diff options
Diffstat (limited to 'src/regexp/syntax/simplify.go')
-rw-r--r-- | src/regexp/syntax/simplify.go | 151 |
1 files changed, 151 insertions, 0 deletions
diff --git a/src/regexp/syntax/simplify.go b/src/regexp/syntax/simplify.go new file mode 100644 index 0000000..e439325 --- /dev/null +++ b/src/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 +} |