why only first letter is returned by match function?

Question

If metacharacter ? matches the preceding element zero or one time, then

why

"ab".match(/a?/)

returns ["a"],

but

"ab".match(/b?/)

returns [""]

?

Answer 1

Because that's the first match. The regex tries at first to match at position 0, where regex#1 does match an a and regex#2 does match the empty string. Then it attempts to match at position 1, where regex#1 does match the empty string and regex#2 does match the letter b. At last, it tries to match at position 3, where both regexes match the empty string.

Compare the returned matches with a global flag:

> "ab".match(/a?/)
["a"]
> "ab".match(/a?/g)
["a", "", ""]
> "ab".match(/b?/)
[""]
> "ab".match(/b?/g)
["", "b", ""]

why not [""] is returned in first case?

Due to the mechanisms of backtracking. When attempting to match at some position, the engine will try to greedily¹ test all letters of the regex against the letters of the string. When it reaches the end of the regex with that method, a match succeeded. When a letter doesn't fit in, it tries to go back in the regex to see whether any omissions can be made - when using modifiers such as * or ? - or alternatives (|) need to be considered, and then continues from there.

Example: Match /b?/ at position 0 of "ab":

//   - "": ✓
/b/  - "a": ×
/b?/ - "": ✓ - succeed (end of regex)
  ^ means here that the "b" token is omitted

Example: Match /a?/ at position 0 of "ab":

//  - "": ✓
/a/ - "a": ✓ - succeed (end of regex)

Example: Match /ab?(bc)?/ at position 0 of "abc"

//        - "": ✓
/a/       - "a": ✓
/ab/      - "ab": ✓
/ab(b)/   - "abc": ×
/ab(bc)?/ - "ab": ✓ - succeed (end of regex)

_{1: Usually, at least. Many regex flavours also provide quantifiers that are lazy or possessive if you want to control the exact matching behaviour. For example, /ab??(bc)?/ matches abc in "abc"}