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// 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
}