Solaris libumem why not show memory leak for first dynamic allocation - debugging

Question

Say
void main()
{
void *buff;
buff = malloc(128);
buff = malloc(60);
buff = malloc(30);
buff = malloc(16);
free(buff);
sleep(180);
}
ulib mem in solaris10 show only 60 bytes and 30 bytes as leak , why it not show 128 bytes also leaked?

Answer 1

Three memory leaks are detected by mdb and libumem with your code here:
cat > leak.c <<%
int main() { void *buff; buff = malloc(128); buff = malloc(60); buff = malloc(30); buff = malloc(16); free(buff); sleep(180); }
%
gcc -g leak.c -o leak
pkill leak
UMEM_DEBUG=default UMEM_LOGGING=transaction LD_PRELOAD=libumem.so.1 leak &
sleep 5
rm -f core.leak.*
gcore -o core.leak $(pgrep leak)
mdb leak core.leak.* <<%
::findleaks -d
%
gcore: core.leak.1815 dumped
CACHE LEAKED BUFCTL CALLER
0807f010 1 0808b4b8 main+0x29
0807d810 1 0808d088 main+0x39
0807d010 1 08092cd0 main+0x49
------------------------------------------------------------------------
Total 3 buffers, 280 bytes
umem_alloc_160 leak: 1 buffer, 160 bytes
ADDR BUFADDR TIMESTAMP THREAD
CACHE LASTLOG CONTENTS
808b4b8 8088f28 1af921ab662 1
807f010 806c000 0
libumem.so.1`umem_cache_alloc_debug+0x144
libumem.so.1`umem_cache_alloc+0x19a
libumem.so.1`umem_alloc+0xcd
libumem.so.1`malloc+0x2a
main+0x29
_start+0x83
umem_alloc_80 leak: 1 buffer, 80 bytes
ADDR BUFADDR TIMESTAMP THREAD
CACHE LASTLOG CONTENTS
808d088 808cf68 1af921c11eb 1
807d810 806c064 0
libumem.so.1`umem_cache_alloc_debug+0x144
libumem.so.1`umem_cache_alloc+0x19a
libumem.so.1`umem_alloc+0xcd
libumem.so.1`malloc+0x2a
main+0x39
_start+0x83
umem_alloc_40 leak: 1 buffer, 40 bytes
ADDR BUFADDR TIMESTAMP THREAD
CACHE LASTLOG CONTENTS
8092cd0 808efc8 1af921f2bd2 1
807d010 806c0c8 0
libumem.so.1`umem_cache_alloc_debug+0x144
libumem.so.1`umem_cache_alloc+0x19a
libumem.so.1`umem_alloc+0xcd
libumem.so.1`malloc+0x2a
main+0x49
_start+0x83