Find the shortest sequence of MIPS instructions that extracts bits 16 down to 11 from register $t0 and uses the value of this field to replace bits 31 down to 26 in register $t1 without changing the other 26 bits of register $t1.

To extract bit 16 down to 11, we have AND the value with 0x0001F800 (the binary loactions 16 down to 11 are 1 and remaining are 0).

Now rotate this ANDded result 15 times to left to align with bits 31 to 26.

Now the bits 31 down to 26 in the $t1 are to be nulled and ORed with previous result obtained from $t0.

0x0001F800 is greater than 16 bits, not possible to pass as immediate value.

So we shift this right by 1 so that it fits in 16 bits. so we load 0xFC00 in R0 and shift left 1 to restore the actual 16 bit value

AND R0,R0,$0 ; And the contents of R0 with 0 to clear R0

AND R1,R1,$0 ; And the contents of R0 with 0 to clear R1

ADDI R0,R0,0xFC00 ; Load R0 with 0xFC00

SLL R0,R0,1 ;

AND R0,R0,$t0 ;

SLL R0,R0,0xF ;

ADDI R1,R1,0xFC00 ; Load R1 with 0xFC00

SLL R1,R1,0x10 ; Rotate R1 16 to left -> R1 = 0xFC000000

NOR R1,R1,$0 ; Invert all the bits in R1 so R1 = 0x03

AND R1,R1,$t1 ; Null the bits 31 to 26 in $t1 and dont touch other bits

OR $t1,R0,R1 ; Replace the contents of $t1 with the modifed value

answer:

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explanation:

answer:

50 vo; ts

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