是否可以强制std :: make_shared使用类的new运算符？这与另一个
SO question有关.根据那个问题,std :: make_shared使用自定义分配器：
From the standard (§20.7.2.2.6 shared_ptr creation ):
Effects: Allocates memory suitable for an object of type T and constructs an object in that memory via the placement new expression@H_404_6@ ::new (pv) T(std::forward(args)…).
因此,我认为我可以使用自定义展示位置新运算符,但这似乎是错误的
// std::cout
#include <iostream>
// std::make_shared
#include <memory>
// Track what we're making
struct Foo {
Foo() {
std::cout << "Foo constructor" << std::endl;
}
Foo(Foo const & foo) {
std::cout << "Foo copy constructor" << std::endl;
}
Foo(Foo && foo) {
std::cout << "Foo move constructor" << std::endl;
}
Foo & operator = (Foo const & foo) {
std::cout << "Foo copy assignment" << std::endl;
return *this;
}
Foo & operator = (Foo && foo) {
std::cout << "Foo move assignment" << std::endl;
return *this;
}
void * operator new(std::size_t size) throw(std::bad_alloc) {
std::cout << "Foo new" << std::endl;
return ::operator new(size);
}
void * operator new(std::size_t size,void * p) throw() {
std::cout << "Foo placement new" << std::endl;
return ::operator new(size,p);
}
void* operator new (std::size_t size,std::nothrow_t const & nothrow_value)
throw()
{
std::cout << "Foo nonthrowing new" << std::endl;
return ::operator new(size,nothrow_value);
}
void operator delete(void * p,std::size_t size) {
std::cout << "Foo delete" << std::endl;
::operator delete(p);
}
~Foo() {
std::cout << "Foo destructor" << std::endl;
}
};
int main() {
std::cout << "---Creating foo" << std::endl;
auto foo = std::make_shared <Foo> ();
std::cout << "---Creating foo2" << std::endl;
auto foo2 = std::shared_ptr <Foo> (new Foo());
std::cout << "---Creating foo3" << std::endl;
auto foo3 = std::allocate_shared <Foo> (std::allocator <Foo>());
std::cout << "---fin" << std::endl;
}
这使
---Creating foo Foo constructor ---Creating foo2 Foo new Foo constructor ---Creating foo3 Foo constructor ---fin Foo destructor Foo destructor Foo delete Foo destructor
还有一个尝试强制一个分配器,它将通过调用std :: allocate_shared来调用自定义新运算符.在任何情况下,有没有办法让std :: make_shared调用自定义新运算符而不定义一个全新的分配器？
