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# Changing a bit value in a binary file

Posted on 1999-01-05
Medium Priority
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Hi,

I have pulled out my hair trying to figure this out.  I've got the following code which finds a unique number within a bitmap file, and lets the calling method know if the bit representation is turned on or off:

byte valByte;

// param is a 6-digit number string
uint16 bitNum = atoi ((char *) &param);
uint32 byteOffset = bitNum / 8;
uint16 bitOffset  = bitNum % 8;
// Error checking deleted
// Error checking deleted
if (valByte & (int)pow (2, bitOffset) && readMap)
// This bit is turned on
else
// This bit is turned off

I now need to be able to flip the bit value.  I was thinking I could try something like this:

if ((valByte & (int)pow (2, bitOffset)) == 0) {
// If the bit is off, turn it on
(valByte & (int)pow (2, bitOffset)) = 1;
} else {
// The card must be on, turn it off
(valByte & (int)pow (2, bitOffset)) = 0;
};
// Now save the byte back to the file

But for obvious reasons (no Lvalue), this won't work.

For information, this code will only be run on PC-based (8-bit byte based) machines.  Any help would be immensely appreciated!!
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Question by:cgoldfarb
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LVL 85

Expert Comment

ID: 1181605
valByte ^= (1<<bitOffset);
0

LVL 7

Expert Comment

ID: 1181606
ozo means to replace all pow (2, bitOffset) with (1<<bitOffset) and for flip the bit value, use
valByte ^= (1<<bitOffset); this single line does the job of your if/else block.
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Author Comment

ID: 1181607
Thank you both for your help.  It's exactly what I was looking for.  Will hopefully return the favor...
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Accepted Solution

erick1217 earned 600 total points
ID: 1181608
try to turning on
valByte |= (int)pow (2, bitOffset);

and to turning off

valByte &= 256 - (int)pow (2, bitOffset);

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LVL 85

Expert Comment

ID: 1181609
255-pow may work better than 256-pow
but it's still pretty silly to use transcendental functions to approximate the << operator
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Expert Comment

ID: 1181610
sorry ozo but the power of bit 4 (by example) is 8
and 256 - 8 = 248
valByte &= 248 will set the bit 4 to zero
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LVL 85

Expert Comment

ID: 1181611
valByte &= 248 will set bits 0, 1, and 2 to zero
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Expert Comment

ID: 1181612
yes.. you're right... excuse me...
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Author Comment

ID: 1181613
erick1217's answer was good, although ozo and faster hit the nail on the head.  They really deserve the points, as they were the first to respond.
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