method that convert decimal to binary:
string toBinary(unsigned int n) {
char binary[33] = {0}; // char array with 0 value in each index
int ix = 32;
do {
binary[--ix] = '0' + n % 2; // adding remainder to specific index
n /= 2; // dividing n with 2
} while (n); // loop until n is not equal to 0
// return a string
return (binary + ix); // but unable to understand this line
}
Can anyone please explain what's happening right here return (binary + ix);
ix is an index into the char array. The function creates the binary string starting with its rightmost bit, near the end of the array, and working its way towards the beginning of the array, creating each bit one at a time.
Therefore, when the last bit is created, ix points to the index of the first, most-significant bit. It's not always going to be at the beginning of the array: specifically it won't be if there were fewer than 32 bits.
"binary + ix" adds the index to the first bit to the start of the char buffer, calculating a pointer to the first bit. Since the function returns a std::string, this is fed to std::string's constructor, which takes the pointer to a literal string and constructs a full std::string object from it, implicitly.
Since ix gets decremented only as long as the loop runs (which changes depending on the magnitude of n), this will truncate the string to not include all of the leading zeros that would be there otherwise.
Also note that you can do this: How to print (using cout) the way a number is stored in memory?
intis not necessarily 32 bit, which is assumed here. bigger numbers will have a strange effect here. The solution is to usesizeof(int)*CHAR_BITinstead of 32.nis already binary. All you're doing is changing to a printable representation.