Difference between global operator new and malloc

2019-01-13 18:41发布

C++ has several functions to acquire dynamic storage, most of which differ in some fundamental way. Several more are usually added by the OS.

Two of these are of special interest due to their portability and similarity: malloc and ::operator new.

Are there any differences (w.r.t. the standard and implementation) between the global void* operator new(size_t, ::std::nothrow&) and void* malloc(size_t)?

Since there seems to be some confusion what I am talking about, consider the following two calls:

void* p = ::std::malloc(10);
void* q = ::operator new(10, ::std::nothrow);

The obvious and trivial difference is how to deallocate the memory:

::std::free(p);
::operator delete(q);

Note: This question is not a duplicate of e.g. What is the difference between new/delete and malloc/free? since it talks about using the global operator new that does not actually perform any ctor calls.

3条回答
别忘想泡老子
2楼-- · 2019-01-13 18:56

There are two differences I can think of:

  1. Which function you must use to deallocate the memory, operator delete vs. free().

  2. A C++ program can legally provide its own version of ::operator new and that version is guaranteed to be called by new expressions. It's not possible to override malloc with your own version.

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走好不送
3楼-- · 2019-01-13 19:11

The macroscopic difference I can see without further research is that the throwing variant of the global ::new operator throws std::bad_alloc if the allocation cannot be done, whereas malloc returns NULL. But I do believe that most of the differences listed here apply to the global ::new operator, even if the topic is about new.

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该账号已被封号
4楼-- · 2019-01-13 19:15

The main differences, aside from syntax and free vs. delete, are

  1. you can portably replace ::operator new;
  2. malloc comes with realloc, for which new has no equivalent;
  3. new has the concept of a new_handler, for which there is no malloc equivalent.

(Replacing malloc opens up a can of worms. It can be done, but not portably, because it requires knowledge of the linker.)

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