Im trying to remove pairs from vector with remove_if, but im getting errors
bool MyClass::isSingleTag(const pair<int,string> & val) {
string tag = val.second;
int posImg, posBr;
posImg = tag.find("<img");
posBr = tag.find("<br");
if (posImg == -1 && posBr == -1) {
return false;
} else {
return true;
}
}
void MyClass::deleteSingleTags() {
vector<pair<int,string>>::iterator last_iter;
last_iter = remove_if(allTags.begin(), allTags.end(), &MyClass::isSingleTag);
allTags.erase(last_iter, allTags.end());
}
Errors: http://pastebin.com/1FCWRVDG
A pointer to a member function isn't callable without an object of the class the function is member of.
Make isSingleTag static - taking its address results in a plain function pointer. Alternatively, make it a free function, since it looks like it's got no bussiness being a member function in the first place (it doesn't access any other members, does it?).
The other option (for when you legitimately need to be a member function) is to bind it to a class object, using std::bind:
MyClass obj;
auto func = std::bind(&MyClass::isSingleTag, obj);
Now func is a callable that you can pass to the algorithm.
Related
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 have this pre-defined function.
void attack(std::vector<GameObject*> objects, unsigned damage) {
for (GameObject* object : objects) {
object->takeDamage(damage);
auto isDead = object->isDead();
objects.erase(std::remove_if(objects.begin(),objects.end(),isDead), objects.end());
}
}
This is my isDead function
bool isDead() const {
if (destructed) {
std::cout << "memory error" << std::endl;
}
return life <= 0;
}
This is the error I keep getting. Have tried a lot of things, but not at all able to figure this one out. Any help appreciated!
error: expression cannot be used as a function
{ return bool(_M_pred(*__it)); }
isDead is a variable in the function. You can't use it as an argument to remove_if.
You can't use a regular member function as argument to std::remove_if either. Use a lambda function instead.
Don't erase objects from a container while you are iterating over it using a range for loop.
Change the argument to attack to be a reference. Otherwise, you will be removing objects from a copy, not the original container.
Here's an updated version of attack:
void attack(std::vector<GameObject*>& objects, unsigned damage)
{
for (GameObject* object : objects)
{
object->takeDamage(damage);
}
objects.erase(std::remove_if(objects.begin(),objects.end(), [](GameObject* object){return object->isDead();}), objects.end());
}
isDead() is a member function of one of your classes, which is exactly why it doesn't work: you did not supply this pointer (object instance) for it to be called on. Oh, and the predicate for remove_if must have exactly one argument of the type objects::value_type.
Do this instead:
objects.erase(std::remove_if(objects.begin(),objects.end(),[](GameObject* object){return object->isDead()), objects.end());
I am implementing a member function called CurrentUser. It will take a username as parameter and return the User instance object which matches the given username. Below is the code
User& UserDB::currentUser(string username){
// userlists is a instance member which is list of user objects
for(list<User>::iterator i = userlists.begin(); i != userlists.end(); ++i)
{
if(*i.getName().compare(username)==0){
return *i;
}
}
return null;
}
Not sure if it is the correct way to do so. Correct me if it is wrong. Thanks!
Update:
hey guys thanks for your advice, i figure out a way to do so by returning a User pointer. Here is the code.
User* UserDB::currentUser(string username){
for(list<User>::iterator i = userlists.begin(); i != userlists.end(); ++i)
{
if(i->getName().compare(username)==0){
return i;
}
}
return null;
}
there are a couple of ways to do this cleanly.
You can of course return a pointer, but that will be surprising to people as it is more normal to return a reference or an object. pointers present object consumers with a number of problems, e.g.:
what should I conclude if it's null?
should I delete it?
and so on.
Returning a reference to something or a something removes these ambiguities.
Having said that, references cannot be empty, so the function must return something. If it does not find the item it's looking for, it must indicate that to the caller. One way is an exception (i.e. it was logically incorrect to ask for that item). However, if the item not being there is a normal occurrence, then you don't want to force your consumers to handle exceptions - that's bad form too.
So the answer is to return an object that encapsulates an optional reference.
A good example of this is boost::optional<User&> but if you don't want to include boost it's fairly simple to roll your own:
struct optional_user
{
using element_type = User;
using reference_type = element_type&;
optional_user() : _p(nullptr) {}
optional_user(reference_type r)
: _p(std::addressof(r))
{}
bool valid() const { return bool(_p); }
// compares to true if the user is present, false otherwise
operator bool() const { return valid(); }
reference_type value() const {
assert(_p);
return *_p;
}
// can be used anywhere a User& is required
operator reference_type () const {
return value();
}
private:
element_type* _p = nullptr;
};
now your function becomes:
optional_user UserDB::currentUser(string username)
{
typedef list<User>::iterator Iter;
for(Iter i = userlists.begin(); i != userlists.end(); ++i)
{
if(i->getName().compare(username)==0)
{
return optional_user(*i);
}
}
// return an indicator that the user is not present
return optional_user();
}
and your call site becomes:
optional_user = users.currentUser("bob");
if (optional_user) {
do_something_with(optional_user /* .value() */);
}
If you want to specifically return a reference, the item you are referring to must have a lifetime after execution leaves the function.
There are several alternatives:
static local variable in function
using dynamic memory
variable declared outside function
Pass by non-const reference
Here is an example of #1:
const std::string& Get_Model_Name(void)
{
static const std::string model_name = "Accord";
return model_name;
}
Other alternatives are to return a variable by value (copy). This doesn't use references. A copy is returned.
For example:
std::string Get_Manufacturer_Name(void)
{
return std::string("Honda");
}
You may also consider passing by parameter and modifying the parameter:
void Get_Lunch_Special_Name(std::string& entree_name)
{
entree_name = std::string("Beef Wellington");
}
So I am unsure why this wont work, ive tried some googling, i just cant find out what the problem is
void Player::Cmd(std::vector<std::string> &tokens)
{
std::string str = tokens[0];
std::map<std::string, void (Player::*)()>::iterator it = playerCommands.find(str);
Func fun;
if (it != playerCommands.end())
{
fun = it->second; //i tried it->second(); same issue
fun(); //error C2064: term does not evaluate to a
//function taking 0 arguments
}
else
{
std::cout << "What? \n";
}
}
git hub for the project
https://github.com/lordkuragari/TextRPG
Contrary to your belief, your map doesn't hold function pointers. So you cannot call the elements in the map.
Rather, your map contains pointers to member functions. Non-static member functions aren't functions and cannot be called; rather, they have to be invoked on an object. You can invoke a member function on an object given by a pointer p via a function pointer ptfm like this:
(p->*ptmf)();
In your case, presumably you want to use p = this and ptfm = fun, so it'd be:
(this->*fun)();
Or, without the local variable:
(this->*it->second)();
In C++17 you can also use std::invoke(it->second, this).