warning: 'void operator delete(void*, std::size_t)' called on unallocated object

user_5072601 0
Viewed 670

I am just playing with the placement-new operator. below code compiles and runs without any error on gcc version 11.2.But I am getting one warning as,

warning: 'void operator delete(void*, std::size_t)' called on unallocated object 'buf' [-Wfree-nonheap-object]x86-64 gcc 11.2 #1

Please see the code which I tried,

#include<iostream>
class sample
{
    public:
    sample()
    {
        std::cout<<"entered constructor"<<std::endl;
    }

    ~sample()
    {
        std::cout<<"entered destructor"<<std::endl;
    }
};

int main()
{
    unsigned char buf[4];
    sample* k = new(buf) sample; 
    delete k;
    return 0;
}

Output as follows(I am using compiler explorer for the same)

Program returned: 139

Program stdout

entered constructor

entered destructor

Program stderr

free(): invalid pointer

dynamic-memory-allocation placement-new c++
1 Answers

Using delete on anything other than pointer returned by an allocating new expression results in undefined beheaviour.

Placement-new doesn't allocate anything. You may not delete a pointer returned by placement new. In order to destroy an object created using placement new, you must call the destructor explicitly, or you can use std::destroy_at:

sample* k = new(buf) sample; 
std::destroy_at(k);
// or
// k->~sample();
user_20793030
Related