Question

2. Write the hexadecimal numbers in the registers of $a0, $a1, $a2, $a3 after the following codes running:

ori $a0, $0, 11

ori $a1, $0, 19

addi $a1, $a1, -7

slt $t2, $a1, $a0

beq $t2, $0, label

addi $a2, $a1, 0

sub $a3, $a1,$a0

j end_1

label: ori $a2, $a0, 0

add $a3, $a1, $a0

end_1: xor $t2, $a1, $a0

*Values in $a0, $a1, $a2, $a3 after the above instructions are executed.

Answer 1

Ans->Values of registers in hexadecimal are :

$a0 =B  

$a1=C

$a2=B

$a3=17

$t2=7

ori $a0, $0, 11   // do or of 0 , 11 and store the value in a0 register

ori $a1, $0, 19 // do or of 0 , 19 and store the value in a1 register

addi $a1, $a1, -7 // add the value of a1 register and -7 , and store the result in a1 register

slt $t2, $a1, $a0 // store 1 in t2 register if value in a1 register is less than value in a0 register ,else 0

beq $t2, $0, label // if t2 is equal to 0 then move to label else continue executing next instructions

addi $a2, $a1, 0 // add the value of a1 register and 0 ..and store the result in a2 ==>a2=a1+a0

sub $a3, $a1,$a0 // subtract the value of a0 register from a1 register and store the value in a3==>a3=a1-a0

j end_1 //this means jump to end_1 label

label: ori $a2, $a0, 0 // do or of value in a0 register and 0 ..and store the value in a2

add $a3, $a1, $a0 // add values of register a1 and a0 ..store the result in a3

end_1: xor $t2, $a1, $a0 // do xor of values in a1 and a0 ...store the result in t2

Execution of program:(Assume value stored in decimal)

after first step $a0=11

after second step $a1=19

after third step $a1=19-7=12

after fourth step $t2=0 ==>since $a1>$a0

after fifth step control goes to label ==>since condition is true

after label instruction execution $a2=11

after add instruction execution $a3=$a1+$a0= 12+11=23

after execution of end_1 label $t2=$a1 ^ $a0 = 12 ^11=7

Hence values in decimal are

$a0 =11  

$a1=12

$a2=11

$a3=23

$t2=7

Converting into hexadecimal we get

$a0 =B  

$a1=C

$a2=B

$a3=17

$t2=7