Suppose AX =FFFE and BX=1234
now if we writecmp ax,bx
(bx will be subtracted from ax and the approprite flages will be updated)
now the binary representation of the numbers in ax and bx is given by
AX = 1111 1111 1111 1110
BX= 0001 0010 0011 0100
As bx will be subtracted from ax so we have to negate bx (as Result= ax+(-bx))
so the negated bx (2's complement of bx ) is given by.
BX= 1110 1101 1100 1100
Now we add both ax and bx (as subtraction is implemented by addition in computer)
AX= 1111 1111 1111 1110
BX= 1110 1101 1100 1100
------------------------------------
1 1110 1101 开发者_C百科 1100 1010
Now as you can see the result is of 17 bits now the 17th bit should go into carry flage, but when i checked it the carry flag is 0 that is CF=0 why?
Found a link here: http://oopweb.com/Assembly/Documents/ArtOfAssembly/Volume/Chapter_6/CH06-2.html
It is as I expected. The carry flag is set only if "borrow" was required. When doing subtraction, you set the carry flag prior to doing the "subtract" and the new carry flag tells if you had to borrow. Your example omitted the addition of 1 for the pre-set carry flag in bit 17 which will cause no carry in the result.
Think of the carry flag as a borrow bit when subtraction is performed. It is initialized to 1 so the operation is a−b−C
, i.e., a + not(b) + C
instead of a + not(b) + 1
as you described. In other words, the carry is inverted for subtract so it can be used to do multiple-precision subtract.
精彩评论