java - How this bitshift to build the number works? -

i saw method used faster reading of positive value of long.

public static long readlong(inputstream in) throws ioexception {     long n = 0;     int c = in.read();     while (c < 48 || c > 57) {         c = in.read();     }     while (c >= 48 && c <= 57) {         n = (n<<3) + (n<<1) + (c-'0');         c = in.read();     }      return n; } 

while understand of part, i'm not able this:
bit shifting odd number build number n = (n<<3) + (n<<1) + (c-'0');

why ignore 3rd bit , how it's able build it?

if of explain me in simple way, helpful.
thanks!

n << i n * 2^i. so,

(n<<3) + (n<<1) = (n * 2^3) + (n * 2^1) = n * (2^3 + 2^1) = n * 10 

basically, means shift value of n 1 digit left.

adding c + '0' means adding last digit integer value of c.