1 files changed, 151 insertions, 0 deletions
diff --git a/src/pkg/exp/regexp/syntax/simplify.go b/src/pkg/exp/regexp/syntax/simplify.go
new file mode 100644
index 000000000..72390417b
--- /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
+}