The following function:
int numOnesInBinary(int number) {
int numOnes = 0;
while (number != 0) {
if ((number & 1) == 1) {
numOnes++;
}
number >>= 1;
}
return numOnes;
}
will only work for positive numbers, because in the case of a negative number, it always add a 1 to the leftmost bit when doing the >> operation. In Java we can use >>> instead, but how can we do it in C++? I read in a book that we can use unsigned integers in C++, but I don't see how since unsigned integers cannot represent negative numbers.
Cast number
to unsigned int and perform your counting on that:
int numOnesInBinary(int number) {
int numOnes = 0;
unsigned int unumber = static_cast<unsigned int>(number);
while (unumber != 0) {
if ((unumber & 1) == 1) {
numOnes++;
}
unumber >>= 1;
}
return numOnes;
}
Unsigned integer lets to count bits in a simple loop.
Casting from signed to unsigned makes invalid result, if we speak about the values:
char c = -127;
unsigned char u = (unsigned char)c; // 129
But if we speak only about the form, it's not changed:
1 0 0 0 0 0 0 1 == decimal signed -127
1 0 0 0 0 0 0 1 == decimal unsigned 129
So casting to unsigned is just a hack.
// How about this method, as hinted in "C Book" by K & R.
// Of course better methods are found in MIT hackmem
// http://www.inwap.com/pdp10/hbaker/hakmem/hakmem.html
//
int numOnesInBinary(int number) {
int numOnes = 0;
// Loop around and repeatedly clear the LSB by number &= (number -1).
for (; number; numOnes++, number &= (number -1));
return numOnes;
}
count represents the number of set bits in the integer n if size of integer is 32 bits then
int count =0;
int n = -25 //(given)
for(int k=0;k<32;k++){
if ((n >> k) & 1){
count++;
}
}
return count;
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.