Funny way:
while (x != 0)
{
x = x& (x-1)
count++;
}
1
Anoniem
12 aug 2011
while (number > 0)
{
count += (number & 0x01);
number >>= 1;
}
Anoniem
1 okt 2011
Shifting may not work if the number has a sign bit... will keep pushing in ones.
num&num-1 should work
Anoniem
30 mei 2009
better solution:
byte input;
int count = 0;
while (input)
{
++count;
input >> 1;
}
Anoniem
6 mei 2009
byte input;
int bits = 0;
for(int i = 0; i > i) & 0x01;
}
Anoniem
9 jul 2009
The *ONLY* portable way to answer this question such that it is correct for *ALL* ANSI C compilers on *ALL* hardware platforms is as follows:
#include
#include
int main()
{
printf("Number of bits per byte on this machine with this ANSI C compiler = %d\n", CHAR_BIT);
return 0;
}
Anoniem
17 okt 2009
i think the question was written wrong by the person
i checked on google and the question is
Count the "on" bits in a byte?
Anoniem
17 okt 2009
one way to solve this
char c;
int bits=0;
for (int i=0;i>1;
}
return bits
Anoniem
30 mei 2009
better solution:
byte input;
int count = 0;
while (input)
{
++count;
cout >> 1;
}
Anoniem
5 mei 2009
Its 8 of course, but I WISH I had that question... lol