A good answer might be:

sll   $11,$11,26    # left shift $11 by 26 places

Program So Far

Arithmetic gives us the same answer. Bit position 5 contains the right-most bit of the six-bit opcode. It must be shifted into position 31, and 31-5 = 26.

Here is a run of the program so far. The source code is seen at the right in the window.

Great! Now we have the correct bit pattern placed in the correct 6-bit field. Only one problem: it's in the wrong register. The instruction is supposed to be in register $25.

QUESTION 18:

Think of an assembly instruction that will put the opcode into register $25. Here is a start:

or   $25,$___,$___    # move opcode into target register