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

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) (?...): 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 ?.