COMP326/526 Tutorial Software Pipelining

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1 13 Dec 2005 From a previous year; assignment #3 Question 1 e. asked for the given loop to be software pipelined. Here is an explanation on how to develop it. First the original loop: loop: f4, 0(r1) multd f4, f4, f0 multd f6, f6, f2 addd f4, f4, f6 sd 0(r1), f4 First step is to get rid of the false dependences (output and anti-dependences) that can pose a problem in the rearrangement of the instructions. It will not always be necessary to rename all the registers in use, after doing the process a few times it will be easier to see which ones are absolutely necessary. The revised loop will look like this: loop f4, 0(r1) Page 1 of 6

2 The next step is to unroll the loop to get a sense of how the instructions will be rearranged. loop f4, 0(r1) f4, 0(r1) f4, 0(r1) Page 2 of 6

3 The next step is to pick out the instructions in reverse order in order to develop the sliding pattern that is shown in the textbook figure loop f4, 0(r1) f4, 0(r1) sd 0(r1), f12 f4, 0(r1) sd 0(r1), f12 Page 3 of 6

4 The idea with software pipelining is to have a number of iterations of the loop being in the process of executing at the same time by using the loop as a way of minimizing the stalls that wou normally occur. What is important in understanding this is the code that is taken out of the loop to get the technique into operation. The final result shou still accomplish the same function and the final data shou still be the same as the original program. Note that the instructions that are // are now just comments and will in fact be executed in the main loop, the first part is the startup code. // iteration #A f4, 0(r1) // // // // iteration #B f4, -8(r1) f6, -8(r2) // // // // // iteration #C f4, -16(r1) // f6, -16(r2) // // // // // Page 4 of 6

5 Then the loop proper gets made up from the instructions removed in reverse order and the loop is loop sd 24(r1), f12 multd f10,f6, f2 f4, 0(r1) The iterations are specified starting at A above, if you follow the final code through its iterations you will see how the sequence of events develop better. For the finished loop the iterations will start at one. Iteration 1: Iteration A will complete its sd instruction and will be finished with the pipe. Iteration B will perform the addd instruction. Iteration C will perform the multd instructions (and also the instructions afterwards. These ones can go together) Iteration 2: Iteration B will perform the sd instruction and will be finished with the pipe. Etc Iteraction C will perform the addd instruction. Iteration D will perform the multd and instructions. Notice that the consistency of the data is maintained in the startup of the loop and as the loop progresses the data all falls into place where it is supposed to be. Then after the loop is completed some final cleanup of instructions will also be required. One significant advantage is that the RAW hazards are all maximized by spanning the iteration of the loop, in the example above the register is not used for reading until the instruction just ahead of the instruction where it was written. Another point to notice, the count for the loop can be tricky to evaluate. It shou be multiplied by the number of instructions that are reversed, but be careful because some instructions can be taken as a group which counts as one. As in the above example the two loads are counted as a group and there are in total four groups including the present position. Page 5 of 6

6 So in one iteration there are A. 0 B. multd multd 8 C. addd 16 D. sd 24 Also watch out for the situation of question 4.11 in the textbook where the loads are the result of unrolling the loop first. The loads will still be grouped together but the count will be increased to 16 since we are actually doing 2 loops on each iteration. The placing of the offsets can be a source of trouble too. When the instructions are rearranged and put into different iterations in the same loop you want that they continue to operate on the correct data. So the offsets have to be calculated for the load and store instructions. The code before the loop has to prepare the data for the last instruction (in this case the store) and the load has to obtain the next piece of data that will be processed through the loop. At the end the load has to bring in the very last piece of data. If the store in the final iteration works on the last piece of data then the subsequent load and calculations will be done on external, unknown data which is not good. So at the end the load shou be matching with the same iteration as the loop iteration and the actual calculations and store of that last piece of data has to be done in the post-loop portion. Also, remember that the pre-loop section will have an adjustment to the count value that is counted down by the number of loads that get done in the preloop section. This adjustment of the count will not be evident if you ignore the pre-loop section and is usually unimportant to the part that you do show as your answer. I have done this in extreme detail to get the idea of what is involved in the technique. Once you understand and gain facility you will find that the step-by-step process is not necessary anymore and that any startup or finishing code can be assumed to exist. End of tutorial Page 6 of 6