1// Copyright 2011 The Go Authors. All rights reserved.
2// Use of this source code is governed by a BSD-style
3// license that can be found in the LICENSE file.
4
5package syntax
6
7// Simplify returns a regexp equivalent to re but without counted repetitions
8// and with various other simplifications, such as rewriting /(?:a+)+/ to /a+/.
9// The resulting regexp will execute correctly but its string representation
10// will not produce the same parse tree, because capturing parentheses
11// may have been duplicated or removed. For example, the simplified form
12// for /(x){1,2}/ is /(x)(x)?/ but both parentheses capture as $1.
13// The returned regexp may share structure with or be the original.
14func (re *Regexp) Simplify() *Regexp {
15	if re == nil {
16		return nil
17	}
18	switch re.Op {
19	case OpCapture, OpConcat, OpAlternate:
20		// Simplify children, building new Regexp if children change.
21		nre := re
22		for i, sub := range re.Sub {
23			nsub := sub.Simplify()
24			if nre == re && nsub != sub {
25				// Start a copy.
26				nre = new(Regexp)
27				*nre = *re
28				nre.Rune = nil
29				nre.Sub = append(nre.Sub0[:0], re.Sub[:i]...)
30			}
31			if nre != re {
32				nre.Sub = append(nre.Sub, nsub)
33			}
34		}
35		return nre
36
37	case OpStar, OpPlus, OpQuest:
38		sub := re.Sub[0].Simplify()
39		return simplify1(re.Op, re.Flags, sub, re)
40
41	case OpRepeat:
42		// Special special case: x{0} matches the empty string
43		// and doesn't even need to consider x.
44		if re.Min == 0 && re.Max == 0 {
45			return &Regexp{Op: OpEmptyMatch}
46		}
47
48		// The fun begins.
49		sub := re.Sub[0].Simplify()
50
51		// x{n,} means at least n matches of x.
52		if re.Max == -1 {
53			// Special case: x{0,} is x*.
54			if re.Min == 0 {
55				return simplify1(OpStar, re.Flags, sub, nil)
56			}
57
58			// Special case: x{1,} is x+.
59			if re.Min == 1 {
60				return simplify1(OpPlus, re.Flags, sub, nil)
61			}
62
63			// General case: x{4,} is xxxx+.
64			nre := &Regexp{Op: OpConcat}
65			nre.Sub = nre.Sub0[:0]
66			for i := 0; i < re.Min-1; i++ {
67				nre.Sub = append(nre.Sub, sub)
68			}
69			nre.Sub = append(nre.Sub, simplify1(OpPlus, re.Flags, sub, nil))
70			return nre
71		}
72
73		// Special case x{0} handled above.
74
75		// Special case: x{1} is just x.
76		if re.Min == 1 && re.Max == 1 {
77			return sub
78		}
79
80		// General case: x{n,m} means n copies of x and m copies of x?
81		// The machine will do less work if we nest the final m copies,
82		// so that x{2,5} = xx(x(x(x)?)?)?
83
84		// Build leading prefix: xx.
85		var prefix *Regexp
86		if re.Min > 0 {
87			prefix = &Regexp{Op: OpConcat}
88			prefix.Sub = prefix.Sub0[:0]
89			for i := 0; i < re.Min; i++ {
90				prefix.Sub = append(prefix.Sub, sub)
91			}
92		}
93
94		// Build and attach suffix: (x(x(x)?)?)?
95		if re.Max > re.Min {
96			suffix := simplify1(OpQuest, re.Flags, sub, nil)
97			for i := re.Min + 1; i < re.Max; i++ {
98				nre2 := &Regexp{Op: OpConcat}
99				nre2.Sub = append(nre2.Sub0[:0], sub, suffix)
100				suffix = simplify1(OpQuest, re.Flags, nre2, nil)
101			}
102			if prefix == nil {
103				return suffix
104			}
105			prefix.Sub = append(prefix.Sub, suffix)
106		}
107		if prefix != nil {
108			return prefix
109		}
110
111		// Some degenerate case like min > max or min < max < 0.
112		// Handle as impossible match.
113		return &Regexp{Op: OpNoMatch}
114	}
115
116	return re
117}
118
119// simplify1 implements Simplify for the unary OpStar,
120// OpPlus, and OpQuest operators. It returns the simple regexp
121// equivalent to
122//
123//	Regexp{Op: op, Flags: flags, Sub: {sub}}
124//
125// under the assumption that sub is already simple, and
126// without first allocating that structure. If the regexp
127// to be returned turns out to be equivalent to re, simplify1
128// returns re instead.
129//
130// simplify1 is factored out of Simplify because the implementation
131// for other operators generates these unary expressions.
132// Letting them call simplify1 makes sure the expressions they
133// generate are simple.
134func simplify1(op Op, flags Flags, sub, re *Regexp) *Regexp {
135	// Special case: repeat the empty string as much as
136	// you want, but it's still the empty string.
137	if sub.Op == OpEmptyMatch {
138		return sub
139	}
140	// The operators are idempotent if the flags match.
141	if op == sub.Op && flags&NonGreedy == sub.Flags&NonGreedy {
142		return sub
143	}
144	if re != nil && re.Op == op && re.Flags&NonGreedy == flags&NonGreedy && sub == re.Sub[0] {
145		return re
146	}
147
148	re = &Regexp{Op: op, Flags: flags}
149	re.Sub = append(re.Sub0[:0], sub)
150	return re
151}
152