Hmm, things seems to be more suble :-)
I have tried a trick with using int *a; fun(int* &a); and it indeed
works but the address is trimmed no matter what type "a" is (int*,
long*, void*) and this causes problems.
21 int a1=0, a2=0;
(gdb) s
22 double *a3;
(gdb)
23 a3=NULL;
(gdb)
f1 (pf11=0x7fffffffe71c, pf12=0x7fffffffe718, pf13=0x7fffffffe710) at
pointertest.c:17
17 return f2(pf11, pf12, pf13);
(gdb)
f2 (pf21=0x7fffffffe71c, pf22=0x7fffffffe718, pf23=0x7fffffffe710) at
pointertest.c:6
6 stuff=calloc(1,sizeof(int));
(gdb)
7 if (!stuff) exit(-1);
(gdb)
8 *stuff=0xDEADBEEF;
(gdb)
9 printf("stuff[@%X]=%X\n", stuff, *stuff);
(gdb) p/x stuff
$1 = 0x800a04090
(gdb) p/x *stuff
$2 = 0xdeadbeef
(gdb) s
stuff[@A04090]=DEADBEEF
10 *pf21=stuff;
(gdb)
11 *pf22=*stuff;
The most important lines below:
(gdb)
12 *pf23=&stuff;
(gdb)
13 return 0;
(gdb) p/x pf23
$3 = 0x7fffffffe710
(gdb) p/x *pf23
$4 = 0xffffe6b0
(gdb) p/x &stuff
$5 = 0x7fffffffe6b0
&stuff is the address of the memory location. I want to write that
location to my pointer, but its trimmed although variable IS a
pointer. Maybe this is because of a type....
The most interesting part is the "*variable=value" construct because
without this we cannot get data out of the function. Even if we change
the address contained within a pointer variable we will get the data
from an old address!!! Note that below I did not use *variable=value
but simple variable=value, variable changed but it still points to the
old memory location (0x00 instead 0xdeadbeef)
(gdb) p/x &stuff
$1 = 0x7fffffffe6b0
(gdb) p/x stuff
$2 = 0x800a04090
(gdb) p/x *stuff
$3 = 0xdeadbeef
(gdb) s
stuff[@A04090]=DEADBEEF
10 *pf21=stuff;
(gdb)
11 *pf22=*stuff;
(gdb)
12 pf23=&stuff;
(gdb)
13 return 0;
(gdb) p/x &pf23
$4 = 0x7fffffffe698
(gdb) p/x pf23
$5 = 0x7fffffffe6b0
(gdb) p/x *pf23
$6 = 0x0
I think this is some memory corruption protection mechanism
implemented by a compiler...
Well I have to live with double pointers ;-P
--
CeDeROM, SQ7MHZ, http://www.tomek.cedro.info
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