Does ::operator new(size_t)
call malloc()
internally, or does it use system calls / OS-specific library calls directly? What does the C++ standard say?
In this answer it says that:
malloc()
is guaranteed to return an address aligned for any standard type.::operator new(n)
is only guaranteed to return an address aligned for any standard type no larger thann
, and ifT
isn't a character type then newT[n]
is only required to return an address aligned forT
.
And that suggests that new()
cannot be required to call malloc()
.
Note: There's an SO question about everything operator new
does other than allocation.
malloc(): It is a C library function that can also be used in C++, while the “new” operator is specific for C++ only. Both malloc() and new are used to allocate the memory dynamically in heap. But “new” does call the constructor of a class whereas “malloc()” does not.
Struct std::alloc::System The default memory allocator provided by the operating system. This is based on malloc on Unix platforms and HeapAlloc on Windows, plus related functions. It can also be used directly to allocate memory independently of whatever global allocator has been selected for a Rust program.
The malloc() function in C++ allocates a block of uninitialized memory to a pointer. It is defined in the cstdlib header file.
The operator new (or better the void* operator new(size_t) variant) just allocate memory, but does not do any object construction. The new keyword calls the operator new function, but then calls the object constructor.
The details of how operator new
is implemented are property of a particular implementation of standard library - not even a compiler or operation system. I am familiar with one (gnu) and aware of 3 others - CLang, Apache and MSFT. All of them are using malloc()
within operator new
, because it just makes a life of library developer so much easier.
If malloc()
were not used, said developer would have to reimplement a lot in terms of memory allocation, and sprinkle the code heavily with OS-dependent logic to actually request memory. No one wants to do this when malloc()
is already there. But by no means they are obliged to use it.
It can, and usually it does.
On Windows (more specificly on VC++), the chain of calls looks like
operator new
calls malloc
calls HeapAlloc
HeapAlloc
is a Windows API function in for allocating memory from specific heap.
when the process goes up, it allocate a heap (the CRT heap) in which all the standard allocation takes memory.
No, it isn't obligated to call malloc. it is up to the library developers/end user developer to decide from where they want their memory from.
For example, I can create a mono-threaded program. usually the heap allocator locks the heap lock when allocation/deallocation takes place, in order to prevent fatal race-condition on the heap.
but if my program is monothreaded, I don't have the problem.
I may choose to create my own heap with WinApi HeapCreate
and pass HEAP_NO_SERIALIZE
which makes the heap skip the lock. then I can use operator new
with plain HeapAlloc
. this is a case where I can make new
work with different function then malloc
.
Another low level approach which is sometimes* is done is to allocate huge memory block with VirtualAlloc
, then pass a re-calculated memory address anytime someone calls new
.
(all of these approches are done pretty rarely, and from my experiance they bring minimal improvment to the execution time)
Yes, it may call malloc - under windows with VS and standard runtime library it does call malloc
.
You are allowed to overload new operator and call your own allocation function. In application I work on, we have custom malloc from Doug Lea with lots of customizations for embeded systems. Windows calls malloc because it calls HeapAlloc, which is standard heap allocation function under windows. It also allows debugging allocation errors with CrtDbg api.
To make answer more formal I have looked up the standard and in §18.6.1.1, I found that new
Executes a loop: Within the loop, the function first attempts to allocate the requested storage. Whether the attempt involves a call to the Standard C library function malloc is unspecified.
so wheteher malloc is used is unspecified - it might use it or not.
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