Malloc, return address order.

[Manu]

Hello,

Can we say that the return addresses from the various malloc function
calls, in a program, will always be in a predefined order (increasing
or decreasing, depeding on how the heap is managed) ?

regards
Manu

[Keith Thompson]

"Manu" <ma************@gmail.com> writes:

Can we say that the return addresses from the various malloc function
calls, in a program, will always be in a predefined order (increasing
or decreasing, depeding on how the heap is managed) ?

No.

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Richard Tobin]

In article <11*********************@p10g2000cwp.googlegroups. com>,
Manu <ma************@gmail.com> wrote:

Can we say that the return addresses from the various malloc function
calls, in a program, will always be in a predefined order (increasing
or decreasing, depeding on how the heap is managed) ?

No. Even supposing your implementation happened to allocate memory
sequentially (which is not required, and many implementations don't),
what do you think would happen after you freed some memory? The point
of freeing it is so that it can be used again.

-- Richard

[Nils O. Selåsdal]

Manu wrote:

Hello,

Can we say that the return addresses from the various malloc function
calls, in a program, will always be in a predefined order (increasing
or decreasing, depeding on how the heap is managed) ?

Not at all.

[Rod Pemberton]

"Keith Thompson" <ks***@mib.org> wrote in message
news:ln************@nuthaus.mib.org...

"Manu" <ma************@gmail.com> writes:

Can we say that the return addresses from the various malloc function
calls, in a program, will always be in a predefined order (increasing
or decreasing, depeding on how the heap is managed) ?

No.

What no OT complaint about 'heap'?

Specifically, memory paging and memory or heap reuse almost guarantees that
the answer is "No."
Rod Pemberton