0x03

0x03

[dec] [home] [inc]

[See this riddle on the book]

sub      rdx,rax
sbb      rcx,rcx
and      rcx,rdx
add      rax,rcx

See explanation

min(a, b)

This snippet finds the minimum between the two values stored in rax and rdx and stores the result in rcx. This is a branchless implementation, and as such it offers performance benefits on architectures with prefetching and branch prediction (like modern x86 CPUs). SSE introduces dedicated instructions, see the section "Unsigned integer compares" in the SSE: mind the gap post.

How does this work?

sub rdx,rax subtracts the value contained in rax from the value contained in rdx, and stores the result in rdx. It also sets CF (carry flag) to 1 if there was carry, or in other words if rax is greater than rdx.

sbb rcx,rcx sets all the bits of rcx to the value of CF after the preceding instruction. In other words it is set to either all zeros, or all ones. We have seen how sbb does it in the previous riddle.

and rcx,rdx sets rcx to 0 if CF was zero, or to the current value of rdx if CF was 1, using rcx as a mask, with the value that was set at the previous step. Remember that at this stage rdx contains the difference between the original values of rdx and rax, this will be relevant in the next instruction.

add rax,rcx sums rax and rcx and sets the result into rax. At this point rcx contains either the initial value of rax, or the value of rdx as seen above, so adding it up yields either the initial value of rax, or the initial value of rdx

Examples

This was the theory, but let's also see the practice with two examples.

Assume that the initial value of rax is 3, and the initial value of rdx is 5. As said above this snippet computes the minimum of the two values, and puts it in eax.

• step 1: sub rdx,rax. This sets rdx to rdx - rax = 5 - 3 = 2, and the carry flag CF is set to 0. Now we have rax = 3 and rdx = 2
• step 2: sbb rcx,rcx. This sets rcx to 0, because CF is 0
• step 3: and rcx,rdx. This uses rcx as a mask for rdx, in fact leaving rcx set to 0
• step 4: add rax,rcx. This sets rax to the sum of the two operands, i.e. 3 + 0 = 3, that is the original value of rax.

Let's see the second example. We assume that rax is set to 5 and rdx is set to 3. This will trigger the opposite behaviour of the previous example, so we expect that the final value in rax will be the initial value of rdx, 3.

• step 1: sub rdx,rax. This sets rdx to rdx - rax = 3 - 5 = -2, and the carry flag CF is set to 1. Now we have rdx = -2 (or 0xfffffffffffffffe), and rax = 5
• step 2: sbb rcx,rcx. This sets rcx to 0xffffffffffffffff, because CF is 1
• step 3: and rcx,rdx. This uses rcx as a mask for rdx, in fact setting rcx to the value of rdx, that is 3
• step 4: add rax,rcx. This sets rax to the sum of the two operands, i.e. 5 + (-2) = 3, that is the original value of rdx.

[dec] [home] [inc]