x86 中如何使用进位和溢出标志来计算乘法

当两个数字相乘溢出寄存器时，如何使用两个标志来正确计算答案？

E.g. if al holds 0xff并乘以0x2，导致溢出到ax，这些标志对此有何帮助？

Multiplication on x86/x64 never overflows when using the one operand form.
This is because mul and its sibling imul produce an output twice as wide as their operands¹.
In your example, multiplying by al produces an output in ax and no overflow is generated.

The CF and OF当结果无法适合操作数大小时设置。
这可用于执行饱和乘法，例如：

;Unsigned

mul ebx
sbb edx, edx      ;EDX = CF replicated along all the 32 bits
or eax, edx       ;EAX = 0ff..ffh if overflow, EAX*EBX otherwise

;Signed (perhaps not the most efficient way)

imul ebx
cmovc eax, 7fffffffh  ;Signed positive max if overflow.   (CMOV-immediate doesn't really exist, but imagine register sources)
cmovnc edx, 0         ; don't modify EAX for the non-overflow case.
sar   edx, 31         ; EDX = all-ones if overflow && negative
xor   eax, edx        ; if negative && overflow
                      ;    flip 7fffffff (INT_MIN) to 80000000 (INT_MIN)
                      ; else xor with 0 is a no-op

（当前的 Rust 编译器还实现a.saturating_mul(b)对于 i32 和 i64，使用 imul 中的标志，但设置不同：https://rust.godbolt.org/z/ab3jMjzbv)

However to implement a multi-precision multiplication, say 64x64-bit, those flags are not needed, in fact, denoting with 2³² with k we have:

(a·k+b) × (c·k+d) = a·c·k² + (a·d+b·c)·k + b·d

其中 32 位产品产生 64 位结果，添加如下

.----.----.
|   b·d   |
'----'----'
       +
     .----.----.
     | a·d+b·c |
     '----'----'
            +
          .----.----.
          |   a·c   |
          '----'----'

 =

.----.----.----.----.
|    128-bit result |
'----'----'----'----' 

¹ And this suffices to prevent overflow.