Computer System Architecture - Central Processing Unit
Consider the following program segment. Here R1, R2
and R3 are the general purpose registers.
Assume that the content of memory location 3000 is
10 and the content of the register R3 is 2000. The
content of each of the memory locations from 2000 to
2010 is 100. The program is loaded from the memory
location 1000. All the numbers are in decimal.
- Option : A
- Explanation : The given program stores results from memory
location 2000 to 2009.
It stores values like 110,109,108.....100.
No change at location 2010 So at 2010 value is 100
Consider the following program segment. Here R1, R2
and R3 are the general purpose registers.
Assume that the content of memory location 3000 is
10 and the content of the register R3 is 2000. The
content of each of the memory locations from 2000 to
2010 is 100. The program is loaded from the memory
location 1000. All the numbers are in decimal.
- Option : B
- Explanation : 100110110 (in 2's complement ignore carry bit so
ignore 1 at msb)
After carry bit is ignored the value at msb is 0 means number is positive so sign bit is 0.
Overflow flag is set only if the X-OR between the carry-into the sign bit and carry -out of the sign bit is 1.
Therefore,
carry flag = 1,
overflow flag = 0,
sign bit = 0
- Option : B
- Explanation : An antidependency also known as write-after-read
(WAR) dependency.
S1 is False, I2 and I5 they both write R7. So, No WAR.
S2 is true, There is an anti-dependence between instructions I2 and I4.
I2 reads R3 and I4 writes it.
S3 is false.depends on how many instruction between the antidependence instructions.
- Option : D
- Explanation : To achieve speedup we can transfer directly like
the code given below.
R1 ￫ R2
R1 ￫ R3
R1 ￫ M[100]
