I have a situation where I need to assign a unique ptr value from within a lambda function.
std::unique_ptr<SomeType> unique_ptr_obj;
// Lambda below has fixed return type.
bool var = ()[unique_ptr_obj=std::move(unique_ptr_obj)]-> bool {
unique_ptr_obj = GetUniqueObject();
return true
} ();
// Should be able to use unique_ptr_obj
UseUniqueObject(unique_ptr_obj.get());
However, as expected unique_ptr_obj is nullptr as it was moved into lambda. Is there a way I can populate unique_ptr_obj from within a lambda and be able to reuse it later ?
Any suggestions on how to accomplish this ? Should I convert unique_ptr_obj to shared_ptr ?
You should change the declaration of your lambda to capture unique_ptr_obj by reference:
bool var = [&unique_ptr_obj]() -> bool {
// Whatever the next line does, now it changes that variable by reference.
// Otherwise you were changing a local copy.
unique_ptr_obj = GetUniqueObject();
return true;
} ();
You don't want to share ownership. Or maybe you do, but it won't help with the lambda assigning something to unique_ptr_obj, hence using a shared_ptr is not the solution.
You also do not want to move from unique_ptr_obj. Sloppy speaking, moving from something means leaving it in an empty state.
If you want a function to modify its argument then you pass by reference. If you want a lambda to modify something in the outer scope you let it capture it by reference.
This is the same whether its an int or a unique_ptr:
int x = 0;
bool value = [&x]() { x = 42; return true; } ();
// ^^ capture x by reference
assert(x == 42);
Related
I want to return a boolean or success/failure enum from the function and modify an argument by reference. However, I want to construct a reference in the calling function instead of copying the value.
I have some container (say 'example_q' of type std::queue). queue.front() will return a reference to the value stored in the queue. I can make a copy of that reference (example A) or I can take a reference of that reference (example B), allowing the value to stay in the queue but be utilized outside of it.
A)
int a = example_q.front();
B)
int& b = example_q.front();
Using this difference I could also return the queued value:
A)
int get_front()
{
int a = example_q.front();
return a;
}
B)
int& get_front()
{
return example_q.front();
}
Using option 'B' I can avoid unnecessary copies without moving the data out of the queue via std::move() semantics.
My question is, can I do 'B' via an argument passed by reference? Would I need to use std::move()/rvalues/&& somehow?
void get_front(int& int_ref)
{
// somehow don't copy the value into referenced int_ref, but construct
// a reference in the caller based on an input argument?
int_ref = example_q.front();
}
The problem this would solve is making API match other functions that modify reference arguments but return a success/failure value, ie:
if(q.get_front(referrence_magic_here))
{
...
}
I could reverse the order to get the desired result, IE:
int& get_front(bool& success)
{
...
}
But I'd rather keep the pattern of my API as well as being able to do it via a single line in the if() statement if possible.
Perhaps something like:
bool get_front(int&& int_rvalue)
{
...
int_rvalue = example_q.front();
...
return true_or_false;
}
void calling_func()
{
...
if(get_front(int& magical_ref))
{
... //use magical_ref here?
}
...
}
No, you can't do that.
Other than in its initialiser, a reference behaves like the thing it refers to. By passing it as a function argument, you "hide" the initialiser from the part that wants to do the assignment. So, the function has no access to the referencey behaviour of the thing.
You will have to use pointers if you want to do that:
void get_front(int*& int_ptr)
{
int_ptr = &example_q.front();
}
int* ptr = nullptr;
get_front(ptr);
// optional:
int& ref = *ptr;
(Ew!)
Option B was fine.
This code is invalid C++:
if(get_front(int& magical_ref))
You cannot declare a new variable as you're passing it to a function. And because a reference variable must be declared and initialized at the same time, it wouldn't be possible to have a reference be initialized by passing it to a function.
You could however, do this:
if(int &magical_ref = get_front()) {
But note that you'd be checking whether magical_ref is 0 or not, which is different from the condition you have in your example.
If your logic is as simple as comparing the int, you could do:
if (int& magical_ref = get_front(); magical_ref == 42)
You can return a std::tuple<int&, /* status condition */> and check the status. For example:
std::tuple<int&, bool> get_front() {
static int example = 0;
return {example, false};
}
...
// C++17's structured bindings + if statement with initializer
if (auto [ref, success] = get_front(); success) {
ref = 42;
}
Demo
I'm not sure if anyone else ever makes a temporary with a shorter name when it'll be used in many lines of code, for example you need to access something a couple of classes deep and instead of chaining member access operators over and over you'll just use a shorter named temporary. I tried doing something like the following:
struct Car
{
struct
{
struct
{
int height, width, depth;
} physicalInfo;
} information;
} objectWithARatherLongName ;
int main()
{
auto lambda1 = [] { return objectWithARatherLongName.information.physicalInfo.height; };
// Create a temporary for a shorter name
auto& dimens = objectWithARatherLongName.information.physicalInfo;
auto lambda2 = [=] { return dimens.height; };
// Have to capture "dimens" because it's a local variable
// And over and over again, the shorter way is preferred.
}
In this case auto dimens is the short named temporary, but since I want to use it in the lambda I need to capture it, which made me made think I should just use the full name that is a global. Is there a way to shorten this in the way I described but still be using the global variable? I thought about a typedef that will only work with types right? Not with actual objects.
Don't use global variables, but if you must you probably can add one more:
struct Car
{
...
} objectWithARatherLongName ;
auto& dimens = objectWithARatherLongName.information.physicalInfo;
int main() { // ...
What is the idiomatic C++ way of doing this?
I have a method which looks like this:
LargeObject& lookupLargeObject(int id) {
return largeObjects[id];
}
This is wrong, because if you call this with a non-existent id it will create a new instance of large object and put it into the container. I don't want that. I don't want to throw an exception either. I want the return value to signal that object wasn't found (as it is a more or less normal situation).
So my options are either a pointer or an optional. Pointer I understand and like, but it feels like C++ doesn't want to me use pointers any more.
So on to optionals. I will return an optional and then the caller looks like this:
std::optional<LargeObject> oresult = lookupLargeObject(42);
LargeObject result;
if (oresult) {
result = *oresult;
} else {
// deal with it
}
Is this correct? It feels kind of crappy because it seems that I'm creating 2 copies of the LargeObject here? Once when returning the optional and once when extracting it from optional into result. Gotta be a better way?
Since you don't want to return a pointer, but also don't want to throw an exception, and you presumably want reference semantics, the easiest thing to do is to return a std::optional<std::reference_wrapper<LargeObject>>.
The code would look like this:
std::optional<std::reference_wrapper<LargeObject>> lookupLargeObject(int id) {
auto iter = largeObjects.find(id);
if (iter == largeObjects.end()) {
return std::nullopt;
} else {
return std::ref(iter->second);
}
}
With C++17 you can even declare the iter variable inside the if-condition.
Calling the lookup function and using the reference then looks like this (here with variable declaration inside if-condition):
if (auto const lookup_result = lookupLargeObject(42); lookup_result) {
auto& large_object = lookup_result.value().get();
// do something with large_obj
} else {
// deal with it
}
There are two approaches that do not require use of pointers - using a sentinel object, and receiving a reference, instead of returning it.
The first approach relies on designating a special instance of LargeObject an "invalid" one - say, by making a member function called isValid, and returning false for that object. lookupLargeObject would return that object to indicate that the real object was not found:
LargeObject& lookupLargeObject(int id) {
if (largeObjects.find(id) == largeObjects.end()) {
static LargeObject notFound(false);
return notFound;
}
return largeObjects[id];
}
The second approach passes a reference, rather than receiving it back:
bool lookupLargeObject(int id, LargeObject& res) {
if (largeObjects.find(id) == largeObjects.end()) {
return false;
}
res = largeObjects[id];
return true;
}
If default constructed LargeObject is unwanted from lookupLargeObject, regardless of whether it is expensive or it does not make semantic sense, you can use the std:map::at member function.
LargeObject& lookupLargeObject(int id) {
return largeObjects.at(id);
}
If you are willing to live with use of if-else blocks of code in the calling function, I would change the return type of the function to LargeObject*.
LargeObject* lookupLargeObject(int id) {
auto it = largeObjects.find(id);
if ( it == largeObjects.end() )
{
return nullptr;
}
return &(it->second);
}
Then, client code can be:
LargeObject* result = lookupLargeObject(42);
if (result) {
// Use result
} else {
// deal with it
}
In my project there is a vector
std::vector<std::shared_ptr<MovingEntity>>gameObjects;
Which I want to delete elements from if they meet the criteria.
Method to delete elements:
void GameWorld::catchBees()
{
auto q = std::remove_if(bees.begin(), bees.end(), beeToClose);
bees.erase(q);
}
Method beeToClose:
bool GameWorld::beeToClose( const MovingEntity & bee)
{
std::shared_ptr<Beekeeper> keeper = std::static_pointer_cast<Beekeeper>(m_beekeeper);
if (bee.getConstPosition().distanceTo(m_beekeeper->getPosition()) > keeper->getCatchDistance())
{
return true;
}
return false;
}
When I try to compile the code I get some errors which I tried to understand:
'GameWorld::beeToClose': non-standard syntax; use '&' to create a
pointer
Not sure why this message is given
'std::remove_if': no matching overloaded function found
I did not declare beeToClose right?
'q': cannot be used before it is initialized SDLFramework
q is not initialized because:
std::remove_if(bees.begin(), bees.end(), beeToClose);
does not run correct?
How can I remove a std::shared_ptr correctly from a vector correctly when meeting some criteria?
The syntax for forming a pointer to member function is &ClassName::FunctionName. So you need &GameWorld::beeToClose for a pointer to the beeToClose member function. In your case, you should use a lambda from which you call that function
auto q = std::remove_if(bees.begin(), bees.end(),
[&](shared_ptr<MovingEntity> const& bee){ return beeToClose(bee); });
Also, you're using the wrong vector::erase overload, you want the one that erases a range of elements, not the one that erases a single element.
bees.erase(q, bees.end());
The vector contains std::shared_ptr<MovingEntity> elements, so beeToClose() needs to accept a const std::shared_ptr<MovingEntity> & parameter as input, not a const MovingEntity & parameter. Also, beeToClose() appears to be a non-static class method that accesses a non-static class member (m_beekeeper), so you can't just pass beeToClose() directly to std::remove_if() as it does not have access to the calling object's this pointer, but you can wrap it in a lambda to capture the this pointer.
Try this:
void GameWorld::catchBees()
{
auto q = std::remove_if(bees.begin(), bees.end(),
[this](const const std::shared_ptr<MovingEntity> &bee) {
return this->beeToClose(bee);
}
);
bees.erase(q, bees.end());
}
bool GameWorld::beeToClose(const std::shared_ptr<MovingEntity> &bee)
{
std::shared_ptr<Beekeeper> keeper = std::static_pointer_cast<Beekeeper>(m_beekeeper);
return (bee->getConstPosition().distanceTo(m_beekeeper->getPosition()) > keeper->getCatchDistance());
}
You might also consider moving the distance calculation into Beekeeper instead:
bool GameWorld::beeToClose(const std::shared_ptr<MovingEntity> &bee)
{
std::shared_ptr<Beekeeper> keeper = std::static_pointer_cast<Beekeeper>(m_beekeeper);
return !keeper->isInCatchDistance(bee);
}
bool Beekeeper::isInCatchDistance(const std::shared_ptr<MovingEntity> &bee)
{
return (bee->getConstPosition().distanceTo(getPosition()) <= getCatchDistance());
}
I want to be able to share a variable in the containing scope between two lambda functions. I have the following:
void holdAdd(const Rect& rectangle, Hold anonymousHeld, Hold anonymousFinish) {
std::map<int,bool> identifierCollection;
HoldFinish holdFinish = [=](const int& identifier) mutable {
if (identifierCollection.count(identifier) == 0) return;
identifierCollection.erase(identifier);
anonymousFinish();
};
holdCollisionCollection.push_back([=](const int& identifier, const Vec2& point) mutable {
if (rectangle.containsPoint(point)) {
identifierCollection[identifier] = true;
anonymousHeld();
} else {
holdFinish(identifier);
}
});
holdFinishCollection.push_back(holdFinish);
}
I can see in the debugger that holdFinish is pointing to a different implementation of identifierCollection than in the 2nd lambda function.
If I use [=, &identifierCollection] it throws a EXC_BAD_ACCESS whether I use mutable or not.
My experience with other languages that implement inline functions is that this should be possible. For instance in javascript:
var a = 10;
var b = function() {
a += 2;
}
var c = function() {
a += 3;
}
b();
c();
alert(a);
Would alert 15.
What do I have to do to get both lambda functions to reference the same identifierCollection implementation? So that it behaves in the same way as the javascript example.
Unlike in some scripting languages, identifierCollection's lifetime won't be extended simply because you captured it into a closure. So as soon as you change that [=] for a [&] to capture by reference, it's a dangling reference to a local variable that you're capturing.
You'll have to manage the lifetime of identifierCollection yourself; frankly, this sounds like the perfect opportunity for a shared pointer, captured by value into each lambda. The dynamically-allocated map it wraps will literally exist for as long as you need it to.
void holdAdd(const Rect& rectangle, Hold anonymousHeld, Hold anonymousFinish)
{
auto identifierCollection = std::make_shared<std::map<int,bool>>();
HoldFinish holdFinish = [=](const int& identifier) mutable {
if (identifierCollection->count(identifier) == 0) return;
identifierCollection->erase(identifier);
anonymousFinish();
};
holdCollisionCollection.push_back([=](const int& identifier, const Vec2& point) mutable {
if (rectangle.containsPoint(point)) {
(*identifierCollection)[identifier] = true;
anonymousHeld();
} else {
holdFinish(identifier);
}
});
holdFinishCollection.push_back(holdFinish);
}
If you wrap the map in a std::shared_ptr then the lifetime will be managed automatically. Your lambda can then capture by value and it will get a reference to the map whose lifetime remains valid until the lambda function returns.
To do this, change your map definition to:
auto identifierCollection = std::make_shared<std::map<int,bool>>();
And then any calls to member functions of the map need to change from using . to -> (as it is now a pointer).