summaryrefslogtreecommitdiffstats
path: root/src/regexp/syntax/simplify.go
blob: e439325139932c87a2a4367cb58586b1e2b3ebdf (plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
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
}