operators - How does the ? make a quantifier lazy in regex

Question

I've been looking into regex lately and figured that the ? operator makes the *,+, or ? lazy. My question is how does it do that? Is it that *? for example is a special operator, or does the ? have an effect on the * ? In other words, does regex recognize *? as one operator in itself, or does regex recognize *? as the two separate operators * and ? ? If it is the case that *? is being recognized as two separate operators, how does the ? affect the * to make it lazy. If ? means that the * is optional, shouldn't this mean that the * doesn't have to exists at all. If so, then in a statement .*? wouldn't regex just match separate letters and the whole string instead of the shorter string? Please explain, I'm desperate to understand.Many thanks.

Answer 1

? can mean a lot of different things in different contexts.

• Following a normal regex token (a character, a shorthand, a character class, a group...), it means "Match the previous item 0-1 times".
• Following a quantifier like ?, *, +, {n,m}, it takes on a different meaning: "Make the previous quantifier lazy instead of greedy (if that's the default; that can be changed, though - for example in PHP, the /U modifier makes all quantifiers lazy by default, so the additional ? makes them greedy).

• Right after an opening parenthesis, it marks the start of a special construct like for example

  a) (?s): mode modifiers ("turn on dotall mode")
  b) (?:...): make the group non-capturing
  c) (?=...) or (?!...): lookahead assertion
  d) (?<=...) or (?<!...): lookbehind assertion
  e) (?>...): atomic group
  f) (?<foo>...): named capturing group
  g) (?#comment): inline comments, ignored by the regex engine
  h) (?(?=if)then|else): conditionals

and others. Not all constructs are available in all regex flavors.

• Within a character class ([?]), it simply matches a verbatim ?.