The aam instruction is used to adjust the content of the AL and AH registers after the AL register has been used to perform the multiplication of two unpacked BCD bytes. The CPU uses the following simple logic:
al = al mod 10 ah = al/10
Although this instruction should be used immediately after the multiplication instruction, it could be used later as long as no other intervening instruction would have changed the AL register.
Example1: mov al,7 mov ah,3 mul ah ;al = 21 = 15h, ah = 0 aam ;al = 21mod10 = 1, ah = 21/10 = 2
In fact, this instruction could be used without any preceeding multiplication. It will convert any value in AL according to the logic described above.
Example2: mov al,73 aam ;al = 73mod10 = 3, ah = 73/10 = 7
The logic of the aam instruction is based on the assumption that it would follow the multiplication of two unpacked BCD bytes. Since neither byte should exceed a value of 9, the multiplication result should never exceed a value of 81 and the AH register would always become 0 after the multiplication. This may be the reason why the Intel manual describes the logic as acting on the AX register instead of effectively acting only on the AL register as mentioned here. This can be verified on a debugger with the next example which is designed to show what could be expected when the instruction is not used as intended.
Example3: mov al,209 mov ah,137 mul ah ;ax = 28633 = 6FD9h (ah = 6Fh, al = D9h = 217) aam ;al = 217mod10 = 7, ah = 217/10 = 21Note: The ASCII value of numerical digits must first be converted to their binary value before a multiplication is performed. The content of the upper 4 bits of the ASCII values would affect the final result in AL.
Although the content of the AH register may be modified by the aaa and aas instructions, the resulting content of that register is rarely used when doing strictly additions or subtractions. However, with the aam instruction, the resulting content of the AH register is the actual carryover from one multiplication which must be added to the result of the next multiplication. And, following that addition, the aaa instruction is necessary; any correction of the digit in AL will then be reflected by the automatic increment of that overflow in the AH register.
The next example will show how such multiplications can be coded. It's designed to compute the factorial of 30, i.e. 1*2*3*4...*30. The exact answer is:
or approximately 2.65*1032. It could be expanded with some minor modifications to compute the factorial of any number, only limited by the available memory (and patience for very large numbers).
Because numbers will be increasing gradually, all BCD numbers are kept in reverse order. It makes for easier coding by simply appending new numbers at the end of the existing ones.
.data size1 dd 1 ;current number of digits in answer size2 dd ? ;current number of digits in multiplier ;this variable is not used in this code counter dd 1 ;initialize with a 1 multiplier db 4 dup(0) answer db 1,47 dup(0) ;initialize with a 1 workbuf1 db 48 dup(0) ;adequate for current code workbuf2 db 48 dup(0) ; idem .code start: ; Because the value of the counter will not exceed 81, the aam instruction ; will be used to obtain its unpacked BCD digits. inc counter mov edi,offset multiplier mov eax,counter aam mov [edi],al ;store 1st multiplier digit or ah,ah jz singledigit ; If the multiplier has only 1 digit, the multiplication result ; can overwrite the previous answer directly. ; Otherwise, the result of individual multiplications must be added before ; being transferred to the answer buffer. inc edi mov [edi],ah ;store 2nd multiplier digit mov ecx,size1 mov edi,offset workbuf1 mov esi,offset answer mov ebx,offset multiplier mov dl,0 ;DL used to keep carryovers @@: lodsb ;get next digit from answer mul byte ptr[ebx] ;multiply it by 1st digit of multiplier aam ;convert the result add al,dl ;add the previous carryover aaa ;convert the result of this addition mov dl,ah ;save the new carryover stosb ;store the digit dec ecx jnz @B ;continue until all digits have been processed mov [edi],dl ;store the last carryover inc ebx ;point to the next multiplier digit mov ecx,size1 mov edi,offset workbuf2+1 ;aligns the two buffers for later addition mov esi,offset answer mov dl,0 ;DL used to keep carryovers inc size1 ;the new answer will have at least 1 more digit @@: lodsb ;get next digit from answer mul byte ptr[ebx] ;multiply it by 2nd digit of multiplier aam ;convert the result add al,dl ;add the previous carryover aaa ;convert the result of this addition mov dl,ah ;save the new carryover stosb ;store the digit dec ecx jnz @B ;continue until all digits have been processed or dl,dl jz @F mov [edi],dl ;store the last carryover inc size1 @@: ; The two multiplication results must now be added and the answer ; overwritten with the sum. If there were more than 2 digits in the ; multiplier, the workbuf1 would be overwritten with the sum and ; the result of subsequent multiplications overwriting the workbuf2. ; Care should then be taken to rezero the required front digits ; of that workbuf2 buffer. mov ebx,offset workbuf1 mov esi,offset workbuf2 mov edi,offset answer mov ecx,size1 clc @@: lodsb adc al,[ebx] aaa stosb inc ebx dec ecx jnz @B jnc nextone mov byte ptr[edi],1 inc size1 jmp nextone singledigit: mov edi,offset answer mov ebx,offset multiplier mov ecx,size1 mov dl,0 ;DL used to keep carryovers @@: mov al,[edi] ;get next digit from answer mul byte ptr[ebx] ;multiply it by the multiplier digit aam ;convert the result add al,dl ;add the previous carryover aaa ;convert the result of this addition mov dl,ah ;save the new carryover stosb ;store the digit dec ecx jnz @B ;continue until all digits have been processed or dl,dl jz nextone mov [edi],dl ;store the last carryover inc size1 nextone: cmp counter, 30 jb start ; The digits in the answer are in reverse order and in their binary form. ; The following will prepare the data for display as a null-terminated string mov esi,offset answer mov edi,esi add edi,size1 mov byte ptr[edi],0 dec edi @@: mov al,[esi] mov ah,[edi] add ax,3030h mov [esi],ah mov [edi],al inc esi dec edi cmp esi,edi jbe @B ; The answer is now ready for display
The described aam instruction is a special case of the more generalized function available on the CPU. The machine code for aam is:
D4 0AThe imm8 value following the D4 first byte is effectively taken to divide the content of the AL register. This means that the content of AL could be adjusted to two unpacked digits in any other numeric base.
However, no mnemonics are available for bases other than 10. The instruction would then need to be hand coded. For example, if the unpacked digits were needed in base 12, hand coding such as follows would be required: