I wrote a generic class for handling and executing a function pointer. This is a simplified equivalent of std::function and std::bind. To handle member functions I use cast to internal EventHandler::Class type. Question: is it ok to cast it that way? Will it work in all cases when invoking handled function?
template <typename ReturnType, typename... Arguments>
class EventHandler
{
class Class {};
ReturnType (Class::*memberFunction)(Arguments...) = nullptr;
union {
Class *owner;
ReturnType(*function)(Arguments...) = nullptr;
};
public:
EventHandler() = default;
EventHandler(EventHandler &&) = default;
EventHandler(const EventHandler &) = default;
EventHandler &operator=(EventHandler &&) = default;
EventHandler &operator=(const EventHandler &) = default;
EventHandler(ReturnType (*function)(Arguments...)) :
function(function)
{
}
template <typename Owner>
EventHandler(Owner *owner, ReturnType (Owner::*memberFunction)(Arguments...)) :
memberFunction((ReturnType (Class::*)(Arguments...)) memberFunction),
owner((Class *) owner)
{
}
template <typename Owner>
EventHandler(const Owner *owner, ReturnType (Owner::*memberFunction)(Arguments...) const) :
memberFunction((ReturnType (Class::*)(Arguments...)) memberFunction),
owner((Class *) owner)
{
}
ReturnType operator()(Arguments... arguments)
{
return memberFunction ?
(owner ? (owner->*memberFunction)(arguments...) : ReturnType()) :
(function ? function(arguments...) : ReturnType());
}
};
The implementation provides handle for a global function, a member function and a const member function. Obviously there is volatile and const volatile that is not show here for clarity.
EDIT
All the code below is just a representation of all of kinds of supported functions.
class Object
{
public:
double y = 1000;
Object() = default;
Object(double y) : y(y) {}
static void s1(void) { std::cout << "s1()" << std::endl; }
static void s2(int a) { std::cout << "s2(a:" << 10 + a << ")" << std::endl; }
static void s3(int a, float b) { std::cout << "s3(a:" << 10 + a << ", b:" << 10 + b << ")" << std::endl; }
static int s4(void) { std::cout << "s4(): "; return 10 + 4; }
static Object s5(int a) { std::cout << "s5(a:" << 10 + a << "): "; return Object(10 + 5.1); }
static float s6(int a, Object b) { std::cout << "s6(a:" << 10 + a << ", b:" << 10 + b.y << "); "; return 10 + 6.2f; }
void m1(void) { std::cout << "m1()" << std::endl; }
void m2(int a) { std::cout << "m2(a:" << y + a << ")" << std::endl; }
void m3(int a, float b) { std::cout << "m3(a:" << y + a << ", b:" << y + b << ")" << std::endl; }
int m4(void) { std::cout << "m4(): "; return ((int) y) + 4; }
Object m5(int a) { std::cout << "m5(a:" << y + a << "): "; return Object(y + 5.1); }
float m6(int a, Object b) { std::cout << "m6(a:" << y + a << ", b:" << y + b.y << "); "; return ((int) y) + 6.2f; }
void c1(void) const { std::cout << "c1()" << std::endl; }
void c2(int a) const { std::cout << "c2(a:" << y + a << ")" << std::endl; }
void c3(int a, float b) const { std::cout << "c3(a:" << y + a << ", b:" << y + b << ")" << std::endl; }
int c4(void) const { std::cout << "c4(): "; return ((int) y) + 4; }
Object c5(int a) const { std::cout << "c5(a:" << y + a << "): "; return Object(y + 5.1); }
float c6(int a, Object b) const { std::cout << "c6(a:" << y + a << ", b:" << y + b.y << "); "; return ((int) y) + 6.2f; }
};
void f1(void) { std::cout << "f1()" << std::endl; }
void f2(int a) { std::cout << "f2(a:" << a << ")" << std::endl; }
void f3(int a, float b) { std::cout << "f3(a:" << a << ", b:" << b << ")" << std::endl; }
int f4(void) { std::cout << "f4(): "; return 4; }
Object f5(int a) { std::cout << "f5(a:" << a << "): "; return Object(5.1); }
float f6(int a, Object b) { std::cout << "f6(a:" << a << ", b:" << b.y << "); "; return 6.2f; }
Here is the usage example for all of the above functions
int main()
{
std::cout << "=== Global functions" << std::endl;
EventHandler ef1(f1); ef1();
EventHandler ef2(f2); ef2(2);
EventHandler ef3(f3); ef3(3, 3.1f);
EventHandler ef4(f4); std::cout << ef4() << std::endl;
EventHandler ef5(f5); std::cout << ef5(5).y << std::endl;
EventHandler ef6(f6); std::cout << ef6(6, Object(6.1)) << std::endl;
std::cout << std::endl;
std::cout << "=== Member static functions" << std::endl;
EventHandler es1(Object::s1); es1();
EventHandler es2(Object::s2); es2(2);
EventHandler es3(Object::s3); es3(3, 3.1f);
EventHandler es4(Object::s4); std::cout << es4() << std::endl;
EventHandler es5(Object::s5); std::cout << es5(5).y << std::endl;
EventHandler es6(Object::s6); std::cout << es6(6, Object(6.1)) << std::endl;
std::cout << std::endl;
std::cout << "=== Member functions" << std::endl;
Object object(20);
EventHandler em1(&object, &Object::m1); em1();
EventHandler em2(&object, &Object::m2); em2(2);
EventHandler em3(&object, &Object::m3); em3(3, 3.1f);
EventHandler em4(&object, &Object::m4); std::cout << em4() << std::endl;
EventHandler em5(&object, &Object::m5); std::cout << em5(5).y << std::endl;
EventHandler em6(&object, &Object::m6); std::cout << em6(6, Object(6.1)) << std::endl;
std::cout << std::endl;
std::cout << "=== Member const functions" << std::endl;
const Object constObject(30);
EventHandler ec1(&constObject, &Object::c1); ec1();
EventHandler ec2(&constObject, &Object::c2); ec2(2);
EventHandler ec3(&constObject, &Object::c3); ec3(3, 3.1f);
EventHandler ec4(&constObject, &Object::c4); std::cout << ec4() << std::endl;
EventHandler ec5(&constObject, &Object::c5); std::cout << ec5(5).y << std::endl;
EventHandler ec6(&constObject, &Object::c6); std::cout << ec6(6, Object(6.1)) << std::endl;
system("pause");
return 0;
}
Finally - to the point - here an example that shows how much easier in use is the EventHandler I prepared when compared to std::function interface. And actually the reason of such approach.
EventHandler<float, int, Object> example;
example = f6;
example(7, Object(7.1));
example = EventHandler(&object, &Object::m6);;
example(8, Object(8.1));
It’s undefined behavior to call a function through a function pointer(-to-member) of a different type. (Some practical reasons for this rule are that the object’s address might need to be adjusted to call a member function of a base class or that a vtable might be involved.) You can use type erasure to allow calling member functions on objects of different types (which is what std::bind does), or you can (restrict to member functions and) add the class type as a template parameter.
Of course, the usual answer is to just use std::function with a lambda that captures the object in question and calls whatever member function. You can also take the C approach and define various functions with a void* parameter that cast that parameter to a known class type and call the desired member function.
Is only the way to modify, not replace, an object stored as std::any is to declare changeable data mutable? E.g. to avoid creation and copy of class S instances:
#include <iostream>
#include <vector>
#include <any>
#include <string>
struct S {
mutable std::string str;
S(S&& arg) : str(std::move(arg.str)) { std::cout << ">> S moved" << std::endl; }
S(const S& arg) : str(arg.str) { std::cout << ">> S copied" << std::endl; }
S(const char *s) : str(s) { std::cout << ">> S created" << std::endl; }
S& operator= (const S& arg) { str = arg.str; return *this; }
S& operator= (S&& arg) { str = std::move(arg.str); return *this; }
virtual ~S() { std::cout << "<< S destroyed" << std::endl; }
};
int main() {
std::vector<std::any> container;
container.emplace_back(S("Test 1"));
std::any_cast<const S&>(container[0]).str = "Test 2";
for (const auto& a : container) {
std::cout << a.type().name() << ", "
<< std::any_cast<const S&>(a).str << std::endl;
}
}
You can any_cast with non-const reference:
std::any_cast<S&>(container[0]).str = "Test 2";
Demo
or
std::any a = 40;
std::any_cast<int&>(a) += 2;
std::cout << std::any_cast<int>(a) <<std::endl;
Demo
nope.
The issue is here.
std::any_cast<const S&>(container[0]).str = "Test 2";
you are trying to change constant data
I am trying to pass literal to a perfect forwarding setter, but this results in object construction, movement and destruction instead. I am looking for a more efficient way to do this.
struct TStruct {
TStruct(int va) : a(va) {
std::cout << "- TStruct(" << va << ")" << std::endl;
}
TStruct(const TStruct& other) :
a(other.a)
{
std::cout << "- TStruct(const TStruct& )" << std::endl;
}
TStruct(TStruct&& other) :
a(std::exchange(other.a, 0))
{
std::cout << "- TStruct(const TStruct&&)" << std::endl;
}
TStruct& operator=(const TStruct& rhs) {
std::cout << "- TStruct operator= const& " << std::endl;
// check for self-assignment
if(&rhs == this) return *this;
a = rhs.a;
return *this;
}
TStruct& operator=(TStruct&& rhs) {
std::cout << "- TStruct operator=&&" << std::endl;
// check for self-assignment
if(&rhs == this) return *this;
a = std::exchange(rhs.a, 0);
return *this;
}
~TStruct() {
std::cout << "~ TStruct() destructor with " << a << std::endl;
}
int a = 1;
};
struct TPerfectForward {
template<class T>
TPerfectForward(T&& tsv) : ts(std::forward<T>(tsv)) {}
template<class T>
void set(T&& tsv) { ts = std::forward<T>(tsv); }
TStruct ts;
};
std::cout << "TPerfectForward (5)" << std::endl;
TPerfectForward pf(5);
std::cout << "TPerfectForward set(4)" << std::endl;
pf.set(4);
This gives following results with gcc 7.4.0 on Ubuntu:
TPerfectForward (5);
- TStruct(5)
TPerfectForward set(4)
- TStruct(4)
- TStruct operator=&&
~ TStruct() destructor with 0
I would like these results better:
TPerfectForward (5)
- TStruct(5)
TPerfectForward set(4)
- TStruct(4)
Is there a way to avoid calling move operator and destructor when perfect forwarding to a function?
I'm using Visual Studio 2017 and came across a bug which I cannot understand. I have this code as an example:
#include <iostream>
struct Foo
{
int a = 0;
int c;
int *b;
Foo(int n)
{
a = n;
b = &c;
}
};
struct Bar
{
Foo nestedFoo = Foo(0);
Bar(int n)
{
std::cout << "bar constructor nestedFoo before:" << std::endl;
std::cout << &nestedFoo.a << std::endl;
std::cout << nestedFoo.b << std::endl;
nestedFoo = Foo(n);
std::cout << "bar constructor nestedFoo after:" << std::endl;
std::cout << &nestedFoo.a << std::endl;
std::cout << nestedFoo.b << std::endl;
}
};
struct App
{
Bar bar = Bar(2);
};
int main()
{
App app;
Bar bar = Bar(10);
Foo foo = Foo(5);
std::cout << "foo before:" << std::endl;
std::cout << &foo.a << std::endl;
std::cout << foo.b << std::endl;
foo = Foo(20);
std::cout << "foo after:" << std::endl;
std::cout << &foo.a << std::endl;
std::cout << foo.b << std::endl;
std::cout << "bar.nestedFoo:" << std::endl;
std::cout << &bar.nestedFoo.a << std::endl;
std::cout << bar.nestedFoo.b << std::endl;
std::cout << "app.bar.nestedFoo:" << std::endl;
std::cout << &app.bar.nestedFoo.a << std::endl;
std::cout << app.bar.nestedFoo.b << std::endl;
return 0;
}
I expected the pointer and the address of the member to be the same after the reassignment but it seems this is not the case, &Foo.a and Foo.b have different values:
bar constructor nestedFoo before:
0058FEE8
0058FEE8
bar constructor nestedFoo after:
0058FEE8
0058FC24
bar constructor nestedFoo before:
0058FED8
0058FED8
bar constructor nestedFoo after:
0058FED8
0058FD04
foo before:
0058FEC8
0058FEC8
foo after:
0058FEC8
0058FDF8
bar.nestedFoo:
0058FED8
0058FD04
app.bar.nestedFoo:
0058FEE8
0058FC24
At first I thought it only happened when I had Foo as a member of another struct, but it seems to happen even at a top-level assignment. So what am I doing wrong?
Any help will be appreciated, thank you!
Edit: showing that assigning b to &c which is a member that doesn't get (explicitly) reassigned also doesn't work.
I am using "emplace" method to avoid memory copy.
But, when I am using the "emplace" inside a Lambda function. It always call implicit move constructor.
How I can avoid memory copy inside a Lambda function?
This sample program should not print “I am being moved.”
#include <vector>
#include <iostream>
struct A
{
int a;
A(int t) : a(t)
{
std::cout << "I am being constructed.\n";
}
A(A&& other) : a(std::move(other.a))
{
std::cout << "I am being moved.\n";
}
};
std::vector<A> g_a;
int main()
{
std::cout << "emplace_back:\n";
g_a.emplace_back(1);
std::cout << "emplace_back in lambda:\n";
auto f1 = [](int x) { g_a.emplace_back(x); };
f1(2);
std::cout << "\nContents: ";
for (A const& t : g_a)
std::cout << t.a << " ";
std::cout << std::endl;
}
It's not about the lambda function but rather about the vector that reallocates its memory. You can ammend this with std::vector::reserve.
int main() {
g_a.reserve(10);
^^^^^^^^^^^^^^^^
std::cout << "emplace_back:\n";
g_a.emplace_back(1);
std::cout << "emplace_back in lambda:\n";
auto f1 = [](int x) { g_a.emplace_back(x); };
f1(2);
std::cout << "\nContents: ";
for (A const& t : g_a)
std::cout << t.a << " ";
std::cout << std::endl;
}
Live Demo