IMO your interpretation of the order of the arguments to setbits() is misplaced. The arguments to setbits(x, p, n, y) are: x is a number whose "n" bits located at "p" need to match the rightmost "n" bits of y.Originally Posted by Sebastiani
Edit: based on the above, it should be called as setbits(85, 4, 1, 170) where
the bit (of x=85) located at bit position 4 needs to match the LSB of the number y (=170)
called such, setbits() outputs 69
Last edited by itCbitC; 07-08-2009 at 09:44 PM.
>> IMO your interpretation of the order of the arguments to setbits() is misplaced. The arguments to setbits(x, p, n, y) are: x is a number whose "n" bits located at "p" need to match the rightmost "n" bits of y.
Ok, you're right, I misread the requirements there. But there does seem to be a bug somewhere in your code. I ran it again on the values given by the OP and only dwks' gave the correct result.
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
IBTD, but perhaps you can point to where the problem in the code maybe.
>> IBTD, but perhaps you can point to where the problem in the code maybe.
Are you kidding? I don't even understand the problem clearly, to be quite honest!I mean, if you look at the OP's first post it sets the the value of 'a' to 0x01234567, or in (32 bit) binary:
Now if 'p' is 30, then we want a mask for 'a' of:Code:0000 0001 0010 0011 0100 0101 0110 0111
So at the very least, the result is going to be:Code:0011 1111 1111 1111 1111 1111 1111 1111
Where X is a position to be filled by bits from 'b'. How then did they arrive at:Code:XX00 0001 0010 0011 0100 0101 0110 0111
I just don't get it!Code:0101 0011 0000 0011 0100 0101 0110 0111
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
All right then call the function as setbits(290, 7, 2, 87).Are you kidding? I don't even understand the problem clearly, to be quite honest!
x = 290 (100100010)
y = 87 (1010111)
Now the 2 bits of x at position 7 must match the 2 rightmost bits of y.
After setbits() returns the new value of x will be 482 (111100010).
>> Now the 2 bits of x at position 7 must match the 2 rightmost bits of y.
I see. The notation is a bit confusing then. I would consider bit position 7 to be the 8th bit (zero-based index) and the rightmost bits of a number to be the most significant.
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
Just that. It's -1u. -1 converted to type unsigned. It's a quick and dirty way to get all 1's in a number. Not very portable. I suppose using ~0 would be better.Not sure what you intended by the number in red below.
Hooray!I ran it again on the values given by the OP and only dwks' gave the correct result.
Do borrow it!EDIT: Nice syntax highlighting, BTW. Uh, do you mind if I borrow it?That's what it's there for. You can run it online or download the source from dwks.theprogrammingsite.com/myprogs/codeform.htm (it's GPL'd).
If you're running on Windows, I wrote a bunch of other nifty features into codeform. Like, you can make it so that when you copy code starting with "//cf", codeform codeforms the code in the background, and when you paste the code, it's automatically syntax-highlighted.
There was a thread about codeform a while back; look it up (perhaps it's in my signature . . .) if you're interested.
It would make me happy if you "borrowed" it.
dwk
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That's correct since bit positions start at zero, so the 8th bit is at bit position 7.
Nope! endianess isn't the same as bit significance.
In big endian architecture, the most significant byte is stored in the lowest memory location.
In little endian scheme the most significant byte is stored in the highest memory location, as in
Code:int x = 0xa1b2c3d4; /* big endian: MSB a1 is in memory cell numbered 0 while LSB d4 is in 3 */ int x = 0xa1b2c3d4; /* little endian: MSB a1 is in memory cell numbered 3 while LSB d4 is in 0 */
>> There was a thread about codeform a while back; look it up (perhaps it's in my signature . . .) if you're interested.
Ok, thanks.
>> Nope! endianess isn't the same as bit significance.
I wasn't even thinking in terms of endianness, actually. I think I'm just going to plead insanity on this one.
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
Code:int setbits(unsigned , unsigned , unsigned , unsigned) { return (( x| ~( ~0 << p+1 | ~(~0 << p-n+1))) & (((y & ~(~0 << n)) << p-n+1 ) | (~(~0 << p-n+1)) | (~0 << p+1))); }
How is it that a new user is able to bump a five year old thread? Shouldn't they all be auto-locked? (I recall this happening a few months ago, as well.)
