Input an integer containing only 0’s and 1’s( ie. “binary” integer) and print its decimal equivalent. {hint: use the remainder and division operators to pick off the binary numbers digits one at a time from right to left. Just as in the decimal number system, in which the right most digit has a positional value of 1, and the next digit has a positional value of 10, then 100, then 1000, and so on, in the binary number system the rightmost digit has a positional value of 1, the next digit left has a positional value of 2, then 4, then 8, and so on. Thus the decimal number 234 can be interpreted as 4*1+3*10+2*100. The decimal equivalent of binary 1101 is 1*1+0*2+1*4+1*8 or 1+0+4+8 or 13.}

int base_numeric() { works with any base 2-10

int number = 1101; binary number

int base = 2; from base 2

int i = 0;

int r = 0;

while (number > 0) {

int digit = number % 10; extract 1 digit (no matter base)

int mult = (int)pow(base, i++); calculate base ^ i (0, 1, 2,

r += digit * mul

number/=10;

}

return r

}

0