How do canary words allow gcc to detect buffer overflows? - security

Question

I could test using strncpy() with larger source string then the destination:
int main() {
char *ptr = malloc(12);
strcpy(ptr,"hello world!");
return 0;
}
Compiling with the flag -fstack-protector and using the -S option I got:
.file "malloc.c"
.text
.globl main
.type main, #function
main:
.LFB2:
.cfi_startproc
pushq %rbp
.cfi_def_cfa_offset 16
.cfi_offset 6, -16
movq %rsp, %rbp
.cfi_def_cfa_register 6
subq $32, %rsp
movl %edi, -20(%rbp)
movq %rsi, -32(%rbp)
movq %fs:40, %rax
movq %rax, -8(%rbp)
xorl %eax, %eax
movq $0, -16(%rbp)
movl $12, %edi
call malloc
movq %rax, -16(%rbp)
movq -16(%rbp), %rax
movabsq $8022916924116329800, %rdx
movq %rdx, (%rax)
movl $560229490, 8(%rax)
movb $0, 12(%rax)
movl $0, %eax
movq -8(%rbp), %rcx
xorq %fs:40, %rcx
je .L3
call __stack_chk_fail
.L3:
leave
.cfi_def_cfa 7, 8
ret
.cfi_endproc
.LFE2:
.size main, .-main
Could someone explain to me how this works? And why isn't the "canary word" also overwritten by the \0 of the hello world! string?

Answer 1

Could someone explain to me how does this work ?
Canary word is read from fs:40 and store at top of frame here:
movq %fs:40, %rax
movq %rax, -8(%rbp)
It's below the return address so if your code happens to overflow the buffer (which will be below -8(%rbp)), it'll first overwrite the -8(%rbp) location. This will be detected by GCC prior to issuing ret here:
movq -8(%rbp), %rcx
xorq %fs:40, %rcx ; Checks that %fs:40 == -8(%rbp)
je .L3 ; Ok, return
call __stack_chk_fail ; Die
as overwritten contents of -8(%rbp) will likely to be different from proper value (installed from fs:40).
And why is not the canary word also overwritten by the \0 of the hello world!?
Your code has heap overflow, not buffer overflow so SSP can't help...