CODEWITHSUNDEEP
c++bit-manipulation
P. Avery
P. Avery

Reputation: 809

Flags using Bits

I'm trying to do simple bit operations on a 'char' variable; I would like to define 5 constants.

const int a = 0;
const int b = 1;
const int c = 2;
const int d = 3;
const int e = 4;

When I try to set more than one bit of the char, all bits apparently up to the set bit a read as set...here is code I use to set and read bits of the char var:

char var = 0;
var |= c;
var|= d;

BOOL set = false;
if(var & b)
set = true; // reads true
if(var & c)
set = true; // also reads true
if(var & d)
set = true; // also reads true

I read an incomplete thread that says that the operation to set bits may be different for x86...the system I'm using...is that the case here?

Upvotes: 0

Views: 188

Answers (2)

parapura rajkumar
parapura rajkumar

Reputation: 24413

Change your definitions to because they way you have defined it some of them has more than one bit set

const int a = 1 << 0;
const int b = 1 << 1;
const int c = 1 << 2;
const int d = 1 << 3;
const int e = 1 << 4;

This way it is evident that each constant only has 1 bit set.

If you want to learn all about the various bit hacks...

Upvotes: 3

Qaz
Qaz

Reputation: 61970

You're cutting into your other "bits"' space. Examining a couple gives us:

b = 1 = 0001
c = 2 = 0010
d = 3 = 0011 //uh oh, it's b and c put together (ORed)

To get around this, make each one represent a new bit position:

const int a = 0; //or 0x0
const int b = 1; //or 0x1 
const int c = 2; //or 0x2 or 1 << 1
const int d = 4; //or 0x4 or 1 << 2
const int e = 8; //or 0x8 or 1 << 3

You should consider not using 0 if there's a possibility of no bits being set meaning something different, too. The main application for this is to set and check flags, and no flags set definitely shows independence.

Upvotes: 4

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