2N, twice the original integer.

`0110 0001`

= 97_{10}.

`1100 0010`

= 194_{10}.

(However, if a 1-bit is shifted off the left side, then the number is ruined).

sll d,s,shft # $d gets the bits in $s # shifted left logical # by shft positions, # where 0 ≤ shft < 32

A **shift left logical** of one position
moves each bit to the left
by one.
The low-order bit (the right-most bit) is replaced by a zero bit
and the high-order bit (the left-most bit) is discarded.

Shifting by two positions is the same as performing a one-position shift two times. Shifting by zero positions leaves the pattern unchanged. Shifting an N-bit pattern left by N or more positions changes all of the bits to zero.

The picture shows the operation performed on eight bits.
The original pattern is
`1010 0111`

.
The resulting pattern is
`0100 1110`

.

The MIPS processor always performs the operation on a 32-bit register and puts the result in a 32-bit register.

The ALU (arithmetic/logic unit) which does the operation pays no attention to what the bits mean. If the bits represent an unsigned integer, then a left shift is equivalent to multiplying the integer by two.