std::allocator 释放部分内存

Question

I am reading about c++ allocators and the deallocate function has sentence that got my attention:

The argument n must be equal to the first argument of the call to allocate() that originally produced p; otherwise, the behavior is undefined.

Why is that? Why couldn't one deallocate part of the allocated memory, stupid example:

#include <memory>
#include <string>


int main()
{

    std::allocator<std::string> alloc;

    auto const p = alloc.allocate(20);

    alloc.deallocate(p+10, 10);

    return 0;
}

Answer 1

Some allocators might be able to do that. The C++ specification only says what all implementations of the abstract allocator interface, Alloc<T> , are required to do. The committee decided not to require the feature you are asking about.

I don't have the C++ Rationale on this computer, but I suspect that the feature you are asking about is not required because the C memory allocator functions ( malloc and free ) cannot do it, ¹ and the C++ committee wanted it to be possible to implement the C++ library on top of the C library.

¹ yes, realloc can do some cases of this, but not all of them