Hi Ankur,
If pointer to allocated buffer manages to survive in one of the
registers or on stack (after program returns from main), LSan would
consider buffer to be reachable and won't report it. This behavior is
very sensitive to compiler/Glibc version and compilation flags.
On 2/15/19, Ankur Raj <ankur...@gmail.com> wrote:
> Hi Folks ,
> Wondering if someone can explain this
> Adress sanitizer does not report and leak in this code below when compiled
> with O1 , O2 ,,,
>
> But it works as expected when compiled with O0
>
> #include <stdio.h>
> #include <stdlib.h>
>
> int main () {
> volatile char *str;
>
> /* Initial memory allocation */
> str = (char *) malloc(25);
> strcpy(str, "sameple st");
> printf("String = %s, Address = %u\n", str, str);
>
> strcat(str, "append");
> printf("String = %s, Address = %u\n", str, str);
>
>
> return(0);
> }
>
> I am compiling this code as
>
> gcc -fsanitize=address -O0 a.c
> gcc --version
> gcc (GCC) 8.2.1 20190102
>
> I looked at asm code also and clearly malloc has not been removed by
> optimizers
>
> 22 .cfi_startproc
> 23 pushq %rbp
> 24 .cfi_def_cfa_offset 16
> 25 .cfi_offset 6, -16
> 26 movq %rsp, %rbp
> 27 .cfi_def_cfa_register 6
> 28 subq $16, %rsp
> 29 movl $25, %edi
> 30 *call malloc*
> 31 movq %rax, -8(%rbp)
> 32 movq -8(%rbp), %rax
> 33 movl $11, %edx
> 34 movl $.LC0, %esi
> 35 movq %rax, %rdi
> 36 call memcpy
>
> Any clue what is happening here ?
>
>
>
>
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Best regards,
Yuri
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