Witch bit?
Given:
typedef enum
{
FLAG1 = 0,
FLAG2 = 1,
FLAG3 = 2,
FLAG4 = 4
} FLAGS;
int bitField = FLAG1 | FLAG3;
Then:
(bitField & FLAG1) will be non-zero
(bitField & FLAG4) will be zero
typedef enum
{
FLAG1 = 0,
FLAG2 = 1,
FLAG3 = 2,
FLAG4 = 4
} FLAGS;
int bitField = FLAG1 | FLAG3;
Then:
(bitField & FLAG1) will be non-zero
(bitField & FLAG4) will be zero
simple enough, do you know about the bit-wise operators in your language of choice?
here's a quick C++ example:
-Danu
[Edited by - silvermace on November 11, 2004 4:15:13 PM]
here's a quick C++ example:
int myBitField = 0;int myFlag1 = 1; // 0001 base 2int myFlag2 = 2; // 0010 base 2int myFlag3 = 4; // 0100 base 2int myFlag4 = 8; // 1000 base 2myBitField = myFlag2 | myFlag4; // myBitFile => 1010 base 2// test to see if a bit is on:if( myBitField & myFlag2 ) { std::cout << "Flag 2 is enabled." << std::endl;}myBitField 1010myFlag2 0010---------------- 0010 > 0 therefore true-Danu
[Edited by - silvermace on November 11, 2004 4:15:13 PM]
You can use something like this.
#define BIT(x) (1 << (x))
then use it for
bool testBit( int bitField, char bitNum )
{
return ( bitField & BIT(bitNum) );
}
#define BIT(x) (1 << (x))
then use it for
bool testBit( int bitField, char bitNum )
{
return ( bitField & BIT(bitNum) );
}
Quote:Original post by silvermaceint myBitField = 0;int myFlag1 = 1; // 001 base 2int myFlag2 = 2; // 010 base 2int myFlag3 = 3; // 011 base 2int myFlag4 = 4; // 100 base 2
Um, last I checked, most bitfields had non overlapping bits. Aka:
int myBitField = 0;const int myFlag1 = 1; // 0001const int myFlag2 = 2; // 0010const int myFlag3 = 4; // 0100const int myFlag4 = 8; // 1000Given your example,
myBitField = myFlag1 | myFlag2;if (myBitField & myFlag3) //evaluates to true!!!(Didn't sleep enough last night? ;-))
Quote:Original post by Anonymous Poster
Given:
typedef enum
{
FLAG1 = 0,
FLAG2 = 1,
FLAG3 = 2,
FLAG4 = 4
} FLAGS;
int bitField = FLAG1 | FLAG3;
Then:
(bitField & FLAG1) will be non-zero
(bitField & FLAG4) will be zero
Another person who didn't get enough sleep last night!!! The first statement will also be zero. Let's take a quick look at the binary values of the flags:
FLAG1 = 00000000 //Note: no 1 bits!!!
FLAG2 = 00000001
FLAG3 = 00000010
FLAG4 = 00000100
Now, some binary math:
00000000 (FLAG1) | 00000010 (FLAG3)-------------- = 00000010 (-> bitField)Then the test: 00000010 (bitField) & 00000000 (FLAG1)-------------- = 00000000 (0, false)Quote:Original post by Anonymous Poster
You can use something like this.
#define BIT(x) (1 << (x))
then use it for
bool testBit( int bitField, char bitNum )
{
return ( bitField & BIT(bitNum) );
}
Why use a macro if you're just going to use it in a function?
bool testBit( int bitField , char bitNum ){ return ( bitField & (1<<bitNum) );}void setBit( int & bitField , char bitNum ){ bitField |= (1 << bitNum);}void clearBit( int & bitField , char bitNum ){ bitField &= ~(1 << bitNum);}Quote:Original post by silvermace
monkey your right, /me edits.
wierd, in my engine its done right lol
People's brains just tend to glaze over details sometimes ^_^. I've done similar. And the AP including a Flag1 = 0 just goes to show it's not just you XP.
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