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看到了这样一段代码，就是把int 15h, ax=e820h的返回结果转化一下的代码，看不懂
其中bx=16

wrnum:
push si
push dx
push cx
push ax
mov si,num_buf
; extended precision division from section 9.3.5
; of Randall Hyde's "Art of Assembly"
; start: DX=dividend MSW, AX=dividend LSW, BX=divisor
wrnum1:
push ax
mov ax,dx
xor dx,dx
; before div: DX=0, AX=dividend MSW, BX=divisor
; after div: AX=quotient MSW, DX=intermediate remainder
div bx
mov cx,ax
pop ax
; before div: DX=intermediate remainder, AX=dividend LSW, BX=divisor
; after div: AX=quotient LSW, DX=remainder
div bx
; end: DX=quotient MSW, AX=quotient LSW, CX=remainder
xchg dx,cx
add cl,'0'
cmp cl,'9'
jbe wrnum2
add cl,('A'-('9'+1))
wrnum2:
dec si
mov [si],cl
mov cx,ax
or cx,dx
jne wrnum1
call cputs
pop ax
pop cx
pop dx
pop si
ret

这段代码是干什么的？为什么除以16呢？

hi, first of all, the code snippet you have given is doing only the printing the 32-bit number in a specified radix. its not part of any memory detection routine. i think you havent't completely gone through the comments given along with the code.
its just dividing the number with the radix, each time getting the least significant number, converting to printable ascii format in the range '0' - '9' and 'A' - 'Z' (though not limited to 'Z'), storing the result in the reverse order starting from the address num_buf(you can notice a 40 bytes array declared just above the address) and after the conversion is over cputs is called to print the formed string.