Use of TEST instruction with same register

Question

Here is some C, found in a textbook I'm learning:

...
do {
    ...
    n--;
} while (n > 0)
...

I assume n is at %edx.

The assembly code produced is:

testl %edx, %edx 
jle .L5

I understand that jle tests for less than or equal to (SF ^ OF) | ZF. However I am unsure how this instruction corresponds to n > 0. Can anyone explain this?

possible duplicate of [x86 Assembly - 'testl' eax against eax?](http://stackoverflow.com/questions/147173/x86-assembly-testl-eax-against-eax) — interjay, Jun 30 '13 at 12:21
I do not believe so, he is using a different jump instruction. — dgamma3, Jun 30 '13 at 12:29
It's still a comparison with 0, you can use any jump instruction after it. — interjay, Jun 30 '13 at 12:33
@interjay but the explanation given there is not complete in this case. — harold, Jun 30 '13 at 12:34

lurker · Accepted Answer · 2013-06-30T20:18:40.493

Some of this has been covered, but I'll fill in a little more detail.

The general purpose of the test reg,mask instruction tests a register value against a mask (register value is internally ANDed with mask) and then sets the status flags SF, ZF, and PF according to the result. [EDIT per comment from @ChrisDodd] It also unconditionally clears the O (overflow) and C (carry) status bits.[/EDIT]

SF = sign flag (1 if sign bit is set (a 2's complement negative value))
ZF = zero flag (1 if result is 0)
PF = parity flag (1 if result has an even number of 1 bits)

In this specific example (test eax,eax), the instruction is taking eax AND eax. When this is done, the bits will be:

SF = 1 if EAX has a negative value (since sign bit will not change when ANDed with itself)
ZF = 1 if EAX is zero (the only value that yields a zero when ANDed with itself is zero)
PF = 1 if EAX has an even number of 1 bits

In other words, it's a simple test for zero or negative. It's a very common pattern in compiler code generation.

A critical additional fact is that the `test` instruction unconditionally clears the O and C flags. Without that it would not be useful for comparisons against zero. — Chris Dodd, Jun 30 '13 at 19:57
@ChrisDodd good info (+1). Thanks. If you don't mind, I shall add that to my answer for completeness. — lurker, Jun 30 '13 at 20:16

score 1 · Answer 2 · answered Jun 30 '13 at 12:34

TEST "sets the SF, ZF, and PF status flags according to the result." (Intel manual, about TEST).

So SF will reflect whether n was negative, and ZF will reflect whether n was zero.

It sets OF to zero.

So (SF ^ OF)|ZF simplifies to SF | ZF, so in conclusion, the jump will be taken if n <= 0. That seems the wrong way around, so hopefully .L5 is the label after the loop, not the label in front of the loop.

Use of TEST instruction with same register

2 Answers2