How this bit conversion going on?

Question

 private int ControlDecode(byte b) { int itype = -1; int ipacekttype = b & 0xF0; switch (ipacekttype) { case 0x40: if ((b & 0x0F) != 0x0F) itype = 0x45; else itype = 0x4F; break; case 0x50: itype = 0x45; break; case 0x00: itype = 0x00; break; case 0x80: itype = 0x89; break; case 0x90: itype = 0x89; break; default: break; } return itype; }

If b=51, ipacekttype=48( according to code execution) but theoratically when masking we should get 50 as answer can anyone explain me how (b & 0xF0) works here please?

"when masking we should get 50" - How do you figure? decimal 51 = 0011 0011, bitwise and with 0xF0 (which is 1111 0000) = 0011 0000, which is indeed decimal 48. — Glorin Oakenfoot
– Glorin Oakenfoot, Commented Jul 27, 2016 at 13:05

Sergey Kalinichenko · Accepted Answer · 2016-07-27 13:02:24Z

It looks like you are mixing decimal notation and hex notation. When you say b=51, it appears that 51 is decimal, so bn=0x33 in hex. This explains why ipacekttype=48 - again, in decimal, because it's ipacekttype=0x30 in hex, i.e. exactly what it is supposed to be after masking 0x33 with 0xF0.

Collectives™ on Stack Overflow

How this bit conversion going on?

1 Answer 1

Comments

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Comments

Related