Hi, suppose I have a sequence of 64bits, i.e., a uint64_t. The goal is to flip a bit at position p of the bit sequence. How can I do this with C? I've read about masks and stuff, but masks only are fixed, where as I need something more generic.
For example, bit sequence: 110001 and I want to flip bit in the 1 position, I would get 110011. Flipping a bit in the 3rd position, I would get 111001.
Many thanks.
You can use exclusive-OR with a mask to toggle a bit.
You can roll this into a macro, and add bit-shifting to create the mask for the desired position.
Code:#define BIT_FLIP(x,y) /* fill in the blank; 'x' is the binary pattern, 'y' is the bit position */
Take the value "1" and left-shift it to the position you want, then XOR that with the value to flip the bit.
Code://try //{ if (a) do { f( b); } while(1); else do { f(!b); } while(1); //}
Okay, I don't really see any code here.Originally Posted by Matticus
I was leaving the coding up to you. Brewbuck gave you the general algorithm.
Here is what I've done:
For some reason, people seem to only want to give algorithms and not actual code for this problem.Code:uint64_t flip(uint64_t num, int p) { if (num & (1 << p)) { //odd num &= ~(1 << p); return num; } num ^= (1 << p); return num; }
Last edited by dayalsoap; 02-19-2015 at 04:07 PM. Reason: mistake in code
We don't provide code as answers to questions as many or most of the questions involve homework assignments. It is the policy here and with most web sites to help instruct the OP with enough information so the OP can code the answer themselves.
How can anyone learn HOW to program, if we do the coding for them?
Is it working for you?
I'm not sure what you're doing with the "if", but you should be (almost) fine with just the last two lines of that function (hint: Since your mask starts at '1', you need to shift it one less than time than you think).
If you're not good enough at coding to know how to do it, you generally won't get code since the point is for you to figure it out yourself and learn in the process.
If you are good enough at coding to know how to do it, you generally won't get code since you could do it yourself.
This line
Hmm. You are right. No need for the if. I thought this always sets it to 0, but it actually just flips it.Code:num ^= (1 << p);
Last edited by dayalsoap; 02-19-2015 at 04:33 PM.
This is where a pen-and-paper analysis would help you understand the details.
See how I subtracted one of the position variable? And do you understand why?Code:/* num = 1101 // binary p = 3 // position >> num ^= (1 << p); Right side: 0001 << 3 - 1 // minus one because we start at 1, so we shift one less time 0010 // shift one 0100 // shift two ... now our set bit is in the third position Now we do the XOR: 1101 // num 0100 // mask ---- 1001 // result A zero in the mask will not change the value: 0 xor 0 = 0 0 xor 1 = 1 A one in the mask will change the value: 1 xor 0 = 1 1 xor 1 = 0 */
I don't see why I need to subtract one, at all. If I pass in position 0, I don't want to subtract position -1.
You can run the code as I've written with your removal of the if statement and it runs exactly fine.
For some reason, I assumed you were considering the LSB as position 1.
Apart from that, did the analysis I showed help you understand the mechanics of the algorithm?
