How many address bits will be required to address 2K memory? What will be the co

Question

How many address bits will be required to address 2K memory? What will be the content of the register R_1 after the operation R_1 leftarrow R1 + R2^bar + 1 R_ leftarrow R2-1 R1 leftarrow R2^bar +1 If the initial contents of R1 and R2 are the following: R1 = 01001100; R2 = 10010010 Starting from the 8-bit operand 11001010, show the values obtained after sequential application of each shift microoperation, both left and right, until the contents are all zeros. What is the mask and the operation that needed to be used to change the contents of a register whose current content is 01110101 11101101? to invert all the alternate bits (starting from the MSB). to set all the alternate bits to 0 (starting from the MSB). to set all the even bits to 1 (starting from the MSB). Suppose you want to replace the contents of selectived bits of a 16 bit register. What is/are the operations you need to do to replace 7th, 8th, 9th and 11th bits by 0.1, 1, 0 respectively? Convert the following binary numbers to decimal: 1001101, 1010011.101, and 10101110.1001.

Explanation / Answer

1. How many address bits will be required to address 2K memory.
2K is like 2,000 or exactly 2048 memory locations.
So, to address these locations, the number of bits required
is 11. This is because 2^11 = 2048.

2. What will be the content of the register R1:
R1 = 01001100, R2 = 10010010
R1'= 10110011, R2'= 01101101
R1 + R2' + 1 =
01001100
01101101
00000001

10111010

R2 - 1 =
10010010
2's complement of 1 is: 10
And adding this to R2 is:
10010010
00000010

10010100

R2' + 1

01101101
00000001

01101110

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