Is it safe to realloc memory allocated with new?

Question

From what is written here, new allocates in free store while malloc uses heap and the two terms often mean the same thing.

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Answer 1

You can only realloc that which has been allocated via malloc (or family, like calloc).

That's because the underlying data structures that keep track of free and used areas of memory, can be quite different.

It's likely but by no means guaranteed that C++ new and C malloc use the same underlying allocator, in which case realloc could work for both. But formally that's in UB-land. And in practice it's just needlessly risky.

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C++ does not offer functionality corresponding to realloc.

The closest is the automatic reallocation of (the internal buffers of) containers like std::vector.

The C++ containers suffer from being designed in a way that excludes use of realloc.

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Instead of the presented code

int* data = new int[3];
//...
int* mydata = (int*)realloc(data,6*sizeof(int));

… do this:

vector<int> data( 3 );
//...
data.resize( 6 );

However, if you absolutely need the general efficiency of realloc, and if you have to accept new for the original allocation, then your only recourse for efficiency is to use compiler-specific means, knowledge that realloc is safe with this compiler.

Otherwise, if you absolutely need the general efficiency of realloc but is not forced to accept new, then you can use malloc and realloc. Using smart pointers then lets you get much of the same safety as with C++ containers.

Answer 2

The only possibly relevant restriction C++ adds to realloc is that C++'s malloc/calloc/realloc must not be implemented in terms of ::operator new, and its free must not be implemented in terms of ::operator delete (per C++14 [c.malloc]p3-4).

This means the guarantee you are looking for does not exist in C++. It also means, however, that you can implement ::operator new in terms of malloc. And if you do that, then in theory, ::operator new's result can be passed to realloc.

In practice, you should be concerned about the possibility that new's result does not match ::operator new's result. C++ compilers may e.g. combine multiple new expressions to use one single ::operator new call. This is something compilers already did when the standard didn't allow it, IIRC, and the standard now does allow it (per C++14 [expr.new]p10). That means that even if you go this route, you still don't have a guarantee that passing your new pointers to realloc does anything meaningful, even if it's no longer undefined behaviour.

Answer 3

In general, no.

There are a slew of things which must hold to make it safe:

1. Bitwise copying the type and abandoning the source must be safe.
2. The destructor must be trivial, or you must in-place-destruct the elements you want to deallocate.
3. Either the constructor is trivial, or you must in-place-construct the new elements.

Trivial types satisfy the above requirements.

In addition:

1. The new[]-function must pass the request on to malloc without any change, nor do any bookkeeping on the side. You can force this by replacing global new[] and delete[], or the ones in the respective classes.
2. The compiler must not ask for more memory in order to save the number of elements allocated, or anything else.
   There is no way to force that, though a compiler shouldn't save such information if the type has a trivial destructor as a matter of Quality of Implementation.