Can new operator be used as macro?

[JeanDean]

For some debugging purpose I want to use overloaded new instead of
existing MyAlloc() in a huge code base.
But below macro doesnt work.
Is there any way to pass "size" to overloaded new ?
#define MyAlloc(size) new(size, WORK_HEAD )

overloaded new code is :
void* operator new(size_t sz, int headType) {
..
..
..
}

[John Harrison]

JeanDean wrote:

For some debugging purpose I want to use overloaded new instead of
existing MyAlloc() in a huge code base.
But below macro doesnt work.
Is there any way to pass "size" to overloaded new ?
#define MyAlloc(size) new(size, WORK_HEAD )

overloaded new code is :
void* operator new(size_t sz, int headType) {
.
.
.
}

Use operator new (which is a different thing from the new operator), but
why use a macro?

inline void* MyAlloc(size_t size)
{
return ::operator new(size, WORK_HEAD);
}

Look's OK to me, unless I'm misunderstanding you.

john

[JeanDean]

On Mar 2, 4:59 pm, John Harrison <[e-mail]:

JeanDean wrote:

For some debugging purpose I want to use overloaded new instead of
existing MyAlloc() in a huge code base.
But below macro doesnt work.
Is there any way to pass "size" to overloaded new ?

#define MyAlloc(size) new(size, WORK_HEAD )

overloaded new code is :
void* operator new(size_t sz, int headType) {
.
.
.
}

Use operator new (which is a different thing from the new operator), but
why use a macro?

inline void* MyAlloc(size_t size)
{
return ::operator new(size, WORK_HEAD);

}

Look's OK to me, unless I'm misunderstanding you.

john

Thanks a lot for the suggestion