why is the base64 encode java code doing this -

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so i'm trying understand base64 encoding better , came across implementation on wikipedia

private static string base64encode(byte[] in)       {         stringbuffer out = new stringbuffer((in.length * 4) / 3);         int b;         (int = 0; < in.length; += 3)  {             b = (in[i] & 0xfc) >> 2;             out.append(codes.charat(b));             b = (in[i] & 0x03) << 4;             if (i + 1 < in.length)      {                 b |= (in[i + 1] & 0xf0) >> 4;                 out.append(codes.charat(b));                 b = (in[i + 1] & 0x0f) << 2;                 if (i + 2 < in.length)  {                     b |= (in[i + 2] & 0xc0) >> 6;                     out.append(codes.charat(b));                     b = in[i + 2] & 0x3f;                     out.append(codes.charat(b));                 } else  {                     out.append(codes.charat(b));                     out.append('=');                 }             } else      {                 out.append(codes.charat(b));                 out.append("==");             }         }          return out.tostring();     }

and i'm following along , line: 

b = (in[i] & 0xfc) >> 2;

and don't it...why bitwise , 252 number shift right 2...wouldn't same if shifted byte without doing bitwise operation? example:

b = in[i] >> 2;

say in[i] letter e...represented 101 or in binary 01100101. if shift 2 right 011001 or 25. if bitwise & 

01100101 11111100 -------- 01100100

but shift going chop off last 2 anyway...so why bother doing it?

can clarify me please. thanks.

in in[i] >> 2, in[i] converted int first. if negative byte (with high bit set) converted negative int (with now-highest 24 bits set well).

in (in[i] & 0xfc) >> 2, in[i] converted int above, , & 0xfc makes sure bits reset 0.

you're partially right, in (in[i] & 0xff) >> 2 give same result. & 0xff common way convert byte non-negative int in range 0 255.

the way know sure why original developer used 0xfc, , not 0xff, ask them - speculate it's make more obvious bits being used.


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