程序集8086 - 在没有MUL和DIV指令的情况下实现任何乘法和除法

Question

I would like to know if there is a way to perform any multiplication or division without use of MUL or DIV instruction because they require a lot of CPU cycles. Can I exploit SHL or SHR instructions for this target? How can I implement the assembly code?

Answer 1

Just like everything else in assembly there are many ways to do multiplication and division.

1. Do division by multiplying by the reciprocal value.
2. Use shifts and adds/subs instead of multiplication.
3. Use the address calculation options of lea (multiplication only).

Myth busting

because they require a lot of CPU cycles

MUL and IMUL are blazingly fast on modern CPU's, see: http://www.agner.org/optimize/instruction_tables.pdf
DIV and IDIV are and always have been exceedingly slow.

An example for Intel Skylake (page 217):

MUL, IMUL r64: Latency 3 cycles, reciprocal throughput 1 cycle.

Note that this is the maximum latency to multiply two 64 ! bit values.
The CPU can complete one of these multiplications every CPU cycle if all it's doing is multiplications.
If you consider that the above example using shifts and adds to multiply by 7 has a latency of 4 cycles (3 using lea). There is no real way to beat a plain multiply on a modern CPU.

Multiplication by the reciprocal

According to Agner Fog's asm lib instruction page 12 :

Division is slow on most microprocessors. In floating point calculations, we can do multiple divisions with the same divisor faster by multiplying with the reciprocal, for example:

 float a, b, d; a /= d; b /= d; 

can be changed to:

 float a, b, d, r; r = 1.0f / d; a *= r; b *= r; 

If we want to do something similar with integers then we have to scale the reciprocal divisor by 2n and then shift n places to the right after the multiplication.

Multiplying by the reciprocal works well when you need to divide by a constant or if you divide by the same variable many times in a row.
You can find really cool assembly code demonstrating the concept in Agner Fog's assembly library .

Shifts and adds/subs
A shift right is a divide by two shr - ( R educe).
A shift left is a multiply by two shl - ( L arger).
You can add and substract to correct for non-powers of two along the way.

//Multiply by 7
mov ecx,eax
shl eax,3    //*8
sub eax,ecx  //*7

Division other than by powers of 2 using this method gets complex quickly.
You may wonder why I'm doing the operations in a weird order, but I'm trying to make the dependency chain as short as possible to maximize the number of instructions that can be executed in parallel.

Using Lea
Lea is an instruction to calculate address offsets.
It can calculate multiples of 2,3,4,5,8, and 9 in a single instruction.
Like so:

                      //Latency on AMD CPUs (K10 and later, including Jaguar and Zen)
                      //On Intel all take 1 cycle.
lea eax,[eax+eax]     //*2     1 cycle      
lea eax,[eax*2+eax]   //*3     2 cycles
lea eax,[eax*4]       //*4     2 cycles   more efficient: shl eax,2 (1 cycle)
lea eax,[eax*4+eax]   //*5     2 cycles 
lea eax,[eax*8]       //*8     2 cycles   more efficient: shl eax,3 (1 cycle)
lea eax,[eax*8+eax]   //*9     2 cycles

Note however that lea with a multiplier (scale factor) is considered a 'complex' instruction on AMD CPUs from K10 to Zen and has a latency of 2 CPU cycles. On earlier AMD CPUs (k8), lea always has 2-cycle latency even with a simple [reg+reg] or [reg+disp8] addressing mode.

AMD
Agner Fog's instruction tables are wrong for AMD Zen: 3-component or scaled-index LEA is still 2 cycles on Zen (with only 2 per clock throughput instead of 4) according to InstLatx64 ( http://instlatx64.atw.hu/ ). Also, like earlier CPUs, in 64-bit mode lea r32, [r64 + whatever] has 2 cycle latency. So it's actually faster to use lea rdx, [rax+rax] instead of lea edx, [rax+rax] on AMD CPUs, unlike Intel where truncating the result to 32 bits is free.

The *4 and *8 can be done faster using shl because a simple shift takes only a single cycle.

On the plus side, lea does not alter the flags and it allows a free move to another destination register. Because lea can only shift left by 0, 1, 2, or 3 bits (aka multiply by 1, 2, 4, or 8) these are the only breaks you get.

Intel
On Intel CPUs (Sandybridge-family), any 2-component LEA (only one + ) has single-cycle latency. So lea edx, [rax + rax*4] has single-cycle latency, but lea edx, [rax + rax + 12] has 3 cycle latency (and worse throughput). An example of this tradeoff is discussed in detail in C++ code for testing the Collatz conjecture faster than hand-written assembly - why? .

Answer 2

Things like SHL/SHR, SAL/SAR, ADD/SUB are faster than MUL and DIV, but MUL and DIV work better for dynamic numbers. For example, if you know that you just need to divide by two, then it's a single-bit shift right. But if you don't know in advance the number, then you might be tempted to repeatedly SUB the values. For example, To determine AX divided by BX, you could just constantly subtract BX from AX until BX is > AX, keeping track of the count. But if you were dividing by 200, by 1 that would mean 200 loops and SUB operations.

MUL and DIV will work better in most cases when the numbers involved aren't hard-coded and known in advance. The only exceptions I can think of is when you know it's something like a multiple/divide by 2, 4, 8, etc. where the Shift operators will work fine.

Answer 3

Implementing multiplication is easier, if you remember, an shl operation performs the same operation as multiplying the specified operand by two. Shifting to the left two bit positions multiplies the operand by four. Shifting to the left three bit positions multiplies the operand by eight. In general, shifting an operand to the left n bits multiplies it by 2n. Any value can be multiplied by some constant using a series of shifts and adds or shifts and subtractions. For example, to multiply the ax register by ten, you need only multiply it by eight and then add in two times the original value. That is, 10*ax = 8*ax + 2*ax. The code to accomplish this is

            shl     ax, 1           ;Multiply AX by two
            mov     bx, ax          ;Save 2*AX for later
            shl     ax, 1           ;Multiply AX by four
            shl     ax, 1           ;Multiply AX by eight
            add     ax, bx          ;Add in 2*AX to get 10*AX

The ax register (or just about any register, for that matter) can be multiplied by most constant values much faster using shl than by using the mul instruction. This may seem hard to believe since it only takes two instructions to compute this product:

            mov     bx, 10
            mul     bx

However, if you look at the timings, the shift and add example above requires fewer clock cycles on most processors in the 80x86 family than the mul instruction. Of course, the code is somewhat larger (by a few bytes), but the performance improvement is usually worth it. Of course, on the later 80x86 processors, the mul instruction is quite a bit faster than the earlier processors, but the shift and add scheme is generally faster on these processors as well.

You can also use subtraction with shifts to perform a multiplication operation. Consider the following multiplication by seven:

            mov     bx, ax          ;Save AX*1
            shl     ax, 1           ;AX := AX*2
            shl     ax, 1           ;AX := AX*4
            shl     ax, 1           ;AX := AX*8
            sub     ax, bx          ;AX := AX*7

This follows directly from the fact that ax*7 = (ax*8)-ax.

A common error made by beginning assembly language students is subtracting or adding one or two rather than ax*1 or ax*2. The following does not compute ax*7:

            shl     ax, 1
            shl     ax, 1
            shl     ax, 1
            sub     ax, 1

It computes (8*ax)-1, something entirely different (unless, of course, ax = 1). Beware of this pitfall when using shifts, additions, and subtractions to perform multiplication operations.

Division is a bit harder, need to think...

Answer 4

Here is an example:

mov bx, 1000b
shl bx, 5
mov cx, bx
shr cx, 2
add bx, cx
add bx, 1000b