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cmd/compile: optimize noding of long summation expressions

Fixes #16394.

Change-Id: I7108c9e8e67d86678bdb6015f0862e5c92bcf911
Reviewed-on: https://go-review.googlesource.com/76450
Reviewed-by: Robert Griesemer <gri@golang.org>
This commit is contained in:
Matthew Dempsky 2017-11-07 15:55:26 -08:00
parent ed3d672766
commit 506386fd37

View File

@ -537,6 +537,9 @@ func (p *noder) expr(expr syntax.Expr) *Node {
// ntype? Shrug, doesn't matter here.
return p.nod(expr, ODOTTYPE, p.expr(expr.X), p.expr(expr.Type))
case *syntax.Operation:
if expr.Op == syntax.Add && expr.Y != nil {
return p.sum(expr)
}
x := p.expr(expr.X)
if expr.Y == nil {
if expr.Op == syntax.And {
@ -597,6 +600,82 @@ func (p *noder) expr(expr syntax.Expr) *Node {
panic("unhandled Expr")
}
// sum efficiently handles very large summation expressions (such as
// in issue #16394). In particular, it avoids left recursion and
// collapses string literals.
func (p *noder) sum(x syntax.Expr) *Node {
// While we need to handle long sums with asymptotic
// efficiency, the vast majority of sums are very small: ~95%
// have only 2 or 3 operands, and ~99% of string literals are
// never concatenated.
adds := make([]*syntax.Operation, 0, 2)
for {
add, ok := x.(*syntax.Operation)
if !ok || add.Op != syntax.Add || add.Y == nil {
break
}
adds = append(adds, add)
x = add.X
}
// nstr is the current rightmost string literal in the
// summation (if any), and chunks holds its accumulated
// substrings.
//
// Consider the expression x + "a" + "b" + "c" + y. When we
// reach the string literal "a", we assign nstr to point to
// its corresponding Node and initialize chunks to {"a"}.
// Visiting the subsequent string literals "b" and "c", we
// simply append their values to chunks. Finally, when we
// reach the non-constant operand y, we'll join chunks to form
// "abc" and reassign the "a" string literal's value.
//
// N.B., we need to be careful about named string constants
// (indicated by Sym != nil) because 1) we can't modify their
// value, as doing so would affect other uses of the string
// constant, and 2) they may have types, which we need to
// handle correctly. For now, we avoid these problems by
// treating named string constants the same as non-constant
// operands.
var nstr *Node
chunks := make([]string, 0, 1)
n := p.expr(x)
if Isconst(n, CTSTR) && n.Sym == nil {
nstr = n
chunks = append(chunks, nstr.Val().U.(string))
}
for i := len(adds) - 1; i >= 0; i-- {
add := adds[i]
r := p.expr(add.Y)
if Isconst(r, CTSTR) && r.Sym == nil {
if nstr != nil {
// Collapse r into nstr instead of adding to n.
chunks = append(chunks, r.Val().U.(string))
continue
}
nstr = r
chunks = append(chunks, nstr.Val().U.(string))
} else {
if len(chunks) > 1 {
nstr.SetVal(Val{U: strings.Join(chunks, "")})
}
nstr = nil
chunks = chunks[:0]
}
n = p.nod(add, OADD, n, r)
}
if len(chunks) > 1 {
nstr.SetVal(Val{U: strings.Join(chunks, "")})
}
return n
}
func (p *noder) typeExpr(typ syntax.Expr) *Node {
// TODO(mdempsky): Be stricter? typecheck should handle errors anyway.
return p.expr(typ)