How does Integer.parseInt(string) actually work?

Question

Was asked this question recently and did not know the answer. From a high level can someone explain how Java takes a character / String and convert it into an int.

M

Answer 1

Usually this is done like this:

• init result with 0
• for each character in string do this
  • result = result * 10
  • get the digit from the character ('0' is 48 ASCII (or 0x30), so just subtract that from the character ASCII code to get the digit)
  • add the digit to the result
• return result

Edit: This works for any base if you replace 10 with the correct base and adjust the obtaining of the digit from the corresponding character (should work as is for bases lower than 10, but would need a little adjusting for higher bases - like hexadecimal - since letters are separated from numbers by 7 characters).

Edit 2: Char to digit value conversion: characters '0' to '9' have ASCII values 48 to 57 (0x30 to 0x39 in hexa), so in order to convert a character to its digit value a simple subtraction is needed. Usually it's done like this (where ord is the function that gives the ASCII code of the character):

digit = ord(char) - ord('0')

For higher number bases the letters are used as 'digits' (A-F in hexa), but letters start from 65 (0x41 hexa) which means there's a gap that we have to account for:

digit = ord(char) - ord('0')
if digit > 9 then digit -= 7

Example: 'B' is 66, so ord('B') - ord('0') = 18. Since 18 is larger than 9 we subtract 7 and the end result will be 11 - the value of the 'digit' B.

One more thing to note here - this works only for uppercase letters, so the number must be first converted to uppercase.