Re: [Boost-users] A question about shared_ptr(Zhuo Hao)
But consider the following code:
#include<iostream>
using namespace std;
class CMutex{};//Hide all the implementations for convience
template<class Y>
void world(Y * p)
{
}
void hello(int *x){
}
int main(){
hello(0);
world<CMutex>(0);
return 1;
}
It compiles and runs without problems.So you see I can pass zero as a
pointer to Y, and as a pointer to int. That's OK.
I wonder how it is implemented that we cannot pass zero as a pointer to Y in
the constructor of shared_ptr.
Thanks
Zhuo Hao
On 14/07/06, Tommy Hinks
Zhuo Hao wrote:
<snap>
I wonder how it is implemented that we cannot pass zero as a pointer to Y in the constructor of shared_ptr.
Consider: template <class U> class P { public: template <class V> P(V * v) : u(v) {} private: U *u; }; Now "P<CMutex> p1(0);" will not compile but "P<CMutex> p1(new CMutex);" will. -delfin
On 7/16/06, Delfin Rojas
template <class U> class P { public: template <class V> P(V * v) : u(v) {}
private: U *u; };
Now "P<CMutex> p1(0);" will not compile but "P<CMutex> p1(new CMutex);" will.
I believe this is due to the fact that in C++ it is illegal for the compiler
to do more than one implicit cast. In this case it would have to implicitly
cast from <int>(0) to
I think the answer would be: At the point of type deduction ctor does not
know what type V is:
It knows that U is CMutex, but V is unknown. The following sample would
compile just fine:
class A1 {};
class A2 : public A1 {};
A2* a_null_ptr=NULL;
P<A1>(a_null_ptr);
or
P<A1>(static_cast
participants (4)
-
Delfin Rojas -
Jonathan Franklin -
Ovanes Markarian -
Zhuo Hao