Re: default contructing an object of unknown type T
Alf P. Steinbach /Usenet <alf.p.steinbach+usenet@gmail.com>, on
30/07/2010 22:27:08, wrote:
* Francesco S. Carta, on 30.07.2010 21:48:
Alf P. Steinbach /Usenet <alf.p.steinbach+usenet@gmail.com>, on
30/07/2010 21:35:43, wrote:
* Hicham Mouline, on 30.07.2010 12:03:
Hello,
inside a template function, I need an instance of type T that is to be
default constructed.
I would normally write:
template<typename T>
void f(const T&) {
const T t();
}
This then requires T to have a default ctor, which is good.
However this doesn't work for primitive types where default ctor means
zeroing it.
Am I missing something obvious?
Like
template< class T >
struct DefaultConstructed
{
T v;
DefaultConstructed(): v() {}
};
I've also tried something like that, but stomped on the fact that the OP
wants a const object within the template function.
If you try to combine your class with the OP's template I think you'll
stomp on the same problem (trying to allow non-copy-constructible types,
of course, because that requirement has been added afterwards).
If you succeed, please post the solution here. My best shot is:
{
T* pt = new T;
{
const T& t = *pt;
// use t
}
delete pt;
}
I'm unable to see the problem that you allude to, but OK:
<code>
#include <iostream>
using namespace std;
template< class Type >
struct DefaultConstructed
{
Type v;
DefaultConstructed(): v() {}
};
template< class Type >
void f( Type const& )
{
DefaultConstructed< Type > const t;
cout << t.v << endl;
}
int main()
{
f( 42 );
}
</code>
The problem was just in my mind, since I've passed through references,
heading for the solution, when I've built a template class equivalent to
yours, I've unsuccessfully tried to create a const reference to it and I
simply discarded that solution without further investigation... silly me.
The solution to Hicham question is the one you just posted.
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