QUESTION 17:
Fill in the amount by which to shift. (Hint: determine which bit should be the new left-most bit. Count the number of bits between its old and new position).
sll $11,$11,_____ # left shift $11 by ______ places
See below
The ori puts its immediate operand in the low
order bits (the right-most bits) of its destination register.
So now, with the following code, the registers have the
correct patterns but the patterns are not in the correct fields.
or $25,$0,$0 # clear $25 ori $11,$0,0xD # opcode ori $12,$0,0x9 # operand $s ori $13,$0,0x8 # dest. $d ori $14,$0,0x004A # immediate operand
Register $11 contains the opcode in its right-most bits, like this:
31....26 25...21 20...16 15...................0 000000 00000 00000 0000 0000 0000 1101
However, the opcode is required to be in the high-order six bits,
26 through 31.
A sll will shift the contents of $11 into the correct
position.
Fill in the amount by which to shift. (Hint: determine which bit should be the new left-most bit. Count the number of bits between its old and new position).
sll $11,$11,_____ # left shift $11 by ______ places