pretend my char is 85 (01010101)
if I only want to read the last 3 bits (010...) from the char and use it as a separate value, how do I do it?
since 010 the result would be 2, I know the value of the bits are found at 0x80, 0x40, and 0x20.
any one know how to do this?
Combination of bit masking and shifting.
0x80 | 0x40 | 0x20 is 0xE0, just shown like that to see where it comes from.Code:#include <iostream> int main() { int test=85; std::cout << ((test & (0x80 | 0x40 | 0x20)) >> 5) << std::endl; return 0; }
Last edited by SilentStrike; 07-19-2002 at 02:08 PM.
SilentStrike,
How does '>> 5' work?
I understand what you're doing up the point of >> 5. The first part, (test & (0x80 | 0x40 | 0x20)), should return 01000000b. Judging by the output, I'm guessing that '>> 5' shifts everything to the right by 5 bits.
fletch
"Logic is the art of going wrong with confidence."
Morris Kline
You can use bitfields to manipulate specific bits:
Code:#include <iostream> using namespace std; typedef struct { unsigned int bit0:1; unsigned int bit1:1; unsigned int bit2:1; unsigned int bit3:1; unsigned int bit4:1; unsigned int bit5:1; unsigned int bit6:1; unsigned int bit7:1; } bitfield_s; typedef union { bitfield_s bitfield; unsigned char byte; } byte_u; int main () { byte_u var; var.byte = 0xBA; cout << "bit 0: " << var.bitfield.bit0 << endl; cout << "bit 6: " << var.bitfield.bit6 << endl; return 0; }
That's exactly what it does.I understand what you're doing up the point of >> 5. The first part, (test & (0x80 | 0x40 | 0x20)), should return 01000000b. Judging by the output, I'm guessing that '>> 5' shifts everything to the right by 5 bits.
okay, but what if I only want to read the first 3 bits of the byte (0x01, 0x02, and 0x04) and use it as a separate value?
Code:int test1, test2; test1 = 85; test2 = ((test1 & 0xE0) >> 5); std::cout << "test1 = " << test1 << std::endl; // 85 std::cout << "test2 = " << test2 << std::endl; // 2
"Logic is the art of going wrong with confidence."
Morris Kline
