6 Give the 16bit 2s complement form of the following 8bit 2s

6. Give the 16-bit 2\'s complement form of the following 8-bit 2\'s complement numbers:

(a) OX94

(b) OXFF

(c) OX23

(d) OXBC

My understanding of this:

oxFF where F=15 and 15 = 1111 so 15 15 is 1111 1111 and then i thought that the 16 bit extension would either be 0000 0000 1111 1111 or 1111 1111 1111 1111 but I am not sure. Can you clarify this for me? Thank you.

Solution

In 2\'s complement form, MSB represents sign bit. If MSB is 0 then number is +ve and if MSB is 1 then number is -ve.

While changing number from 8 bit to 16-bit, if MSB is 1 then the all new 8 bits are 1 and if MSB is 0 then all new bits will be 0.

Now,

a.) 0x94 = 1001 0100

since MSB is 1 therefore the all new 8 bits are 1, now number is 1111 1111 1001 0100 which is 0xFF94.

b.) 0xFF = 1111 1111

since MSB is 1 therefore the all new 8 bits are 1, now number is 1111 1111 1111 1111 which is 0xFFFF.

c.) 0x23 = 0010 0011

since MSB is 0 therefore the all new 8 bits are 0, now number is 0000 0000 0010 0011 which is 0x0023

d.) 0xBC = 1011 1100

since MSB is 1 then the all new 8 bits are 1, now number is 1111 1111 1011 1100 which is 0xFFBC.

Hope it helps, do give your response.