Placement new and templates

[Kalle Rutanen]

Hello.

The following code compiles fine with VC7. I seemingly call int's
destructor. Why does it compile ? What does the standard say about this
?

#include <new>

using namespace std;

template <typename T>
void destruct(T* that)
{
that->~T();
}

int main()
{
char memory[100];
int* a = 0;

// Can do this...
a = new(memory) int;
destruct(a);

// But can't do this
a = new(memory) int;
//a->~int();

return 0;
}

 

[Victor Bazarov]

The following code compiles fine with VC7. I seemingly call int's
destructor. Why does it compile ? What does the standard say about
this ?

It's explicitly allowed. It's called "pseudo-destructor call". The
Standard defines that in 5.2.4. If your compiler does not allow you
do

int *pa;
...
pa->~int();

then your compiler is broken. For built-in types a pseudo-destructor
call has no effect except to evaluate the expression before the . or
the ->.

V

 

[msalters]

Victor Bazarov wrote:
If your compiler does not allow you

do

int *pa;
...
pa->~int();

then your compiler is broken. For built-in types a pseudo-destructor
call has no effect except to evaluate the expression before the . or
the ->.

So a pedantic compiler may reject this as UB, considering
that you're dereferencing an uinitialized pointer

Of course, normal compilers will just skip this instruction.

Regards,
Michiel Salters

 

[Victor Bazarov]

Victor Bazarov wrote:
If your compiler does not allow you



So a pedantic compiler may reject this as UB, considering
that you're dereferencing an uinitialized pointer

The '...' contains all the necessary code to prevent the UB.

V

 

[Kalle Rutanen]

It's explicitly allowed. It's called "pseudo-destructor call". The

Standard defines that in 5.2.4. If your compiler does not allow you
do

int *pa;
...
pa->~int();

then your compiler is broken. For built-in types a pseudo-destructor
call has no effect except to evaluate the expression before the . or
the ->.

Thank you for a clearing answer.