Understanding Windows stack layout - c

Question

I am following this buffer overflow tutorial: https://insecure.org/stf/smashstack.html
I want to make this program work in Windows
#include <stdio.h>
void f(int x, int y)
{
char buffer1[5];
char buffer2[10];
int *ret = buffer1 + 12;
(*ret) += 7;
}
int main()
{
int x;
x = 10;
f(1, 2);
x = 21;
printf("%d\n", x);
return 0;
}
This program attempts to modify the return address of function f so that this line x = 21; is ignored in main (ie. the program jumps directly to executing printf).
For some reason this trivial buffer overflow attack didn't work in Windows and I am not sure why.
This is how I understand the stack layout after the function f is called in x64 machines
high address
...
return address (4 bytes)
saved frame pointer (4 bytes)
1 (first argument for f; 4 bytes)
2 (second argument for f; 4 bytes)
buffer1 (1*8=8 bytes)
buffer2 (1*12=12 bytes)
...
low address
Using gdb, I get the following disassembly
Dump of assembler code for function main:
0x0000000000401590 <+0>: push %rbp
0x0000000000401591 <+1>: mov %rsp,%rbp
0x0000000000401594 <+4>: sub $0x30,%rsp
0x0000000000401598 <+8>: callq 0x401690 <__main>
0x000000000040159d <+13>: movl $0xa,-0x4(%rbp)
0x00000000004015a4 <+20>: mov $0x2,%edx
0x00000000004015a9 <+25>: mov $0x1,%ecx
0x00000000004015ae <+30>: callq 0x401560 <f>
0x00000000004015b3 <+35>: movl $0x15,-0x4(%rbp)
0x00000000004015ba <+42>: mov -0x4(%rbp),%eax
0x00000000004015bd <+45>: mov %eax,%edx
0x00000000004015bf <+47>: lea 0x2a3a(%rip),%rcx # 0x404000
0x00000000004015c6 <+54>: callq 0x402b70 <printf>
0x00000000004015cb <+59>: mov $0x0,%eax
0x00000000004015d0 <+64>: add $0x30,%rsp
0x00000000004015d4 <+68>: pop %rbp
0x00000000004015d5 <+69>: retq
End of assembler dump.
So in order to get return address in the stack, I must add 4+4+4=12 bytes. From the disassembly, in order to skip x = 21;, I must skip movl $0x15,-0x4(%rbp). So I need to add the additional 42-35=7 bytes to the return address.
Is there anything wrong with my understanding so far?