exp/regexp/syntax: finish Regexp manipulation
Except for the inevitable bug fixes, the Regexp code is done.
R=sam.thorogood, r
CC=golang-dev
https://golang.org/cl/4635082
diff --git a/src/pkg/exp/regexp/syntax/simplify.go b/src/pkg/exp/regexp/syntax/simplify.go
new file mode 100644
index 0000000..7239041
--- /dev/null
+++ b/src/pkg/exp/regexp/syntax/simplify.go
@@ -0,0 +1,151 @@
+// Copyright 2011 The Go Authors. All rights reserved.
+// Use of this source code is governed by a BSD-style
+// license that can be found in the LICENSE file.
+
+package syntax
+
+// Simplify returns a regexp equivalent to re but without counted repetitions
+// and with various other simplifications, such as rewriting /(?:a+)+/ to /a+/.
+// The resulting regexp will execute correctly but its string representation
+// will not produce the same parse tree, because capturing parentheses
+// may have been duplicated or removed. For example, the simplified form
+// for /(x){1,2}/ is /(x)(x)?/ but both parentheses capture as $1.
+// The returned regexp may share structure with or be the original.
+func (re *Regexp) Simplify() *Regexp {
+ if re == nil {
+ return nil
+ }
+ switch re.Op {
+ case OpCapture, OpConcat, OpAlternate:
+ // Simplify children, building new Regexp if children change.
+ nre := re
+ for i, sub := range re.Sub {
+ nsub := sub.Simplify()
+ if nre == re && nsub != sub {
+ // Start a copy.
+ nre = new(Regexp)
+ *nre = *re
+ nre.Rune = nil
+ nre.Sub = append(nre.Sub0[:0], re.Sub[:i]...)
+ }
+ if nre != re {
+ nre.Sub = append(nre.Sub, nsub)
+ }
+ }
+ return nre
+
+ case OpStar, OpPlus, OpQuest:
+ sub := re.Sub[0].Simplify()
+ return simplify1(re.Op, re.Flags, sub, re)
+
+ case OpRepeat:
+ // Special special case: x{0} matches the empty string
+ // and doesn't even need to consider x.
+ if re.Min == 0 && re.Max == 0 {
+ return &Regexp{Op: OpEmptyMatch}
+ }
+
+ // The fun begins.
+ sub := re.Sub[0].Simplify()
+
+ // x{n,} means at least n matches of x.
+ if re.Max == -1 {
+ // Special case: x{0,} is x*.
+ if re.Min == 0 {
+ return simplify1(OpStar, re.Flags, sub, nil)
+ }
+
+ // Special case: x{1,} is x+.
+ if re.Min == 1 {
+ return simplify1(OpPlus, re.Flags, sub, nil)
+ }
+
+ // General case: x{4,} is xxxx+.
+ nre := &Regexp{Op: OpConcat}
+ nre.Sub = nre.Sub0[:0]
+ for i := 0; i < re.Min-1; i++ {
+ nre.Sub = append(nre.Sub, sub)
+ }
+ nre.Sub = append(nre.Sub, simplify1(OpPlus, re.Flags, sub, nil))
+ return nre
+ }
+
+ // Special case x{0} handled above.
+
+ // Special case: x{1} is just x.
+ if re.Min == 1 && re.Max == 1 {
+ return sub
+ }
+
+ // General case: x{n,m} means n copies of x and m copies of x?
+ // The machine will do less work if we nest the final m copies,
+ // so that x{2,5} = xx(x(x(x)?)?)?
+
+ // Build leading prefix: xx.
+ var prefix *Regexp
+ if re.Min > 0 {
+ prefix = &Regexp{Op: OpConcat}
+ prefix.Sub = prefix.Sub0[:0]
+ for i := 0; i < re.Min; i++ {
+ prefix.Sub = append(prefix.Sub, sub)
+ }
+ }
+
+ // Build and attach suffix: (x(x(x)?)?)?
+ if re.Max > re.Min {
+ suffix := simplify1(OpQuest, re.Flags, sub, nil)
+ for i := re.Min + 1; i < re.Max; i++ {
+ nre2 := &Regexp{Op: OpConcat}
+ nre2.Sub = append(nre2.Sub0[:0], sub, suffix)
+ suffix = simplify1(OpQuest, re.Flags, nre2, nil)
+ }
+ if prefix == nil {
+ return suffix
+ }
+ prefix.Sub = append(prefix.Sub, suffix)
+ }
+ if prefix != nil {
+ return prefix
+ }
+
+ // Some degenerate case like min > max or min < max < 0.
+ // Handle as impossible match.
+ return &Regexp{Op: OpNoMatch}
+ }
+
+ return re
+}
+
+// simplify1 implements Simplify for the unary OpStar,
+// OpPlus, and OpQuest operators. It returns the simple regexp
+// equivalent to
+//
+// Regexp{Op: op, Flags: flags, Sub: {sub}}
+//
+// under the assumption that sub is already simple, and
+// without first allocating that structure. If the regexp
+// to be returned turns out to be equivalent to re, simplify1
+// returns re instead.
+//
+// simplify1 is factored out of Simplify because the implementation
+// for other operators generates these unary expressions.
+// Letting them call simplify1 makes sure the expressions they
+// generate are simple.
+func simplify1(op Op, flags Flags, sub, re *Regexp) *Regexp {
+ // Special case: repeat the empty string as much as
+ // you want, but it's still the empty string.
+ if sub.Op == OpEmptyMatch {
+ return sub
+ }
+ // The operators are idempotent if the flags match.
+ if op == sub.Op && flags&NonGreedy == sub.Flags&NonGreedy {
+ return sub
+ }
+ if re != nil && re.Op == op && re.Flags&NonGreedy == flags&NonGreedy && sub == re.Sub[0] {
+ return re
+ }
+
+ re = &Regexp{Op: op, Flags: flags}
+ re.Sub = append(re.Sub0[:0], sub)
+ return re
+}
