Not your average question...

One reason this isn't your average C forum question is that it isn't really about C :)

This is closest really to assembler, so I've posted the thread there. I know you C programmer people tend to work in ASM quite often, and generally work at a lower level, so if you wouldn't mind, take a look would ya? Thanks :)

http://forums.vb-world.net/showthrea...threadid=43228

In case anyone cares, this was my solution (couldn't get the simulator to work so I could test it) and I'm pretty sure it's right.

Code:

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imul1        OPC=0x20                                        // Set OPC as a counter... OPC = 32
imul2        MAR=SP-1; rd                                        // Read first operand (second word on the stack)
imul3        H=TOS                                                // Operands are now in H and MDR, TOS freed to be accumulator
imul4        TOS=0                                                // Clear accumulator.  One 0 just handled.
imul5        MDR=MDR; SRA1; if (notLSB) goto imul17                // Check for 00, else carry on for 10
imul6        TOS=TOS-H; SRA1                                        // 10 - subtract multiplicand, start the shift
imul7        MDR=MDR; RROT                                        // Bitwise rotate on MDR, LSB has the next recode bit
imul8        OPC=OPC-1; if (Z) goto imul18                        // Decrement the counter (OPC), if it's 0 then we've finished
imul9        MDR=MDR; SRA1; if (notLSB) goto imul14                // One 1 just handled, Check for 11/10 and skip one line if 10
imul10        TOS=TOS; SRA1; goto imul10                        // 11 so start a shift then check for 10
imul11        TOS=TOS+H; SRA1                                        // 01 so add multiplicand then shift
imul12        MDR=MDR; RROT                                        // Bitwise rotate on MDR, LSB has the next recode bit
imul13        OPC=OPC-1; if (Z) goto imul19 else goto imul8        // Decrement counter.  Set JAM then set again if counter=zero
imul14        TOS=TOS; SRA1; goto imul15                        // 00 so middle of string of 0s - keep shifting.
imul15        wr; goto imul20                                        // Write MDR to SP-1 (most significant word).  Skip next line.
imul16        wr                                                // Extra end-point, due to the way mic-1 conditional jumps work
imul17        MAR=SP                                                // Prepare to write LSW to top of stack.
imul18        MDR=TOS; wr; goto Main1                                // Write least significant word to top of stack.  End instruction

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That has to be the hardest piece of coursework I have ever had to do. Still, the lecturer said he would give people high marks (80%+) just for attempting this (I picked the hardest question ;) ).