#include using namespace std; // input a num like 11 /* before diving into bit operations we have to discuss what is a bit (binary digit) ? a bit is the most basic unit of information in computing and digital communications it represents a logical state with one of two possible values. it can be represented as :- true/false yes/no on/off +/− "0" or "1" a link explaning what is a bit https://www.techtarget.com/whatis/definition/bit-binary-digit */ /* we use Decimal number system(base 10) requiring 10 different numerals, the digits (0-9). while the binary (base 2 ) requirs (0 and 1) we start from right to left from 2^0 2^1 2^2 and so on 1 means yes 0 means no basicly 1 0 1 1 - - - - 2^3 2^2 2^1 2^0 now this number translate to 1*2^3 +0*2^2 +1*2^1 + 1*2^0 = 11 now moving on to how to convert a decimal number to binary in cpp */ int count_set_bits(int x); int count_set_bits2(int x); void to_binary(int x); int main() { //entering a number for example 11 and converting it to binary int x; cin>>x; to_binary(x); cout<> points to the right left shift << points to the left another question that rose here is the following why are the bit wise operaters here or are these different from them cout<< cin>> https://stackoverflow.com/questions/4854248/why-are-bitwise-shifts-and-used-for-cout-and-cin#:~:text=7-,%3E%3E,and%20%3C%3C%20for%20your%20classes.&text=They%20are%20not%20bitwise%20operators,extraction%20operators%20in%20this%20context. now lets expalin each operater & both cases have to be true to result in true true is 1 false is 0 1 & 1 = 1 1 & 0 = 0 0 & 1 = 0 0 & 0 = 0 | if either is true results in true 1 | 1 = 1 1 | 0 = 1 0 | 1 = 1 0 | 0 = 0 the xor operator results in true the two bits are different. 1 ^ 1 = 0 1 ^ 0 = 1 0 ^ 1 = 1 0 ^ 0 = 0 the right shift and left shift lets sart with left shift lets take the number 11 in binary form 1011 we want to shift it to the left one digit 11<<1 10110 two digits 11<<2 101100 11>>1 shifting to the right one digit is a bit different since we want to shift to the starting point so it is basicly deleting the first digit 1011 becomes 101 now lets implement right shift and left shift */ cout<<"left shift "; cout<<(x<<1)<>1)<0){ //one the bit is 1 we count it then we shift to the right until the number becomes 0 //1011& //0001 //we said that and only return true if 1 1 so only the first digit may return 1 when & to 1 since the rest of his digits are 0 //0001 = 1 if((x&1)==1){ count++; } //we right shift // 101 then 10 then 1 then 0 and we stop x>>=1; } return count; } //this is a really nice and interesting way of umproving the function above for explanation here //https://www.youtube.com/watch?v=u96TGyV73kk&list=PLQXZIFwMtjozwj9qF3h-1pF1kAcUzFya0&index=4 //got tired of writting lol int count_set_bits2(int x){ int count=0; while(x>0){ x=(x&(x-1)); count++; } return count; } void to_binary(int x){ //if the number 1/2=0 0/2=0 in this case we stop the recursive function and start returning res if(x/2!=0){ to_binary(x/2); } //it will return from the last reminder to the first cout<<(x%2); }

explanation of bit operations and bits