What do the regex flags do?

Eight flags, readable back as re.flags in the alphabetical order dgimsuvy.

Three things interviewers look for:

• g and y make the regex object stateful. lastIndex is an own data property that exec and test read and write. A failed match resets it to 0, so reusing one /…/g object across inputs gives alternating results. Without g or y, lastIndex is ignored and never updated.
• u and v change the unit of matching from UTF-16 code unit to code point, so . matches an astral character as one unit. v is the successor of u: it adds set subtraction --, intersection &&, nested classes, and \q{abc} string literals, plus properties of strings such as \p{RGI_Emoji}. u and v together is a SyntaxError. Prefer v in new code.
• Inline modifiers (?i:…), (?-i:…) apply i, m or s only inside the group (ES2025). Other flags cannot be toggled.

// g: shared state across calls
const seen = /b/g;
console.log(['ab', 'ab'].map((s) => seen.test(s)), seen.lastIndex);
// → [ true, false ] 0

// y: anchored at lastIndex, the basis of tokenizers
const tok = /\d+/y;
tok.lastIndex = 2;
console.log(tok.exec('a 42')?.[0], tok.lastIndex, tok.exec('a 42'));
// → 42 4 null

// d: where did the group match?
const m = /(?<year>\d{4})/d.exec('on 2026-10');
console.log(m.indices.groups.year);
// → [ 3, 7 ]

// u: code points; v: set notation
console.log('😀'.match(/./)[0].length, '😀'.match(/./u)[0].length);
// → 1 2
console.log(/^[\p{L}--[a-z]]$/v.test('a'), /^[\p{L}--[a-z]]$/v.test('Z'));
// → false true

// inline modifier: ignore case only inside the group
console.log(/(?i:ab)c/.test('ABc'), /(?i:ab)c/.test('ABC'));
// → true false

• Reusing a g or y regex with test or exec across different strings. lastIndex carries over, so a correct pattern returns false on the second input. Create the regex per use, reset lastIndex = 0, or drop g when you only need a boolean.
• Expecting y to scan ahead like g. With y, a miss at lastIndex fails immediately and resets lastIndex to 0.
• Using . and [^x] on text with emoji or other astral characters without u or v. They match half a surrogate pair, and i uses simpler case folding without u: /\u212a/i.test('k') is false, /\u212a/iu.test('k') is true.
• Combining u and v. It throws at construction. Also, v rejects some patterns u accepts (for example unescaped (, ), [, {, }, /, -, | inside a class), so migration needs a pass over every character class.