Simple boolean operators for bit flags
Solution 1
And the remove two from it? I thought I could do something like this:
flags &= ~(0x1 | 0x2);
to remove those two flags, but apparently they remain there or something either way.
That is the correct way to remove flags. If you printf("%d\n", flags)
after that line, the output should be 4
.
I also do not know how to check if they do NOT exist in the bit flag (so I cannot check if my previous code works), would it be something like this?
if(flags & ~0x2) printf("flag 2 not set");
Nope:
if ((flags & 0x2) == 0)
printf("flag 2 not set");
EDIT:
To test for the presence of multiple flags:
if ((flags & (0x1 | 0x2)) == (0x1 | 0x2))
printf("flags 1 and 2 are set\n");
To test for the absence of multiple flags, just compare to 0 as before:
if ((flags & (0x1 | 0x2)) == 0)
printf("flags 1 and 2 are not set (but maybe only one of them is!)\n");
Solution 2
I'm not sure why you think that clearing operation won't work.
flags &= ~(0x1 | 0x2);
is the correct way to do it. The operation to check if a bit isn't set is:
if (!(flags & 0x2)) ...
The one you have:
if (flags & ~0x2) ...
will be true if any other bit is set, which is probably why you thing the clearing operation isn't working. The problem lies not with the clearing but with the checking.
If you want to check that all bits in a group are set:
if ((flags & (0x2|0x1)) == 0x2|0x1) ...
John
Updated on June 18, 2022Comments
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John almost 2 years
I am attempting to learn more about this to implement in my project.
I currently have got this basically:
unsigned char flags = 0; //8 bits flags |= 0x2; //apply random flag if(flags & 0x2) { printf("Opt 2 set"); }
Now I am wishing to do a little more complex things, what I am wanting to do is apply three flags like this:
flags = (0x1 | 0x2 | 0x4);
And then remove flags
0x1
and0x2
from it? I thought I could do something like this applying bitwise NOT (and bitwise AND to apply it):flags &= ~(0x1 | 0x2);
Apparently they remain there or something either way when I check.
I also do not know how to check if they do NOT exist in the bit flags (so I cannot check if my previous code works), would it be something like this?
if(flags & ~0x2) printf("flag 2 not set");
I can not find any resources from my recent searches that apply to this, I am willing to learn this to teach others, I am really interested. I apologize if this is confusing or simple.
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cdhowie over 13 yearsSure, no problem. Bitwise operators are fun. :)
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John over 13 yearsNow how would I check if two flags were set without boolean and? like
if(flags & (0x1 | 0x2))
but that appears to work even if I only set0x1
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John over 13 yearsThis has answered all my questions plus more, I am so glad for this. Thank you again. :)
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cdhowie over 13 yearsNo problem, that's what I'm here for. ;)