| // 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 |
| } |