Using malloc instead of new, and calling the copy constructor when the object is created

Question

I wanted to try out TBB's scalable_allocator, but was confused when I had to replace some of my code. This is how allocation is done with the allocator:

SomeClass* s = scalable_allocator<SomeClass>().allocate( sizeof(SomeClass) );

EDIT: What's shown above is not how allocation is done with scalable_allocator. As ymett correctly mentioned, allocation is done like this:

int numberOfObjectsToAllocateFor = 1;
SomeClass* s = scalable_allocator<SomeClass>().allocate( numberOfObjectsToAllocateFor );
scalable_allocator<SomeClass>().construct( s, SomeClass());
scalable_allocator<SomeClass>().destroy(s);
scalable_allocator<SomeClass>().deallocate(s, numberOfObjectsToAllocateFor);

It's pretty much like using a malloc:

SomeClass* s = (SomeClass*) malloc (sizeof(SomeClass));

This is the code I wanted to replace:

SomeClass* SomeClass::Clone() const
{
   return new SomeClass(*this);
}//Clone

So tried a program:

#include<iostream>
#include<cstdlib>
using namespace std;

class S
{
        public:
        int i;
        S() {cout<<"constructed"<<endl;}
        ~S() {cout<<"destructed"<<endl;}
        S(const S& s):i(s.i) {}
};

int main()
{
        S* s = (S*) malloc(sizeof(S));
        s = (S*) S();//this is obviously wrong
        free(s);
}

and here I found that calling malloc does not instantiate the object (I've never used malloc earlier). So before figuring out how to pass *this to the copy ctor, I'd like to know how to instantiate the object when working with malloc.

Answer 1

You'll need to use placement new after getting the raw memory from malloc.

void* mem = malloc(sizeof(S));
S* s = new (mem) S(); //this is the so called "placement new"

When you're done with the object you have to make sure to explicitly call its destructor.

s->~S();
free(mem);

Answer 2

Use placement new

#include <memory>
//...
int main()
{
        S* s = (S*) malloc(sizeof(S));
        s = new (s) S();//placement new
        //...
        s->~S();
        free(s);
}

Answer 3

The parameter to allocate() is the number of objects, not the size in bytes. You then call the allocator's construct() function to construct the object.

scalable_allocator<SomeClass> sa;
SomeClass* s = sa.allocate(1);
sa.construct(s, SomeClass());
// ...
sa.destroy(s);
sa.deallocate(s);

If want to use it with a standard library container or other std allocator aware type, simply give it the allocator type.

std::vector<SomeClass, scalable_allocator<SomeClass>> v;