Malloc'd memory cannot be accessed - GDB

Question

I'm trying to call a library function to test something while debugging with GDB. (dlinfo())

It takes a reference to an int and the function writes the result to that address.

I tried calling malloc so I could pass a reference into the function, but when I try and see if malloc worked, I can never access the memory.

(gdb) p (int *) malloc(sizeof(int)) 
$8 = (int *) 0xffffffffa9e83f28
(gdb) p * (int *)0xffffffffa9e83f28 = 5                     
Cannot access memory at address 0xffffffffa9e83f28

Am I doing something wrong?

Thanks in advance

I guess the problem is that it doesn't know the prototype of malloc, therefore it will miscast the address — Antti Haapala -- Слава Україні, Dec 22 '17 at 21:13
https://stackoverflow.com/questions/5122570/why-does-gdb-evaluate-sqrt3-to-0 — Antti Haapala -- Слава Україні, Dec 22 '17 at 21:13
Same principle. GDB thinks the return value of `malloc` is an `int` and not `void *`. But the rest of the 64-bit pointer is lost when GDB coerces it to a 32-bit int. It is then sign-extended to 64-bits which results in a pointer that is in **kernel space**. — Antti Haapala -- Слава Україні, Dec 22 '17 at 21:17
ahhh ok, one sec i'll see if I can fix it, thanks. I thought the repeating FFF's were because I had a lot of free RAM, but I see it's half of them now — user2837329, Dec 22 '17 at 21:20
I fixed it, thanks. Here's the steps I used. (gdb) set $malloc = (void*(*)(long long)) malloc (gdb) set $mem = $malloc(4) (gdb) p $mem $1 = (void *) 0x5591420469b0 (gdb) p * (int *)0x5591420469b0 = 5 $2 = 5 (gdb) — user2837329, Dec 22 '17 at 21:21

Malloc'd memory cannot be accessed - GDB

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