import Foundation
let pattern = #"\{(?:(?:\{(?:(?:\{(?:[^{}])*\})|(?:[^{}]))*\})|(?:[^{}]))*\}"#
let regex = try! NSRegularExpression(pattern: pattern)
let testString = #"""
captures up to singly-nested blocks:
p{1} = \{(?:[^{}])*\}
int main() {
return 0;
}
captures up to doubly-nested:
p{2} = \{(?:(?:p{1})|(?:[^{}]))*\}
...or...
p{2} = \{(?:(?:\{(?:[^{}])*\})|(?:[^{}]))*\}
void more_complex_fxn(int z) {
int x = z / 9;
if (x > 3) {
do_something;
x += 5;
}
}
Following the pattern set above...
up to triply-nested:
p{3} = \{(?:(?:p{2})|(?:[^{}]))*\}
...or...
p{3} = \{(?:(?:\{(?:(?:\{(?:[^{}])*\})|(?:[^{}]))*\})|(?:[^{}]))*\}
double so_many_blocks(double x) {
double y = x;
while(y > 4) {
if (y < 6) {
y -= 0.3;
} else {
y -= 0.5;
}
}
return y;
}
"""#
let stringRange = NSRange(location: 0, length: testString.utf16.count)
let matches = regex.matches(in: testString, range: stringRange)
var result: [[String]] = []
for match in matches {
var groups: [String] = []
for rangeIndex in 1 ..< match.numberOfRanges {
let nsRange = match.range(at: rangeIndex)
guard !NSEqualRanges(nsRange, NSMakeRange(NSNotFound, 0)) else { continue }
let string = (testString as NSString).substring(with: nsRange)
groups.append(string)
}
if !groups.isEmpty {
result.append(groups)
}
}
print(result)
Please keep in mind that these code samples are automatically generated and are not guaranteed to work. If you find any syntax errors, feel free to submit a bug report. For a full regex reference for Swift 5.2, please visit: https://developer.apple.com/documentation/foundation/nsregularexpression