assembly - Why is imul used for multiplying unsigned numbers?

Question

I compiled the following program:

#include <stdint.h>

uint64_t usquare(uint32_t x) {
  return (uint64_t)x * (uint64_t)x;
}

This disassembles to:

 0: 89 f8                   mov    eax,edi
 2: 48 0f af c0             imul   rax,rax
 6: c3                      ret  

But imul is the instruction for multiplying signed numbers. Why is it used by gcc then?

/edit: when using uint64_t the assembly is similar:

0:  48 0f af ff             imul   rdi,rdi
4:  48 89 f8                mov    rax,rdi
7:  c3                      ret  

Answer 1

TL:DR: because it's a faster way of getting the correct result when we don't care about the high half (i.e. the output is only as wide as the 2 inputs). And more flexible register-allocation instead of forced use of RAX and RDX.

If it wasn't usable for this, Intel probably would have added two-operand versions of mul as well. But that wasn't necessary, as this answer explains.

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WARNING This answer is long!

... and it's full of unneeded explanations - but I have always wanted to write something more lengthy about the multiplication.

A bit of theory

When multiplying two number a and b of length n the result is of length 2 n^? and, most importantly, the k-th digit only depends on the lowest k digits (a proof is given in Appendix A).

x86 imul's two forms

The x86 multiplication instruction imul comes in two form: the full form and the partial form.

The first form is of the kind n×n→2 n, meaning that it produces a result twice the size of the operands - we know from the theory why this makes sense.
For example

imul ax         ;16x16->32, Result is dx:ax
imul rax        ;64x64->128, Result is rdx:rax 

The second form is of the kind n×n→n, this necessarily cut out some information.
Particularly, this form takes only the lower n bits of the result.

imul ax, ax          ;16x16->16, Lower WORD of the result is ax
imul rax, rax        ;64x64->64, Lower QWORD of the result is rax 

Only the single operand version is of the first form.

(There's also a 3-operand form, imul r64, r/m64, imm8/32, which allows you to copy-and-multiply by a constant in one instruction. It has no implicit operands and again doesn't write the high half anywhere so we can just treat it as equivalent to the imul r64, r/m64 dst *= src form.)

The two instructions: imul vs mul

Regardless of the form used, the processor always calculates the result with a size twice the operands' (i.e. like the first form).
In order to be able to do that, the operands are first converted from their size n to size 2 n (e.g. from 64 to 128 bits).
See Appendix B for more on this.

The multiplication is done and the full, or partial, result is stored in the destination.

The difference between imul and mul is in how the operands are converted.
Since the size is extended, this particular type of conversion is called extension.

The mul instruction simply fills the upper part with zeros - it zero extends.
The imul instruction replicate the high-order bit (the first from the left) - this is called sign extension and it has the interesting property of transforming a two's complement signed number of n bits into a signed number of 2 n bits with the same sign and modulus (i.e. it does the right thing, it is left to the reader to found a counter-example for the zero-extension case).

     How mul extends              How imul extends       
       and operand                  and operand

     +----+       +----+          +----+       +----+
     |0...|       |1...|          |0...|       |1...|
     +----+       +----+          +----+       +----+  

+----+----+  +----+----+     +----+----+  +----+----+
|0000|0...|  |0000|1...|     |0000|0...|  |1111|1...|
+----+----+  +----+----+     +----+----+  +----+----+

The thesis

The difference between imul and mul is noticeable only from the (n+1)-th bit onward.
For a 32-bit operand, it means that only the upper 32-bit part of the full result will eventually be different.

This is easy to see as the lower n bits are the same for both instructions and as we know from the theory the first n bits of the result only depends on the first n bits of the operands.

Thus the thesis: The result of the partial form of imul is identical to that of mul.

Then why imul exits?

Original 8086 only had one-operand versions of mul and imul. Later versions of x86 added more flexible two and three operand versions of imul only, intended for the common use-case where you don't want the double-width result.

They only write one output register, which for modern x86 means they can decode to a single uop: https://agner.org/optimize/. (In modern x86 microarchitectures, each uop can write at most register.) One-operand imul r32 is 3 uops on Intel CPUs: presumably one to multiply, another to split the 64-bit product into 2 halves and write the low half, and another to do the same for the high half. imul r64 is 2 uops; presumably the 128-bit result comes out of the multiplier already split in 64-bit halves.

mul still only exists in the very ancient one-operand form with fixed registers as part of the interface.

imul sets the flags according to a signed multiplication - CF and OF are set if the partial result has discarded any significant information (the technical condition being: the sign extension of the partial result is different from the full result) such in case of overflow.
This is also why the two and three operand forms are not called mul, which otherwise would have been a perfectly fit name.

The practice

To test all this in practice we can ask a compiler^[live] for the assembly of the following program

#include <stdint.h>

uint64_t foo(uint32_t a)
{
    return a*(uint64_t)a;
}

While we know that for 64-bit target the code generated uses imul because a unint64_t fits a register and thus a 64×64→64 multiplication is available as imul <reg64>, <reg64>

foo(unsigned int):
        mov     eax, edi        ;edi = a
        imul    rax, rax        ;64x64->64
        ret

in 32-bit code there is no such multiplication using imul.
A imul <reg32> or imul <reg32>, <reg32>, <reg32> is necessary but that would produce a full result! And a full signed result is not generally equal to a full unsigned result.
Infact, the compiler reverts back to mul:

foo(unsigned int):
        mov     eax, DWORD PTR [esp+4]
        mul     eax
        ret

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Appendix A

Without loss of generality, we can assume base 2 and that the numbers are n + 1 bits long (so that the indices run from 0 to n) - then

c = a·b = ∑_i=0..n (a_i·2ⁱ) · ∑_j=0..n(b_j·2^j) = ∑_i=0..n [a_i·∑_j=0..n (b_j·2^i+j)] (by the distributive property)

we see that the k-th digit of the result is the sum of all the addends such that i + j = k plus an eventual carry

c_k = ∑_{i,j=0..n; i+j=k} a_i·b_j·2^i+j + C_k

The term C_k is the carry and, as it propagates towards higher bits, it depends only on the lower bits.
The second term cannot have a a_i or b_j with i or j > k as if the first were true then i = k + e, for a positive, non null, e and thus j = k - i = k - k -e = -e
But j cannot be negative!
The second case is similar and left to the reader.

Appendix B

As BeeOnRope pointed out in the comments the processor probably doesn't compute a full result if only the partial result is needed.

You probably means that this is only a way of thinking about it, conceptually. The processor does not necessarily do a full 128-bit multiplication when you use the 64x64 -> 64 form. Indeed, the truncated form takes only 1 uop on recent Intel, but the full form takes 2 uops, so some extra work is being done

Comment from BeeOnRope

Also, the sign extension is probably conceptually too

Similarly the sign extension may happens "conceptually", but probably not in hardware. They won't have the extra wires and transistors just to do the sign or zero extension, which would add a lot of bulk to an already huge multiplier, but will use some other tricks to do the multiplication "as if" that had happened.

Comment from BeeOnRope

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^? Binary numbers of length n are in the order of magnitude of 2ⁿ, thus the multiplication of two such numbers is in the order of magnitude 2ⁿ · 2ⁿ = 2ⁿ⁺ⁿ = 2^{2 n}. Just like a number of length 2 n.