在 C++ 中将二进制数转换为十进制数

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

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 2

'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 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;
}