I have a vector of shared_ptr, I want to combine boost shared_ptr and bind together.
My question is very similar to this, except that instead of "&MyClass::ReferenceFn" I would like to call "&Element::Fn".
Here is a similar piece of code:
typedef boost::shared_ptr< Vertex > vertex_ptr;
std::set<vertex_ptr> vertices;
void B::convert()
{
...
if( std::find_if(boost::make_indirect_iterator(vertices.begin()),
boost::make_indirect_iterator(vertices.end() ), boost::bind( &Vertex::id, boost::ref(*this), _1 ) == (*it)->id() ) == vertices.end() )
}
here is the error:
no matching function for call to ‘bind(<unresolved overloaded function type>, const boost::reference_wrapper<B>, boost::arg<1>&)’
NOTE: I am limited to use the C++03.
To call a member function for each object stored in a collection, you need to use a placeholder as the first bound argument of the boost::bind:
boost::bind(&Vertex::id, _1) == (*it)->id())
// ~^~
This way, each argument a, will be bound to a member function pointer, and called as (a.*&Vertex::id)().
However, seeing that the error message says unresolved overloaded function type, it draws a conclusion that your class Vertex can have multiple overloads of member function id. As such, the compiler is unable to tell which one it should pass as the argument of boost::bind. To solve this, use an explicit cast to a member function pointer (the asterisk after colon indicates it's a pointer to a member):
boost::bind(static_cast<int(Vertex::*)()const>(&Vertex::id), _1) == (*it)->id())
// ~~~~~~~~~~~~~~~~~~~~^
In the case class Vertex has multiple overloads, say:
int id() const { return 0; }
void id(int i) { }
you will use the first one for binding.
Related
This code does not compile.
std::remove_if( m_selectedFields.begin(),
m_selectedFields.end(),
[](const FieldWin *obj, FieldWin *m_draggingField)
{
obj == m_draggingField;
} );
The m_selectedFields is a vector and its a member of the class, defined as std::vector<Foo *> m_selectedFields.
The m_draggingField is also a member of the class, defined as Foo m_draggingField.
What am I doing wrong?
The errors are:
error C2064: term does not evaluate to a function taking 1 arguments
note: class does not define an 'operator()' or a user defined conversion operator to a pointer-to-function or reference-to-function that takes appropriate number of arguments
I think lambda function should have only one argument, but then how do I pass the element that needs to be removed?
You need to "capture" this or m_draggingField in the lambda:
std::remove_if( m_selectedFields.begin(),
m_selectedFields.end(),
[this](const FieldWin *obj)
{
obj == m_draggingField;
} );
But if your predicate is really just equality, you can simplify:
std::remove(m_selectedFields.begin(), m_selectedFields.end(), m_draggingField);
Also note that remove and remove_if just reorder elements in the container, you still need to call erase as shown in any tutorial for these functions.
I would like to pass function pointer as a function parameter.
Here is my code:
void AuthServerOpcodes::ValidateAndSetServerOpcode(ServerOpcode serverOpcode, void(*handlerFunc(std::vector<std::byte> data))) {}
Here is the function I would like to pass as second parameter in ValidateAndSetServerOpcode:
void AuthServerOpcodes::Test(std::vector<std::byte> data) {
std::cout << "all good" << std:end
}
Here is how I try to pass it:
ValidateAndSetServerOpcode(SMSG_LOGIN_REQUEST, &Test);
However this seems to be not the correct way. When I try to do it in that way I get error:
Cannot initialize a parameter of type 'void (*(*)
(std::vector<std::byte>))' with an rvalue of type 'void
(AuthServerOpcodes::*)(std::vector<std::byte>)': different return type
('void (*)' vs 'void')
Why is that and how can I fix it?
Pointers to member must be qualified with the class type, so you need to get the pointer you'll need to use
ValidateAndSetServerOpcode(SMSG_LOGIN_REQUEST, &AuthServerOpcodes::Test);
But it looks like you've tried that in the previous edit, so I guess you've called the function pointer to member incorrectly. You didn't show a minimal, reproducible example so I can't help you more, please create one. Anyway I've created a compiled example on Compiler Explorer
typedef void (AuthServerOpcodes::*HandlerFunc)(std::vector<std::byte> &);
void AuthServerOpcodes::ValidateAndSetServerOpcode(ServerOpcode serverOpcode,
HandlerFunc handlerFunc)
{
std::vector<std::byte> myVector;
(this->*handlerFunc)(myVector); // call the hander
}
void FreeStandingFunction(AuthServerOpcodes& opc,
AuthServerOpcodes::HandlerFunc handlerFunc,
std::vector<std::byte> &data)
{
(opc.*handlerFunc)(data);
}
As you can see the pointer to member must be called with ->* or .* and the whole dereferencing must be wrapped inside () because those operators has lower precedence than the function call operator ()
See also Function pointer to member function
Some off-topic note:
Don't use lines that are too long like that
Don't pass vectors by values unless you really need to preserve the outside value. Always pass by reference with const std::vector<>& (or remove const to modify the outside variable)
Use '\n' instead of std::endl
You can't do that.
There is no function pointer to that function, because it is a member function.
You can instead pass a pointer-to-member-function, or better yet a std::function bound to a lambda that captures the this pointer.
It's just a type mismatch, your function is a method of the AccountManager class,
so it has this signature similar to:
static void Login(AccountManager *this, std::vector<..> data);
You can either detach function from class, change your type definition of handlerFunc or consider different techniques like std::mem_fn or std:bind
https://en.cppreference.com/w/cpp/utility/functional/mem_fn
https://en.cppreference.com/w/cpp/utility/functional/bind
I am trying to set a delegate for a function and have the 2 following classes to achieve that.
On the bottom is the error I'm getting. How do I handle it?
Class A
typedef void (*SocketEventString) (String);
class SocketIO
{
public:
SocketIO();
void onMessage(SocketEventString _cb);
private:
SocketEventString _onMessage;
};
Class B
class BoardManager
{
public:
BoardManager();
void handleAction(String action);
SocketIO io;
};
//Constructor
BoardManager::BoardManager() {
io.onMessage( std::bind( &BoardManager::handleAction, this, std::placeholders::_1 ) );
}
ERROR
sketch/BoardManager.cpp: In member function 'void BoardManager::initializeSocketIO()':
BoardManager.cpp:68: error: no matching function for call to 'SocketIO::onMessage(std::_Bind_helper<false, void (BoardManager::*)(String), BoardManager* const, const std::_Placeholder<1>&>::type)'
io.onMessage( std::bind( &BoardManager::handleAction, this, std::placeholders::_1 ) );
^
sketch/BoardManager.cpp:68:90: note: candidate is:
In file included from sketch/BoardManager.h:10:0,
from sketch/BoardManager.cpp:8:
sketch/SocketIO.h:25:18: note: void SocketIO::onMessage(SocketEventString)
void onMessage(SocketEventString _cb);
The std::bind function return an object that is not compatible or convertible to a pointer to a non-member function.
Instead use std::function:
using SocketEventString = std::function<void(String)>;
With the definition
typedef void (*SocketEventString) (String);
you say that SocketEventString is a pointer to a non-member function (i.e. a function not a member in a class or struct) that takes one argument of type String and returns no value.
The std::bind function returns an object of an unknown class. That object is not the same a the pointer-type you define SocketEventString to be.
The two types (SocketEventString and the object returned by std::bind) are not compatible. You can not convert from one of the types to the other.
The compiler tell you this, because it tries to find a function SocketIO::onMessage which takes the type of the object returned by std::bind and don't find any such overload.
Instead of the SocketEventString type you have defined, you need to use type that is compatible with the object returned by std::bind. That's what I have shown above in my answer, defined SocketEventString to be a different type, a type that is compatible with the type returned by std::bind.
Firstly, you can't use a C function pointer for a C++ function binding like that. Essentially, when you use bind it captures some variables to be used in the function call (such as this), so you need to use std::function which handles capturing variables if you want to bind a member function (because member functions at the very least need the this pointer captured). Also, in my opinion, std::bind is fairly ugly, and I recommend getting familiar the new C++ lambdas.
BoardManager::BoardManager() {
io.onMessage( [&]( String action ) {
handleAction( action );
});
}
I'm working with std::bind but I still don't get how it works when we use it with member class functions.
If we have the following function:
double my_divide (double x, double y) {return x/y;}
I understand perfectly well the next lines of code:
auto fn_half = std::bind (my_divide,_1,2); // returns x/2
std::cout << fn_half(10) << '\n'; // 5
But now, with the following code where we have a bind to member function I have some questions.
struct Foo {
void print_sum(int n1, int n2)
{
std::cout << n1+n2 << '\n';
}
int data = 10;
};
Foo foo;
auto f = std::bind(&Foo::print_sum, &foo, 95, _1);
f(5);
Why is the first argument a reference? I'd like to get a theoretical explanation.
The second argument is a reference to the object and it's for me the most complicated part to understand. I think it's because std::bind needs a context, am I right? Is always like this? Has std::bind some sort of implementation to require a reference when the first argument is a member function?
When you say "the first argument is a reference" you surely meant to say "the first argument is a pointer": the & operator takes the address of an object, yielding a pointer.
Before answering this question, let's briefly step back and look at your first use of std::bind() when you use
std::bind(my_divide, 2, 2)
you provide a function. When a function is passed anywhere it decays into a pointer. The above expression is equivalent to this one, explicitly taking the address
std::bind(&my_divide, 2, 2)
The first argument to std::bind() is an object identifying how to call a function. In the above case it is a pointer to function with type double(*)(double, double). Any other callable object with a suitable function call operator would do, too.
Since member functions are quite common, std::bind() provides support for dealing with pointer to member functions. When you use &print_sum you just get a pointer to a member function, i.e., an entity of type void (Foo::*)(int, int). While function names implicitly decay to pointers to functions, i.e., the & can be omitted, the same is not true for member functions (or data members, for that matter): to get a pointer to a member function it is necessary to use the &.
Note that a pointer to member is specific to a class but it can be used with any object that class. That is, it is independent of any particular object. C++ doesn't have a direct way to get a member function directly bound to an object (I think in C# you can obtain functions directly bound to an object by using an object with an applied member name; however, it is 10+ years since I last programmed a bit of C#).
Internally, std::bind() detects that a pointer to a member function is passed and most likely turns it into a callable objects, e.g., by use std::mem_fn() with its first argument. Since a non-static member function needs an object, the first argument to the resolution callable object is either a reference or a [smart] pointer to an object of the appropriate class.
To use a pointer to member function an object is needed. When using a pointer to member with std::bind() the second argument to std::bind() correspondingly needs to specify when the object is coming from. In your example
std::bind(&Foo::print_sum, &foo, 95, _1)
the resulting callable object uses &foo, i.e., a pointer to foo (of type Foo*) as the object. std::bind() is smart enough to use anything which looks like a pointer, anything convertible to a reference of the appropriate type (like std::reference_wrapper<Foo>), or a [copy] of an object as the object when the first argument is a pointer to member.
I suspect, you have never seen a pointer to member - otherwise it would be quite clear. Here is a simple example:
#include <iostream>
struct Foo {
int value;
void f() { std::cout << "f(" << this->value << ")\n"; }
void g() { std::cout << "g(" << this->value << ")\n"; }
};
void apply(Foo* foo1, Foo* foo2, void (Foo::*fun)()) {
(foo1->*fun)(); // call fun on the object foo1
(foo2->*fun)(); // call fun on the object foo2
}
int main() {
Foo foo1{1};
Foo foo2{2};
apply(&foo1, &foo2, &Foo::f);
apply(&foo1, &foo2, &Foo::g);
}
The function apply() simply gets two pointers to Foo objects and a pointer to a member function. It calls the member function pointed to with each of the objects. This funny ->* operator is applying a pointer to a member to a pointer to an object. There is also a .* operator which applies a pointer to a member to an object (or, as they behave just like objects, a reference to an object). Since a pointer to a member function needs an object, it is necessary to use this operator which asks for an object. Internally, std::bind() arranges the same to happen.
When apply() is called with the two pointers and &Foo::f it behaves exactly the same as if the member f() would be called on the respective objects. Likewise when calling apply() with the two pointers and &Foo::g it behaves exactly the same as if the member g() would be called on the respective objects (the semantic behavior is the same but the compiler is likely to have a much harder time inlining functions and typically fails doing so when pointers to members are involved).
From std::bind docs:
bind( F&& f, Args&&... args ); where f is a Callable, in your case that is a pointer to member function. This kind of pointers has some special syntax compared to pointers to usual functions:
typedef void (Foo::*FooMemberPtr)(int, int);
// obtain the pointer to a member function
FooMemberPtr a = &Foo::print_sum; //instead of just a = my_divide
// use it
(foo.*a)(1, 2) //instead of a(1, 2)
std::bind(and std::invoke in general) covers all these cases in a uniform way. If f is a pointer-to-member of Foo, then the first Arg provided to bind is expected to be an instance of Foo (bind(&Foo::print_sum, foo, ...) also works, but foo is copied) or a pointer to Foo, like in example you had.
Here is some more reading about pointers to members, and 1 and 2 gives full information about what bind expects and how it invokes stored function.
You also can use lambdas instead std::bind, which could be more clear:
auto f = [&](int n) { return foo.print_sum(95, n); }
I'm trying to write predicate function for use with STL algorithms. I see that they are two ways to define a predicate:
(1) Use a simple function as below:
bool isEven(unsigned int i) { return (i % 2 == 0); }
std::find_if(itBegin, itEnd, isEven);
(2) Use the operator() function as below:
class checker {
public:
bool operator()(unsigned int i) { return (i % 2 == 0); }
};
std::find_if(itBegin, itEnd, checker);
I have more use for the second type as I usually would like to create a predicate object with some members in it and use them in the algorithm. When I add the same isEven function inside checker and use it as a predicate, I get an error:
3. Syntax which gives error:
class checker {
public:
bool isEven(unsigned int i)
{ return (i%2 == 0); }
};
checker c;
std::find_if(itBegin, itEnd, c.isEven);
Calling c.isEven gives an error during compilation saying undefined reference to some function. Can someone explain why 3. is giving error? Also, I would appreciate any pointers to read about predicate and iterator basics.
A pointer to a member function requires an instance to be called on, and you are passing only the member function pointer to std::find_if (actually your syntax is incorrect, so it doesn't work at all; the correct syntax is std::find_if(itBegin, itEnd, &checker::isEven) which then still doesn't work for the reasons I gave).
The find_if function expects to be able to call the function using a single parameter (the object to test), but it actually needs two to call a member function: the instance this pointer and the object to compare.
Overloading operator() allows you to pass both the instance and the function object at the same time, because they're now the same thing. With a member function pointer you must pass two pieces of information to a function that expects only one.
There is a way to do this using std::bind (which requires the <functional> header):
checker c;
std::find_if(itBegin, itEnd, std::bind(&checker::isEven, &c, std::placeholders::_1));
If your compiler doesn't support std::bind, you can also use boost::bind for this. Though there's no real advantage to doing this over just overloading operator().
To elaborate a bit more, std::find_if expects a function pointer matching the signature bool (*pred)(unsigned int) or something that behaves that way. It doesn't actually need to be a function pointer, because the type of the predicate is bound by the template. Anything that behaves like a bool (*pred)(unsigned int) is acceptable, which is why functors work: they can be called with a single parameter and return a bool.
As others have pointed out, the type of checker::isEven is bool (checker::*pred)(unsigned int) which doesn't behave like the original function pointer, because it needs an instance of checker to be called on.
A pointer to a member function can be conceptually considered as a regular function pointer that takes an additional argument, the this pointer (e.g. bool (*pred)(checker*, unsigned int)). You can actually generate a wrapper that can be called that way using std::mem_fn(&checker::isEven) (also from <functional>). That still doesn't help you, because now you have a function object that must be called with two parameters rather than only one, which std::find_if still doesn't like.
Using std::bind treats the pointer to a member function as if it was a function taking the this pointer as its first argument. The arguments passed to std::bind specify that the first argument should always be &c, and the second argument should bind to the first argument of the newly returned function object. This function object is a wrapper that can be called with one argument, and can therefore be used with std::find_if.
Although the return type of std::bind is unspecified, you can convert it to a std::function<bool(unsigned int)> (in this particular case) if you need to refer to the bound function object explicitly rather than passing it straight to another function like I did in my example.
I guess it's because the type of c.isEven() is,
bool (checker::*)(unsigned int) // member function of class
which may not be expected by find_if(). std::find_if should be expecting either a function pointer (bool (*)(unsigned int)) or a function object.
Edit: Another constraint: A non-static member function pointer must be called by the class object. In your case, even if you succeed to pass the member function then still find_if() will not have any information about any checker object; so it doesn't make sense to have find_if() overloaded for accepting a member function pointer argument.
Note: In general c.isEven is not the right way to pass member function pointer; it should be passed as, &checker::isEven.
checker::isEven is not a function; it is a member function. And you cannot call a non-static member function without a reference to a checker object. So you can't just use a member function in any old place that you could pass a function pointer. Member pointers have special syntax that requires more than just () to call.
That's why functors use operator(); this makes the object callable without having to use a member function pointer.
I prefer functors (function objects) because make your program more readable and, more importantly, expressing the intent clearly.
This is my favorite example:
template <typename N>
struct multiplies
{
N operator() (const N& x, const N& y) { return x * y; }
};
vector<int> nums{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
// Example accumulate with transparent operator functor
double result = accumulate(cbegin(nums), cend(nums), 1.1, multiplies<>());
Note: In recent years we've got a lambda expression support.
// Same example with lambda expression
double result = accumulate(cbegin(nums), cend(nums), 1.1,
[](double x, double y) { return x * y; });
The example given says you should use the call operator (operator()) whereas in your example you've called your function isEven. Try re-writing it as:
class checker {
public:
bool operator()(unsigned int i)
{ return (i%2 == 0); }
};