C++ simple program help!?

Question

There's a question on my homework assignment asking me to have someone put a string of 8 zero and ones as a binary number.
I'm then suppose to convert that binary integer into it's base 10 representation, so pretty much binary to decimal.

So far all I managed to get down is have user input the string and store it as an array. I know the formula would be something like
sum += (pow(2,j) * arr[i])
Any help would be much appreciated, thanks!

Answer 1

First off, do you know how to calculate a base 10 integer from a binary string? (That's where you start.) You go through the binary string, look for the zero or ones, and raise 2 to the ith power if a 1 is present, where i is the loop iteration between 0 and 7 (Remember integers are represented as 2^(n-1) - 1 (- 1 from the power to include negative numbers, - 1 from total number to include 0).



So with a bit of psuedocode:

int i = 0, num = 0;

char bin[8] = '01010101';



for(i from 0 to 7 incrementing by 1)

---if bin[i] == '1'

------num += Math.pow(2, i);

---else if bin[i] == '0'

------continue loop; //handle cases where user enters something other than 0 or 1



print num;



Try to code this on your own. It's not that difficult and there are many solutions. Post your code here if you get stuck. Good luck and have fun.

Answer 2

Jonathan's answer and algorithm are good: you don't need pow, just shift bits into place. Here's another way to do it (and I always recommend including user input validation):



#include

#include

#include

#include



using namespace std;



bool isValidBinaryStr(const string&);

int str2bin(const char *);

const size_t bitsPerInt = sizeof(int) * CHAR_BIT;



int main(int argc, char *argv[]) {

    string in;



    while (true) {

        cout << endl << "Enter up to " << bitsPerInt << " 1s and 0s: ";

        getline(cin,in);

        if ((in.size() > 0) && (in.size() <= bitsPerInt) && (isValidBinaryStr(in) == true)) {

            cout << str2bin(in.c_str()) << endl;

        }

    }

    return 0;

}



//

// Convert string of bits to an integer

//

int str2bin(const char *s) {

    if (*s == '\0') return 0;

    return ((*s - '0') << ((strchr(s, '\0') - s) - 1)) | str2bin(s + 1);

}





//

// Return true if all characters of s are either '1' or '0'.

//

bool isValidBinaryStr(const string &s) {

    string::const_iterator i;

    bool valid = true;



    for (i = s.begin(); (i != s.end()) && (valid == true); i++) {

        valid = ((*i == '1') || (*i == '0'));

    }

    return valid;

}





#if 0



Sample run:



Enter up to 32 1s and 0s: 10

2



Enter up to 32 1s and 0s: 1111

15



Enter up to 32 1s and 0s: 101001110010

2674



Enter up to 32 1s and 0s: 11010110

214



Enter up to 32 1s and 0s: 11111111111111111111111111111111

-1



#endif

Answer 3

Try something like the following. No need for the pow() function:



#include

#include

using namespace std;

int main( ) {

    string q;

    while ( getline( cin, q ) && !q.empty( ) ) {

        unsigned int result= 0;

        const string::size_type len= q.size();

        for ( string::size_type i= 0; i < len; ++i ) {

            result <<= 1;

            if ( q[i] == '1' ) ++result;

        }

        cout << result << endl;

    }

    return 0;

}



The above uses one of several pre-C++11 style approaches to looping over strings. And it assumes every character is a 0 unless it is exactly a 1. You could add error checking if you want that.