What will be the hexadecimal value in the register ax (32-bit) after executing the following instructions ? mov al, 15 mov ah, 15 xor al, al mov cl, 3 shr ax, cl Codes :
01E0 h
Decimal 15 is 0F h, al is the low byte of ax and ah is the high byte.
mov al, 15; mov ah, 15 \(\Rightarrow ax = 0F0F\,h\)
xor al, al clears the low byte \(\Rightarrow ax = 0F00\,h = 0000\,1111\,0000\,0000_{2}\)
shr ax, 3 shifts right by 3 bits \(\Rightarrow 0000\,0001\,1110\,0000_{2} = 01E0\,h\)
Check: \(0F00\,h = 3840\) and \(3840 / 8 = 480 = 01E0\,h\).
Hence, 01E0 h.
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mov ax, 0h
mov cx, 0A h
doloop:
dec ax
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