Q:malloc

From: Michael Schuster (schusterSoccer_at_enertex.de)
Date: 05/27/04 

Date: Thu, 27 May 2004 08:49:13 +0200

Hi,
I'm having a question referring malloc, refering to the docs of libc,
I found:
"The address of a block returned by `malloc' or `realloc' in the GNU
system is always a multiple of eight (or sixteen on 64-bit systems)"

Does this mean, if I compile:
        int *m1;
        m1=malloc(sizeof(int))
that the system occupates at least 8 (16 on a 64 Bit system) Words of
memory ?

So if my next instruction would be:
        int *m2;
        m2=malloc(sizeof(int))
my system would occupy already at least 16 (32) Bytes?

Also does the pointer allocation itself (int *m1 or int *m2) occupy
sizeof(int*) Bytes or at least 8 (or 16) Words?

I encountered the problem with my programm using a lot of void* - pointers
with only 1 element allocated with malloc. Due to my expirience (gcc (GCC)
3.3.3 (SuSE 9.1 Linux) ) with /proc/PID/status or top every malloc of an
void* needs at least 16 Bytes (or 4 32-Bit words). So in my programm the
system allocates the same space for
        int *m1;
        m1=malloc(sizeof(int))
and
        int *m1;
        m1=malloc(sizeof(int)*3)

Thanks for any comment!
Michael 

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