signed and unsigned arithmetic implementation on x86

Question

C language has signed and unsigned types like char and int. I am not sure, how it is implemented on assembly level, for example it seems to me that multiplication of signed

Answer 1

A little supplement for cmp and sub. We know cmp is considered as non-destructive sub, so let's focus on sub.

When a x86 cpu does a sub instruction, for example,

sub eax, ebx

How does the cpu know if either values of eax or ebx are signed or unsigned? For example, consider a 4 bit width number in two's complement:

eax: 0b0001
ebx: 0b1111

In either signed or unsigned, value of eax will be interpreted as 1(dec), which is fine.

However, if ebx is unsigned, it will be interpreted as 15(dec)， result becomes:

ebx:15(dec) - eax: 1(dec) = 14(dec) = 0b1110 (two's complement)

If ebx is signed, then results becomes:

ebx: -1(dec) - eax: 1(dec) = -2(dec) = 0b1110 (two's complement)

Even though for both signed or unsigned, the encode of their results in two's complement are same: 0b1110.

But one is positive: 14(dec), the other is negative: -2(dec), then comes back our question: how does the cpu tell which to which?

The answer is the cpu will evaluate both, from: http://x86.renejeschke.de/html/file_module_x86_id_308.html

It evaluates the result for both signed and unsigned integer operands and sets the OF and CF flags to indicate an overflow in the signed or unsigned result, respectively. The SF flag indicates the sign of the signed result.

For this specific example, when the cpu sees the result: 0b1110, it will set the SF flag to 1, because it's -2(dec) if 0b1110 is interpreted as a negative number.

Then it depends on the following instructions if they need to use the SF flag or simply ignore it.

Answer 2

Most of the modern processors support signed and unsigned arithmetic. For those arithmetic which is not supported, we need to emulate the arithmetic.

Quoting from this answer for X86 architecture

Firstly, x86 has native support for the two's complement representation of signed numbers. You can use other representations but this would require more instructions and generally be a waste of processor time.

What do I mean by "native support"? Basically I mean that there are a set of instructions you use for unsigned numbers and another set that you use for signed numbers. Unsigned numbers can sit in the same registers as signed numbers, and indeed you can mix signed and unsigned instructions without worrying the processor. It's up to the compiler (or assembly programmer) to keep track of whether a number is signed or not, and use the appropriate instructions.

Firstly, two's complement numbers have the property that addition and subtraction is just the same as for unsigned numbers. It makes no difference whether the numbers are positive or negative. (So you just go ahead and ADD and SUB your numbers without a worry.)

The differences start to show when it comes to comparisons. x86 has a simple way of differentiating them: above/below indicates an unsigned comparison and greater/less than indicates a signed comparison. (E.g. JAE means "Jump if above or equal" and is unsigned.)

There are also two sets of multiplication and division instructions to deal with signed and unsigned integers.

Lastly: if you want to check for, say, overflow, you would do it differently for signed and for unsigned numbers.

Answer 3

If you look at the various multiplication instructions of x86, looking only at 32bit variants and ignoring BMI2, you will find these:

• imul r/m32 (32x32->64 signed multiply)
• imul r32, r/m32 (32x32->32 multiply) *
• imul r32, r/m32, imm (32x32->32 multiply) *
• mul r/m32 (32x32->64 unsigned multiply)

Notice that only the "widening" multiply has an unsigned counterpart. The two forms in the middle, marked with an asterisk, are both signed and unsigned multiplication, because for the case where you don't get that extra "upper part", that's the same thing.

The "widening" multiplications have no direct equivalent in C, but compilers can (and often do) use those forms anyway.

For example, if you compile this:

uint32_t test(uint32_t a, uint32_t b)
{
    return a * b;
}

int32_t test(int32_t a, int32_t b)
{
    return a * b;
}

With GCC or some other relatively reasonable compiler, you'd get something like this:

test(unsigned int, unsigned int):
    mov eax, edi
    imul    eax, esi
    ret
test(int, int):
    mov eax, edi
    imul    eax, esi
    ret

(actual GCC output with -O1)

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So signedness doesn't matter for multiplication (at least not for the kind of multiplication you use in C) and for some other operations, namely:

• addition and subtraction
• bitwise AND, OR, XOR, NOT
• negation
• left shift
• comparing for equality

x86 doesn't offer separate signed/unsigned versions for those, because there's no difference anyway.

But for some operations there is a difference, for example:

• division (idiv vs div)
• remainder (also idiv vs div)
• right shift (sar vs shr) (but beware of signed right shift in C)
• comparing for bigger than / smaller than

But that last one is special, x86 doesn't have separate versions for signed and unsigned of this either, instead it has one operation (cmp, which is really just a nondestructive sub) that does both at once, and gives several results (multiple bits in "the flags" are affected). Later instructions that actually use those flags (branches, conditional moves, setcc) then choose which flags they care about. So for example,

cmp a, b
jg somewhere

Will go somewhere if a is "signed greater than" b.

cmp a, b
jb somewhere

Would go somewhere if a is "unsigned below" b.

See Assembly - JG/JNLE/JL/JNGE after CMP for more about the flags and branches.

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This won't be a formal proof that signed and unsigned multiplication are the same, I'll just try to give you insight into why they should be the same.

Consider 4-bit 2's-complement integers. The weights their individual bits are, from lsb to msb, 1, 2, 4, and -8. When you multiply two of those numbers, you can decompose one of them into 4 parts corresponding to its bits, for example:

0011 (decompose this one to keep it interesting)
0010
---- *
0010 (from the bit with weight 1)
0100 (from the bit with weight 2, so shifted left 1)
---- +
0110

2 * 3 = 6 so everything checks out. That's just regular long multiplication that most people learn in school, only binary, which makes it a lot easier since you don't have to multiply by a decimal digit, you only have to multiply by 0 or 1, and shift.

Anyway, now take a negative number. The weight of the sign bit is -8, so at one point you will make a partial product -8 * something. A multiplication by 8 is shifting left by 3, so the former lsb is now the msb, and all other bits are 0. Now if you negate that (it was -8 after all, not 8), nothing happens. Zero is obviously unchanged, but so is 8, and in general the number with only the msb set:

-1000 = ~1000 + 1 = 0111 + 1 = 1000

So you've done the same thing you would have done if the weight of the msb was 8 (as in the unsigned case) instead of -8.