemory is the version of ReallocMem compatible with C++.
Is there an alternative that does zero the newly allocated memory?
The simple answer is no. The only raw memory reallocator is ReallocMem. There is nothing else. The design of a re-allocator is always to preserve the contents that were there before. You will have to write your own routine.
David is right but using WinAp it's possible: you can use Global Alloc and GlobalReAlloc using the GMEM_ZEROINIT flag.

Keeping a strong pointer to a local object

I encountered this thing in a book which I am reading and it got me thinking:
"When you allocate a block, it is created on the stack. This means that, even if you were to keep a strong reference to it, calling it later would result in a crash because the memory would be destroyed as soon as you leave the method in which it was defined."
I thought if I have a strong pointer to something, it is kept alive?
Does this mean this does not apply for objects allocated on the stack?
I am trying to think of an example without using blocks...(e.g., of pointer - maybe an ivar- pointing to a stack allocated object which gets destroyed even though the pointer is alive)
Objects are never allocated on the stack in Objective-C. Blocks are special however, since they are stack allocated. So if you want to retain a pointer to a block, you must first copy it by using Block_copy and use the copy, then release it with Block_release. This must be done if the block is to be used after the scope it was declared in is destroyed. More on the matter here: https://developer.apple.com/library/mac/documentation/cocoa/Conceptual/Blocks/Articles/bxUsing.html (under "Copying Blocks"). Yet again though, this does not apply to regular objects.
Blocks can be messaged like objects. To move them from the stack to the heap, just "copy" them.
void (^stackBlock)() = [^(){
NSLog(#"Hello world");
} copy];

Defining Objective-C blocks as properties - best practice

I've recently come across an Apple document that shows the following property declaration for a block:
#interface XYZObject : NSObject
#property (copy) void (^blockProperty)(void);
#end
Also, this article states:
Note: You should specify copy as the property attribute, because a block needs to be copied to keep track of its captured state outside of the original scope. This isn’t something you need to worry about when using Automatic Reference Counting, as it will happen automatically, but it’s best practice for the property attribute to show the resultant behavior. For more information, see Blocks Programming Topics.
I also read the suggested Blocks Programming Topics but haven't found anything relevant there.
I'm still curious as to why defining a block property as "copy" is best practice. If you have a good answer, please try to distinguish between ARC and MRC differences if there are any.
Thank you
By default blocks are created on the stack. Meaning they only exist in the scope they have been created in.
In case you want to access them later they have to be copied to the heap by sending a copy message to the block object. ARC will do this for you as soon as it detects a block needs to be accessed outside the scope its created in. As a best practise you declare any block property as copy because that's the way it should be under automatic memory management.
Read Stack and Heap Objects in Objective-C by Mike Ash for more info on stack vs. heap.
Blocks are, by default, allocated on the stack. This is an optimization, since stack allocation is much cheaper than heap allocation. Stack allocation means that, by default again, a block will cease to exist when the scope in which it is declared exits. So a block property with retain semantics will result in a dangling pointer to a block that doesn't exist anymore.
To move a block from the stack to the heap (and thus give it normal Objective-C memory management semantics and an extended lifetime), you must copy the block via [theBlock copy], Block_copy(theBlock), etc. Once on the heap, the block's lifetime can be managed as needed by retaining/releasing it. (Yes, this applies in ARC too, you just don't have to call -retain/-release yourself.)
So you want to declare block properties with copy semantics so the block is copied when the property is set, avoiding a dangling pointer to a stack-based block.
The "best practices" you refer to simply say, "Seeing as ARC is going to magically copy your block no matter what you write here, it's best you explicitly write 'copy' so as not to confuse future generations looking at your code."
Explanation follows:
Typically, you shouldn’t need to copy (or retain) a block. You only need to make a copy when you expect the block to be used after destruction of the scope within which it was declared. Copying moves a block to the heap.
–Blocks Programming Topics: Using Blocks, Copying Blocks
Clearly, assigning a block to a property means it could be used after the scope it was declared in has been destroyed. Thus, according to Blocks Programming Topics that block should be copied to the heap with Block_copy.
But ARC takes care of this for you:
Blocks “just work” when you pass blocks up the stack in ARC mode, such as in a return. You don’t have to call Block Copy any more.
–Transitioning to ARC
Note that this isn't about the retain semantics the block. There's simply no way for the block's context to exist without being moved off the (soon-to-be-popped) stack and on to the heap. So regardless of what attributes you qualify your #property with, ARC is still going to copy the block.

Resources