Translation from binary into decimal numbers in C++ - c++

Question

I tried to build a function that calculates a binary number stored in a string into a decimal number stored in a long long. I'm thinking that my code should work but it doesn't.
In this example for the binary number 101110111 the decimal number is 375. But my output is completely confusing.
Here is my code:
#include <string>
#include <stdio.h>
#include <math.h>
#include <iostream>
#include <string.h>
int main() {
std::string stringNumber = "101110111";
const char *array = stringNumber.c_str();
int subtrahend = 1;
int potency = 0;
long long result = 0;
for(int i = 0; i < strlen(array); i++) {
result += pow(array[strlen(array) - subtrahend] * 2, potency);
subtrahend++;
potency++;
std::cout << result << std::endl;
}
}
Here is the output:
1
99
9703
894439
93131255
9132339223
894974720087
76039722530902
8583669948348758
What I'm doing wrong here?

Answer 1

'1' != 1 as mentioned in the comments by #churill. '1' == 49. If you are on linux type man ascii in terminal to get the ascii table.
Try this, it is the same code. I just used the stringNumber directly instead of using const char* to it. And I subtracted '0' from the current index. '0' == 48, so if you subtract it, you get the actual 1 or 0 integer value:
auto sz = stringNumber.size();
for(int i = 0; i < sz; i++) {
result += pow((stringNumber[sz - subtrahend] - '0') * 2, potency);
subtrahend++;
potency++;
std::cout << result << std::endl;
}
Moreover, use the methods provided by std::string like .size() instead of doing strlen() on every iteration. Much faster.
In a production environment, I would highly recommend using std::bitset instead of rolling your own solution:
std::string stringNumber = "1111";
std::bitset<64> bits(stringNumber);
bits.to_ulong();

Answer 2

You're forgetting to convert your digits into integers. Plus you really don't need to use C strings.
Here's a better version of the code
int main() {
std::string stringNumber = "101110111";
int subtrahend = 1;
int potency = 0;
long long result = 0;
for(int i = 0; i < stringNumber.size(); i++) {
result += pow(2*(stringNumber[stringNumber.size() - subtrahend] - '0'), potency);
subtrahend++;
potency++;
std::cout << result << std::endl;
}
}
Subtracting '0' from the string digits converts the digit into an integer.
Now for extra credit write a version that doesn't use pow (hint: potency *= 2; instead of potency++;)

Answer 3

c++ way
#include <string>
#include <math.h>
#include <iostream>
using namespace std;
int main() {
std::string stringNumber = "101110111";
long long result = 0;
uint string_length = stringNumber.length();
for(int i = 0; i <string_length; i++) {
if(stringNumber[i]=='1')
{
long pose_value = pow(2, string_length-1-i);
result += pose_value;
}
}
std::cout << result << std::endl;
}