PocketMatrix

Does anyone understand this stuff i can count strait up in hex and i know the binary number system from my cisco class pretty well. But i dont understand how to use hex as flags and i really dont understand the pupose or bitwise operators i know what they do but why use them?

Numerical representation of FALSE is zero. Numerical representation of TRUE is either one or minus one, or any other non-zero standard.<br><br>When you're doing:<br><br>if(flag){}<br><br>you're really checking if the flag variable is non-zero, upon which the if statement is executed.<br>This is equivalent to the following statement:<br><br>if( flag != 0 )<br><br>In C, true and false are defined as:<br><br>#define TRUE 1<br>#define FALSE 0<br><br>whereas there is a defined type Boolean<br><br>typedef int Boolean;<br><br>Bitwise operators are used for accessing certain bits in a numerical entity. It's used in different areas, some of them being areas where you need to access only a certain bit (hardware registers), or bit per bit accessing a file format structure, or transformation of input data, or for example conversion from big endian to little endian format. <br><br>Ofcourse, there are millions of instances where you use bitwise operations, so I suggest you get a hang of it pretty soon <br><br>Say, you have a longword (32bits) and you only want the lowest 8 bits (byte). Look at this C snippet showing the & (AND) operator:<br><br>long a = 0xBAADF00D;<br>unsigned char b;<br><br>b = a & 0xFF;      // 0x000000FF<br><br><br>Variable b will now contain 0x0D.<br><br>>> is the bitwise shift right operator. It will shift the bitmask as many positions as you want to the right. << does the same to the left.<br><br>b = (a & 0xFF0000) >> 16;<br><br>will give b the value 0xAD. First the paranthesis will result in 0x00AD0000 and then the shift to right will result in 0x000000AD.<br><br><< and >> also have the effect of multiplying or diving by the power of 2 multiples.<br><br>a = 6;<br>a = a << 1; // times 2<br><br>a equals now 12;<br><br>a= 5;<br>a = a << 2; // times 4<br><br>a equals now 20.<br><br>a = 16;<br>a = a >> 3; // divided by 8<br><br>a equals 2.<br><br>the | (OR) operator is used for setting bits (AND was used for clearing bits):<br><br>unsigned int a = 0xFF00;<br>a = a | 0xAC;<br><br>a equals now 0xFFAC.<br><br>Bits that are already set, remain set.<br><br>The ~ (NOT) operator isused to reverse the bits from zero to 1 and viceversa.<br><br>a = 0x00FF00FF;<br>a = ~a<br><br>a equals 0xFF00FF00.<br><br>Let me see if you understood these things...<br>What is the value of a after these statements:<br><br>long a = 0x55AA55AA;<br>long b = 0xBAADF00D;<br><br>a = (a & b) | (~a >> 16);<br><br>Hope I haven't been a pain in the ass <br>

and bitwise shifting is faster than multiplication or division<br><br>(I had to add something!) ;D

BITFLAGS:<br>  Say you want to return information or set information, but only use one variable.  Well, it's like this. <br>===============================================<br>  #define DEF_COLOR       0x1 //BIN = 00000001<br>  #define DEF_BW          0x0 //BIN = 00000000<br>  #define DEF_FULLSCREEN  0x2 //BIN = 00000010<br>  #define DEF_SPRITE      0x0 //BIN = 00000000<br>  #define DEF_OPT1        0x4 //BIN = 00000100<br><br>  Then you can say <br>    INT x = DEF_COLOR | DEF_FULLSCREEN<br>  And you get<br>     x = 00000011<br>  So if you check it you see that<br>     x & DEF_FULLSCREEN > 0 //so the flag is on<br>     AND<br>     x & DEF_COLOR > 0 //so the flag is on<br>===============================================<br>BITWISE OPERATORS (SHIFTING)<br>  First off & and | are used to do BITWISE AND and OR, see above and you can see what this is about.<br>  Next is the shift, either << or >>.  Why?  Well, say you are converting WORD into two differnt values (Windows does this with lParams and wParams all the time).  Well you can do this.<br>  BYTE x = lParam<br>  BYTE y = lParam << 4<br>==================================================

Thanks guys between this and some online articles i think im about there, boy ill tell you ive known c for like 2 years and never understood this stuff, and i even know how to write windows apps in c with strait api, thanks to vb for teaching me api and you guys always knock vb. Now that im making the transition from C to C++ i guess i had better learn it.

No one showed how to clear a bit(s).<br><br>[fixed]<br>#define FLAG_1 0x0001<br>#define FLAG_2 0x0010<br><br>//Set both flag 1 and flag 2<br>int flags=FLAG_1|FLAG_2;<br><br>//Clear flag 1<br>flags&=~FLAG_1<br>[/fixed]<br><br>Dan East

Another thing to add; if you are using bits for flags, make sure that each flag is only one bit. You mentioned that you know hex, so you probably already understand:<br>[fixed]<br>Hex  Binary<br>0    0000<br>1    0001<br>2    0010<br>3    0011<br>4    0100<br>5    0101<br>6    0110<br>7    0111<br>8    1000<br>9    1001<br>A    1010<br>B    1011<br>C    1100<br>D    1101<br>E    1110<br>F    1111<br>[/fixed]<br><br>So you can see that hex values 1 2 4 and 8 are particularly valuable, because each of those hex numbers can be represented by a single bit. Those are the hex values you usually use for flags (unless you know what you are doing and use a value that sets 2 or more flags at a time).<br><br>Finally, do not start using binary flags in place of boolean type / operators, just because you can cram 32 true / false flags into a single int. Generally flags are used to pass many boolean values to a function, instead of passing a zillion parameters. It takes a lot more processing to extract / set binary flags as opposed to checking / setting the value of an entire int. So in other words, don't use flags in areas where your code is iterated often (games).<br><br>Dan East

Why does it feel like I am back in school?  I haven't read this much about Hex and Binary in about 10 years....<br>And people think that 4GLs are taking over!  HA!<br><br>I love this....