How to compare a signed value and an unsigned value in x86 assembly
Solution 1
You're probably using one of the unsigned variants like:
cmp eax, ebx
jb lesser
There are equivalents for checking signed numbers against each other, such as:
cmp eax, ebx
jl lesser
This link gives a good run down on the jump variations, including their signed-ness and the flags they check, partially copied here for self-containment:
Instruction Jump if ... Signed? Flags
----------- ----------- -------- -----
JO overflow OF=1
JNO not overflow OF=0
JS sign SF=1
JNS not sign SF=0
JE/JZ equal
zero ZF=1
JNE/JNZ not-equal
not-zero ZF=0
JB/JNAE/JC below
not-above-or-equal
carry unsigned CF=1
JNB/JAE/JNC not-below
above-or-equal
no-carry unsigned CF=0
JBE/JNA below-or-equal
not-above unsigned CF=1 or ZF=1
JA/JNBE above
not-below-or-equal unsigned CF=0 and ZF=0
JL/JNGE less
not-greater-or-equal signed SF<>OF
JGE/JNL greater-or-equal
not-less signed SF=OF
JLE/JNG less-or-equal
not-greater signed ZF=1 or SF<>OF
JG/JNLE greater
not-less-or-equal signed ZF=0 and SF=OF
JP/JPE parity
parity-even PF=1
JNP/JPO not-parity
parity-odd PF=0
JCXZ/JECXZ CX register is zero
ECX register is zero
Solution 2
You can't directly compare two numbers that have different signs. Actually most software languages have that flow. C and C++ specifically mention that in their documentation and in most cases will generate a warning when you use a signed and an unsigned integer in the same expression which then can result in an unknown sign.
The only way is to first check whether the signed number is negative, if so then you know it is smaller. Then you can compare the two numbers as unsigned integers.
; is eax < ebx (eax signed, ebx unsigned)
cmp eax, $0
jl less
cmp eax, ebx
jc less
Side note: it is obviously possible to compare two numbers signed if their size is less than the maximum size supported by the processor. In that case you extend the bits of the signed and unsigned appropriately, then you can compare as if both values were signed.
Assuming you wanted to compare two bytes al and bl, then you could have something like this:
movsx ax, al
xor bh, bh ; or movzx bx, bl
cmp ax, bx
jl less
(Note, I do not guarantee that jl is correct, it may be jle or jnl...)
Roy Li
Updated on June 08, 2022Comments
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Roy Li almost 2 years
I am having trouble finding a way to compare a positive number and a negative number in x86 assembly code.
For example: when I compare -1 and 1 I always get -1 as greater. I know that it's because 2's complement format makes the -1 bigger than 1 in underlying binary.
But can anyone provide a snippet of x86 assembly to compare positive number with a negative one and get it mathematically correct? (e.g 1 > -1)
Thanks!