Homework about bit sequence in C - c

Question

I have a homework. The question is:
Write a function that takes as parameter a single integer, int input, and returns an unsigned character
such that:
a- if input is negative or larger than 11,111,111 or contains a digit that is not 0 or 1; then the function will
print an error message (such as “invalid input”) and return 0,
b- otherwise; the function will assume that the base-10 value of input represents a bit sequence and return the magnitude-only bit model correspondent of this sequence.
For Example: If input is 1011, return value is 11 and If input is 1110, return value is 14
This is my work for a, and I am stuck on b. How can I get bit sequence for given integer input?
int main()
{
int input = 0;
printf("Please type an integer number less than 11,111,111.\n");
scanf("%d",&input);
if(input < 0 || input > 11111111)
{
printf("Invalid Input\n");
system("PAUSE");
return 0;
}
for (int i = 0; i < 8; i++)
{
int writtenInput = input;
int single_digit = writtenInput%10;
if(single_digit == 0 || single_digit == 1)
{
writtenInput /= 10;
}
else
{
printf("Your digit contains a number that does not 0 or 1. it is invalid input\n");
system("PAUSE");
return 0;
}
}
printf("Written integer is %d\n",input);
system("PAUSE");
return 0;
}

Answer 1

The bit that you are missing is the base conversion. To interpret a number in base B, what you need to do is multiply digit N times B^N (assuming that you start counting digits from the least significative). For example, in base 16, A108 = (10)*16^3 + 1*16^2 + 0*16^1 + 8*16^0. Where in your base is 2 (binary).
Alternatively, you can avoid the exponentiation if you rewrite the expression as:
hex A008 = ((((10*16) + 1)*16 +0)*16 + 8
Which would be simpler if your input was stated in terms of an array of possibly unknown length as it is easily convertible into a loop of only additions and multiplications.
In the particular case of binary, you can use another direct solution, for each digit that is non-zero, set the corresponding bit in an integer type large enough (in your case, unsigned char suffices), and the value of the variable at the end of the loop will be the result of the conversion.

Answer 2

First off, the existing part of your code needs work. It doesn't report non-binary digits correctly. The question is "Are there any digits that are neither 0 nor 1". To answer this question in the negative, you need to check every single digit. Your code can terminate early. I also suggest renaming the function to something that clearly tells you what it does. For example haveNonBinaryDigit. The term check doesn't tell you what you should expect on the return value.
As for the second part, read up on the binary representation. It is fairly similar to decimal, except that instead of each digit being weighted by 1, 10, 100, .., 10^x, they are weighted by 1, 2, 4, ..., 2^n. Also the digits can only have values 0 and 1.

Answer 3

Your current code has a problem: it will terminate as soon as it finds one single binary bit (at the end) and call the entire integer correct.
I would take a different approach to this problem. What you want to do is get your hands on every single digit from the integer seperately. You made a good start in doing so by using modulo. Say you had this code:
for (int i = 0; i < 8; i++)
{
char single_digit = input%10;
input /= 10;
}
This makes it easy to start working with the digits. You would be checking 8 bits because that seems to be the maximum allowed (11,111,111). Check if each digit is either 0 or 1. Then you can start pushing it in to an unsigned character using bitwise operations. Shift every digit to the left by i, then use a bitwise OR.