这是如何工作的：chr（（$ number >> 6）+192）.chr（（$ number＆63）+128）;

Question

Can you please explain how this line of code is equivalent to the next code:

<?php
$string = chr( ( $number >> 6 ) + 192 ).chr( ( $number & 63 ) + 128 );
?>

Its equivalent to :

if ( $number >=128 && $number <=2047 ){

   $byte1 = 192 + (int)($number / 64); //= 192 + ( $number >> 6 )
   $byte2 = 128 + ($number % 64);      //= 128 + ( $number & 63 )
   $utf = chr($byte1).chr($byte2);
 }

for example entering number 1989 both produces ߅

These codes are used for converting UNICODE Entities back to original UTF-8 characters.

Answer 1

The code on top uses binary operators. >> is right shift operator. It shifts the bit in the number to the right (towards more significant bits).

So 11110000 >> 2 = 00111100

It's equivalent to division by powers of 2 $number >> $n is the same as $number / pow(2,$n).

The & is the "bitwise and" operator. It compares respective bits on both numbers, and sets in result those, that are 1 in both numbers.

11110000 & 01010101 = 01010000

By and'ing $number with 63 (001111111) you get the remainder of dividing $number by 64 (aka the modulus), which is written $number % 64.