CS302 Spr'99 Lecture Notes
Lecture 10a
Control-flow Graphs

To assign registers on a per-procedure vasis, need to perform liveness analysis on entire procedure, not just basic blocks.

To analyze the properties of entire procedures with multiple basic blocks, we use a control-flow graph.

In simplest form, control flow graph has one node per statement, and an edge from n1 to n2 if control can ever flow directly from statement 1 to statement 2.

We write for the set of predecessors of node n, and for the set of successors.

(Can also build control-flow graphs where each node is a basic block.)

Example routine:

Control-flow graph

Liveness Analysis using Dataflow Analysis

Working from the future to the past, we can determine the edges over which each variable is live.

In the example:

b is live on and on .

a is live from on , on , and on (but not on ).

c is live throughout (including on entry ).

Can see that two registers suffice to hold a,b,c.

Dataflow Equations

We can do liveness analysis (and many other analyses) via dataflow analysis.

A node defines a variable if its corresponding statement assigns to it.

A node uses a variable if its corresponding statement mentions that variable in an expression (e.g., on the rhs of assignment).

For any variable v, define

= set of graph nodes that define v

= set of graph nodes that use v

Similarly, for any node n, define

= set of variables defined by node n;

= set of variables used by node n.

Setting up Equations

A variable is live on an edge if there is a directed path from that edge to a use of the variable that does not go through any def.

A variable is live-in at a node if it is live on any in-edge of that node; it is live-out if it is live on any out-edge.

Then the following equations hold:

We want the least fixed point of these equations: the smallest in and out sets such that the equations hold.

We can find this solution by iteration:

Use equations to add variables to sets, one node at a time.

Repeat until sets don't change any more.

Adding additional variables to the sets is safe, as long as the sets still obey the equations, but inaccurately suggests that more live variables exist than actually do.

Solution

For correctness, order in which we take nodes doesn't matter, but it turns out to be fastest to take them in roughly reverse order:

 1st 2nd 3rd node use def out in out in out in 6 c c c c 5 a c ac ac ac ac ac 4 b a ac bc ac bc ac bc 3 bc c bc bc bc bc bc bc 2 a b bc ac bc ac bc ac 1 a ac c ac c ac c

Implementation issues:

Algorithm always terminates, because each iteration must enlarge at least one set, but sets are limited in size (by total number o variables).

Time complexity is O(N4) worst-case, but between O(N) and O(N2)in practice.

Typically do analysis using entire basic blocks as nodes.

Can compute liveness for all variables in parallel (as here) or independently for each variable, on demand.

Sets can be represented as bit vectors or linked lists; best choice depends on set density.

Static vs. Dynamic Liveness

Consider the following graph:

Is a live-out at node 2? It depends on whether control flow ever reaches node 4.

A smart compiler could answer no.

But no compiler can ever always answer such questions correctly.

This is a consequence of the uncomputability of the Halting Problem.

So we must be content with static liveness, which talks about paths of control-flow edges, and is just a conservative approximation of dynamic liveness, which talks about actual execution paths. Halting Problem

Theorem There is no program H that takes an input any program P and input X, and (without infinite-looping) returns true if P(X) halts and false if P(X) infinite-loops.

Proof Suppose there were such an H. From it, construct the function

Now consider F(F).

If F(F) halts, then, by the definition of H, H(F,F) is true, so the then clause executes, so F(F) does not halt.

But,if F(F)loops forever, then H(F,F) is false, so the else clause is taken, so F(F)halts.

Hence F(F) halts if any only if it doesn't halt.

Since we've reached a contradiction, the initial assumption is wrong: there can be no such H.

Corollary No program H'(P,X,L) can tell, for any program P, input X, and label L within P, whether L is ever reached on an execution of P on X.

Proof If we had H', we could construct H. Consider a program transformation T that, from any program Pconstructs a new program by putting a label L at the end of the program, and changing every halt to goto L. Then H(P,X) = H'(T(P),X,L).

Andrew P. Tolmach
1999-05-24