1. What is 0x1FBEAD58 shifted left by 2 and convert to decimal? 2. What is the v
ID: 2248729 • Letter: 1
Question
1. What is 0x1FBEAD58 shifted left by 2 and convert to decimal?
2. What is the value of the index field if the index field is 6 bits and is to the left of a 5-bit block offset field for the 64bit address 0xFFFFFFFF81054AA0? This is a cache memory problem.
3. Given the MIPS command lw $t0 76($a3), what is the numerical value of the instruction if the contents of register 5 = 0x48ce4568, $6 = 0x52525250 and $7 = 047be4afc? This is equivalent to the machine code value of the instruction.
4. What is 0x12345678 % 32?
5. Given the MIPS command lw $t0 76($a7), what is the calculated address if the contents of register 5 = 0x48ce4568, $6 = 0x52525250 and $7 = 047be4afc?
Explanation / Answer
value in hex = 0x1FBEAD58
in binary = 00011111101111101010110101011000
after left shift by 2
binary value= 01111110111110101011010101100000
in hex = 7EFAB560
in decimal = 2130359648
2. In binary of given 64bit address 0xFFFFFFFF81054AA
block offset = 5 bits = 00000
index field = 6 bits = 010101
in decimal = 21
3.
$a3 = register 7
and register ,$7 = 0x47be4afc
register 8 or $t0 = [0x47be4afc + 76] =[47be4b72]
4 . 0x12345678 % 32 = 0x12345678 modulo 32 = 0x18
5. name : register number
$a7 is not used . it can be $a1 , in that case address is = 0x48ce4568 + 76 = 0x48CE45B4
$a0-$a3 4-7Related Questions
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