如何使用移位和加法在 MIPS 程序集中执行乘法？

Question

Please help, I've spent over 5 hours trying to figure out this simple problem and I cannot for the life of me figure out how the hell you do it.

I am supposed to do some problem like 7 * 5 and return the result in MIPS assembly, but I HAVE to do it using shift and add. I cannot figure out how to use the hint we were given and turn it into MIPS code. Like I understand 100% how the picture below works and the logic behind it, but I CANNOT turn it into MIPS assembly code, thats what I've spent over 5 hours trying to do.

here is what we were given as a hint

I understand how shifting works, I understand how adding works in MIPS, I just don't know how I can do what is shown in that image, in MIPS assembly language. (I don't know how to turn the english into code that does that stuff. MIPS is too barebones for this!)

I SHOULD CLARIFY!: Both numbers are VARIABLES. I don't know how to write the code that takes two differing numbers each time and gets the result, using shift and add. 7*5, 6*9, 17*3, etc. Whatever the user enters for those two numbers as long as they are non-negative, the code must get the result.

Answer 1

I grade school we learned that a * (b + c) = ab + ac right? So 7 * 5 = 7 * (4 + 1) = 7*4 + 7*1 which are both powers of 2...(2^2 and 2^0).

If not a constant does it matter?

     abcd
*    efgh
==========
     abcd  * h
    abcd   * g
   abcd    * f
+ abcd     * e
=============

same stuff we learned in grade school except much simpler as efgh are either zero or one. So if the h bit (bit 0) then we add abcd<<0, if the g bit is set (bit 1) then add abcd<<1 if f is set (bit 2) then add abcd<<2 and so on.

So actually you do/did know how to do it just not how to apply it. Pencil and paper just like grade school, no reason to spend 5 hours...

EDIT

unsigned int a,b,x,y;
for(y=0,x=0;x<32;x++) if(a&(1<<x)) y+=b<<x;


for(y=0,x=1,z=b;x;x<<=1,z<<=1) if(x&a) y+=z;

Answer 2

For the effort shown, I will put here major spoiler, but you will probably regret it, once you will realise how you were running in circles around solution.

s0 = a, s1 = b, s2 = 0 (will be result)
multiply_loop:
  if (s1 == 0) goto output_result
  t1 = s1&1
  if (t1 == 0) goto skip_add
  s2 += s0
skip_add:
  s0 <<= 1;  multiply "a" by 2
  s1 >>= 1;  shift bits of "b" to right by 1 for s1&1 test
  goto multiply_loop

output_result:
  s2 is result of multiplication

Of course this will overflow for big "a"/"b" values, etc.. Try it decipher on paper first, with binary values, how it works internally.