| // 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 | |
| } | |