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c++templatesnew-operatordestructorplacement-new

How to delete object constructed via placement new operator?

char * buf = new char[sizeof(T)];
new (buf) T;
T * t = (T *)buf;
//code...
//here I should destruct *t but as it is argument of template and can be
//instantiated via basic types as well (say int) so such code 
/*t->~T();*/
//is incorrect (maybe correct? Strange, but it works on VS 2005 for basic types.)
//and this code 
/*delete t;*/ 
//crashes the program.
delete [] buf;

So what is correct way to destruct t?

P.S. The code above is only for describing my problem, and have not real relationship with code I'm going to write. So please don't give answers like (Why use placement new instead of non-placement? or something similar)

Solution

  • ...instantiated via basic types as well (say int) so such code
    t->~T(); is incorrect    ...

    Wrong. That code is legal and correct in template code even if T can be a primitive type.

    C++ standard: 5.4.2

    5.2.4 Pseudo destructor call [expr.pseudo]

    1. The use of a pseudo-destructor-name after a dot . or arrow -> operator represents the destructor for the non-class type named by type-name. The result shall only be used as the operand for the function call operator (), and the result of such a call has type void. The only effect is the evaluation of the postfix expression before the dot or arrow.
    2. The left hand side of the dot operator shall be of scalar type. The left hand side of the arrow operator shall be of pointer to scalar type. This scalar type is the object type. The type designated by the pseudo destructor- name shall be the same as the object type. Furthermore, the two type-names in a pseudodestructor- name of the form ::opt nested-name-specifieropt type-name :: ˜ type-name shall designate the same scalar type. The cv-unqualified versions of the object type and of the type designated by the pseudo-destructor-name shall be the same type.