Fast calculation of dominators

Question

It is said that when converting intermediate code to static single assignment form,

It is necessary to calculate the dominators of basic blocks
The somewhat obvious way to do this as a fixed point of equations is slow
To do it fast, you need to use the fairly complicated Lengauer-Tarjan algorithm

I can see the first two points, but I'm not clear about the reasoning behind the third. In particular, is there any reason why you can't do it just in the process of calculating the preorder dominator tree? I've implemented a version of that in JavaScript and it seems to work:

var domPreorder = [];

function doms(b) {
    b.children = [];
    for (var c of b[b.length - 1].targets)
        if (c.i == null) {
            c.i = domPreorder.length
            domPreorder.push(c)
            b.children.push(c)
            c.parent = b
            doms(c)
        }
}

f[0].i = 0
domPreorder.push(f[0])
doms(f[0])

Is there some disadvantage to this method that I'm not seeing?

On a side note, you don't necessarily need to compute the dominators. A relatively recent article presents an [algorithm to construct SSA form directly from bytecode or an AST](http://compilers.cs.uni-saarland.de/papers/bbhlmz13cc.pdf). — Martin Törnwall, May 19 '15 at 14:22

score 3 · Accepted Answer · answered May 19 '15 at 11:40

While calculating dominators quickly and correctly is indeed not trivial, there are much simpler algorithms than Lengauer-Tarjan that are as fast or faster in practice (i.e. on the kinds of control flow graphs that occur in most programs), even though the worst-case complexity sounds scary. See: Cooper, Keith D.; Harvey, Timothy J; and Kennedy, Ken (2001). "A Simple, Fast Dominance Algorithm"

score 1 · Answer 2 · answered May 19 '15 at 10:46

Ah, I see now, simple techniques don't correctly handle control flow graphs that have arbitrarily many forks and rejoins. You need to be able to find bypassing paths, and if you want the code for doing that to be fast, you need to calculate and remember semidominators, and then it looks like you end up with Lengauer-Tarjan or something along similar lines anyway.

Fast calculation of dominators

2 Answers2