How to efficiently use the address register? Address register = contains the address of the operand to fetch from memory.

Transcription

1 Lesson 13 Storage Assignment Optimizations Sequence of accesses is very important Simple Offset Assignment This lesson will focus on: Code size and data segment size How to efficiently use the address register? Address register = contains the address of the operand to fetch from memory. If we store the operands in memory in the same order that they are called, we can just increment the address register. Auto increment mode get the operand and increment in same instruction. It reduces code size and improve efficiency. So we need to place the operands carefully to take full advantage of auto increment. In this example the program shown. The loads and stores are of interest to us here. The compiler puts the operands into the given order to increase the usage of auto increment. auto increment to get the data. And then we can use The final code created is very fast to execute and compact. Notice the dependence of the layout on the stack. Dependences must be maintained to take advantage of auto increment.

2 Access Sequence and Access Graph Access sequence = sequence in which the memory locations are accessed. Example: z = x op y The access sequence is x, y, z. Access graph == nodes are a unique variables Edge e(u,v) exists with weight w(e) if variables u and v are adjacent to each other w(e) times in the access sequence. The directions are not material in this graph. Example: Weight of a Graph and Cost of Assignment The weight of a graph G is the sum of the edges in G(V,E) The cost of a cover G is the sum of the weights of the edges in G but not in C. There can be multiple covers of a graph. We are looking for the one that maximizes the weight. cost(c) = weight(g) weight(c) the cost is edges that are not covered The cost of an assignment is equal to the number of adjacent accesses of variables that are not assigned to adjacent locations.

3 Weight of a Graph and Cost of Assignment Example: Each variable gets a node. An edge is added for each access (regardless of direction A B == B A) Now pick the highest cost edges to cover the graph Cover = there should be no cycles and we can remove uncovered edges and the graph will still be connected. MWPC = maximum weight per cover Pick the cover so that the weight is maximum of all covers. The weight of this cover is 9. The cost is 4. (total of the uncovered edges)

4 What is the layout of the max cover? Heaviest edges are put next to each. So the most used edges are next to each other. The layout can be found by starting at one end of the cover and stepping through it. e f a d c b Or b c d a f e The goal is to put transitions next to each other to take advantage of auto increment. Assumptions in Single Offset Assignment Every data object has a size of one word A single address register AR) is used to address all variables. Auto increment and decrement are used to transition One to one mapping of variables to locations The basic block has a fixed evaluation order (schedule). General Offset Assignment Problem K address registers Ar0 AR(k 1) Given access sequence L, set of variables V number of address registers k Find a partition of V, where m <= k. Such that the total cost of the optimal SOA of the corresponding access subsequences plus the setup cost for using the m register is minimal. General offset Assignment Problem: Assumptions: Fixed setup cost of each additional register Each address register is used to point to a disjoint subset of variables So variables are divided into the address registers, some variables under each register.

5 GOA Example Using SOA the cost is 6, meaning we are using only one address register, it is getting incremented as we move through the graph. Now do with two address registers to reduce the cost. Using 2 address register: AR1 a d e AR2 b c Now determine the cost. For the first address register we get this access sequence: When we focus on the a d e edges, we get a cost of 1. There is a cost of 1 because we will have to do an explicit load of the address register for the transition that is not covered.

6 The cost here is 0. We have no uncovered edges. We use auto increment and decrement to move between variables. Algorithm for GOA Solve the SOA problem If we only have one register, then we are done Otherwise, divide the registers into 2 subsets. Solve the SAO for each group Compare the solutions Choose the solution that is the lower cost The do it again, reducing the remaining subset into subsets. Repeat until we have reached the lowest cost or all the registers. KEY which variables in which subset? Benchmarks show there is increased performance when using GOA.

7 Relax Fixed Evaluation Order Assumption In addition to GOA, what else can be done? The previous solutions were dependent upon the access sequence. While good, it makes the assumption of a fixed evaluation order. To relax the access sequence: We will transform the access sequence Algebraic properties to change access order: Commutativity Associativity A review of using fixed access sequence: Now we can get the code using the AR0 register:

8 Access Graph Algebraic Transformations Commutative transformation on Expression Tree We can change the access sequence using the commutative property: Associative transformation on Expression tree The access order is changed from yzx to xyz.

9 New assignment: The code will now be:

10 The access graph is now: Least Cost Access Sequence This means: Find the MWPC The do algebraic transformations Determine the new cost, if less than the SAO cost, keep the transformation We are looking for the least cost access sequence (LCAS)

11 To find the LCAS : Focus on distinct access transitions which is a new edge that is being added to the access graph. So increase the number of consecutive accesses to a variable to reduce the number of distinct transitions. Reducing the number of distinct transitions means we are reducing the number of edges We also want to increase the occurrences of identical access transitions. Which will increase the weight of covered edges. So we are building candidate access sequences incrementally while pruning the candidate set by retaining those with minimum number of edges. Understand the relationship between the access sequence and the access graph Now we have done some transformations to the access sequence and gotten a new graph

12 The transformations specifically instructions 3 and 4. We used the comm. Property to change the order of access. This transformation simplifies the graph. Pruning Example Goal: to prune the graph through algebraic transformations Focus on statement 1: c = a +b swap to c = b + a Both have same cost, so both are retained. If one had a lower cost, it would be kept and the higher cost one would be pruned. Continue example: f= d + e we can do f = d + e or f = e + d In one case we will get this access sequence. In another case we will get this access sequence.

13 In another case we will get this access sequence. We must also note the frequency of the transitions, we want to keep the edges with the most transitions. Pruning does partial access graphs. Keeps the lowest cost with the highest weights at each step.

14 Some of the possible sequences are shown above. Anytime an access graph is more expensive than the others, it will be pruned from the set. So by the time we are done, only the best sequences are left. Working through the access sequences we get: We can see that one of the sequences has a lower cost than the other, so this is the one we will keep.

15 There is no correlation between cost and access graph topology. Most important at each step prune the high cost sequences. This is no guarantee that you will have the least cost solution using this method, that problem is NP complete. Our goal is to reduce the cost. Benchmark results show us that commutative transformation can lead to significant improvements, up to 13%. Variable Coalescence and Separation Now we will look at different way of simplifying the access graph by putting two variables in the same memory location. Recall from reg. allocation: Basic unit that must be put into the same memory location: web maximal d u chains Variables in the memory can be coalesced (put in the same memory location) When can we co locate two variables in the same location? Wehn the webs do not overlap or they do not interfere with each other So we must make sure the live ranges of the two variables do not overlap This will make it easier to make a better layout. In the above example, the last access of b occurs before the first access of d, so the two variables can share a location. To do this we must know the webs used. This is important because we can now coalesce parts of webs. The smaller the web, the more freedom for the code generator.

16 Fact and assumptions: Alias analysis is nec. To determine the variable that might be referenced via pointers. We have to make sure we know if a variable is referenced by a pointer. Multiple aliased variables cannot be tackled. Statically we don t know where the variable is being referenced. These variables will not be coalesced. We need to work on two graphs: access graph (AG) and interference graph (IG). The IG will show us the disjointed webs that can be colocated in the same memory location. So we pick webs that have a high degree on AG and a low degree on IG. Because if we coalesce a web with a high degree AG will mean we can remove a lot of edges from the AG. A low degree in IG will mean we have lots of choices on which webs to coalesce. The weights on the AG can be calculated statically or with profiling information. Variable Coalescence and Separation Example Begin with the following: We will need to insert 6 instructions to load the register. Let s convert the program into a web:

17 The cost has gone up. This is a temporary situation. Now we will coalesce b 2 and c 2 into a new node called X. Coalescing has reduced the cost back to 6. Now do even more coalescing and we get the following:

18 You can see we have a high weight and a low cost. This example shows the benefit of coalescing. Issues on Applying Coalescence What are the difficulties, why is this work novel? Coalescence is not a new technique. But now we are coalescing disjointed webs to the same location. Maximal coalescence may not lead to the best storage assignment. In fact it may worsen it. This happens because coalescence increases the number of neighbors for coalesced nodes. Only two neighbors are allowed for each node on the path cover. Increasing the degree of the nodes, MWPC must work harder to find a node pair. Coalescing is not the best for generating the best access sequence.

19 Example: The nodes c and h were coalesced together, with the resulting edge weights. Note: the weights around c h have gone up. Let s do different coalescing and see the effect on the MWPC: Note: some coalescing leads to a higher MWPC. Coalescence does not change the cost of the graph, it just changes it to different edges. The rightmost solution is the best, it has the least cost among the choices. Coalescence leads to higher degree nodes. Path covers prefer to have only 2 neighbors.

20 Framework for SOA Pre Iteration Rules If we apply pre iteration rules before the coalescence iteration loop, the profitability is always guaranteed. Rule 1: coalesce all degree 0 nodes with any other node. Doing so will not affect the SOA cost. Rule 2: coalesce all degree 1 nodes with it neighbor. If its edge is already on the PC, the SOA cost is not affected. Otherwise we reduce the SOA cost by the weight of this edge. Rule 3: coalesce all degree 2 nodes with the neighbor having a higher weight edge connected to it. The Main SOA Algorithm At each stage the cost is calculated.

21 Benchmark Results Benchmarks show the costs of doing coalescing and program transformations. The combination of the two leads to the best performance improvements.