Now that we grasp what each instruction does let’s analyze the code line by line

Line 1: neg rax

This computes the two’s complement, places the result into rax and updates the Carry Flag (CF) accordingly.

Example 1: Suppose the original value from rax on line 1 was 101 neg rax ; rax is now = !rax+1 = (!101)+1 = 010+1 = 011

; CF=1

Example 2: Suppose the original value from rax on line 1 was 000 neg rax ; rax is now = !rax+1 = (!000)+1 = 111+1 = 000

; CF=0

Line 2: sbb rax, rax

This op performs subtraction between rax and itself + the CF (which is 0 if the original value from rax was 0, or 1 otherwise).

Following Example 1 (original rax value was 101): sbb rax, rax ; This can be expressed as:

; 1. 011 - (011+CF) CF=1 in this case, from prev op

; 2. 011 - (011+1) <--|

; 3. 011 - 100

; 4. 111 A borrow occurred here

Following Example 2 (original rax value was 000): sbb rax, rax ; This can be expressed as:

; 1. 000 - (000+CF) CF=0 in this case, from prev op

; 2. 000 - (000+0) <--|

; 3. 000 - 000

; 4. 000

Line 3: neg rax

Same as in line 1, of course, but just with updated values from previous ops:

Following Example 1 (original rax value was 101):

neg rax ; rax is now = !rax+1 = (!111)+1 = 000+1 = 001

Following Example 2 (original rax value was 000):

neg rax ; rax is now = !rax+1 = (!000)+1 = 111+1 = 000

Conclusion

As a final conclusion, we can say that this piece of code sets rax to 0 if its original value was 0 or to 1 if it was otherwise. I can imagine this to be a sort of integer to boolean conversion. In a high-level programming language this could be expressed as: