 For this example, we're going to work with a couple of unusual instructions. We have a shift-left logical and a jump-register instruction. These are still standard R-type instructions, so in both cases the opcode will be 0, but the rest of this is going to be a little bit unusual. Our shift-left logical instruction has one source register, so you might think that we'd like to put it in the RS field. Unfortunately, this is one odd case where our source register is actually going to end up in the RT field. We won't put anything into the RS field, and A3 is our source register. The destination register is V1, and the shift amount is obviously 2. The function code for a shift-left logical instruction is also 0. Now for our jump-register instruction. This time we only have one source register, which is the return address, so in this case it can go into the RS field. And since that's the only operand I have for my jump-register instruction, I'm just going to fill in zeros for the other fields. Finally, the function code for our jump-register instruction is 8. Next, I'll go through and convert these into binary. I have six bits for my op codes, five bits for each of my register fields. A3 is register 7, V1 is register 3. The return address is register number 31, and zeros are zeros. 2 gets translated into 2, and we have five bits for our shift amount, and then we just have our function field, which is still just six bits. Lastly, I'll convert both of these into hexadecimal. So that's what both of those instructions look like in machine language.