Java converting a string binary to integer without using maths pow

Question

Main:

public class Main{                                      
  public static void main(String[] args){                                       
    System.out.println(Convert.BtoI("101101010101"));                                   
    System.out.println(Convert.BtoI("1011110"));                                                                            
  }                                     
}

Sub:

public class Convert{                                       
  public static int BtoI(String value){                                     
    int no = 0;                                                                         

    for(int i=value.length()-1;i>=0;i--){                                       
      if(value.charAt(i)=='1')                                      
        no += (???) ;                                       
      ++;                                       
    }                                       
    return no;                                      
  }                                     
}

How can I convert a string binary to integer without using maths.pow, just using + - * and /, should I implement another for loop for it is int j = 1;i <= example; i*=2){ ?. I am quite confused and want to learn without the usage of maths.pow or any similar codes.

https://stackoverflow.com/a/10179141/4267015 duplicate her is your answer — Naor Tedgi, Dec 17 '18 at 23:12
In Java `x << y` is effectively equivalent to `Math.pow(2, y)`. — Poohl, Dec 17 '18 at 23:18
Duplicate is correct, however I don't like the accepted answer because sometimes the goal is to re-invent the wheel and not use a built-in routine like `Integer.parse()`. The secret is to at each step multiply by the radix. For base 10 you multiply by 10, and for base 2 you multiply by two. Note in the duplicate link at least two answer shift by 1 at each iteration, which is the same as multiplying by 2. — markspace, Dec 17 '18 at 23:18
@Jean-ClaudeArbaut in case shifting would not require to parse string into binary form first, I would not ask you that question. — Antoniossss, Dec 18 '18 at 09:25

taygetos · Answer 1 · 2018-12-17T23:48:04.727

From the beginning of the string until to the end you can just multiply each character with its power and sum up the total which gives you the base ten value:

public static int binaryToInt(String binaryNum) {
    int pow = 1;
    int total = 0;

    for (int i = binaryNum.length(); i > 0; i--) {
        if (binaryNum.charAt(i-1) == '1')  {
            total += pow;
        }
        pow *= 2;
    }

    return total;
}

But i think this is way more elegant:

String binaryNum = "1001";
int decimalValue = Integer.parseInt(binaryNum, 2);

score 1 · Answer 2 · answered Dec 17 '18 at 23:12

1

How about Integer.parseInt(yourString,2); ?? 2 means you are parsing base2 number.

answered Dec 17 '18 at 23:12

Antoniossss

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This isn't wrong, but I'm pretty sure the OP's goal is to understand how `Integer.parse()` works, so they want to re-implement it themselves. – markspace Dec 17 '18 at 23:21
Tldr and got focus on the topic as it was clear enough. – Antoniossss Dec 17 '18 at 23:32
He / she explicitly says *"just using + - * and /"* and loops. – Stephen C Dec 17 '18 at 23:47
*"Shift is likely better than multiplication"* - I doubt that, and if it was then the JIT compiler would / should make the same optimization! – Stephen C Dec 18 '18 at 00:25

Stephen C · Answer 3 · 2018-12-19T01:06:12.523

Starting from your code + some vital style changes:

public class Convert {                                       
    public static int binaryToInt(String value) {                                     
        int no = 0;                                                                         

        for (int i = 0; i < value.length() - 1; i++) {
            no = no * 2;             // or no *= 2;                                  
            if (value.charAt(i) == '1') {                                     
                no = no + 1;         // or no++                                    
            }                                       
        }                                       
        return no;                                      
    }                                     
}

The meaning of the code I added should be self-evident. If not, I would encourage you to work out what it does as an exercise. (If you are not sure, use a pencil and paper to "hand execute" it ...)

The style changes (roughly in order of importance) are:

Don't start a method name with an uppercase character.
Use camel-case.
Avoid cute / obscure / unnecessary abbreviations. The next guy reading your code should not have to use a magic decoder ring to understand your method names.
Put spaces around operators.
4 character indentation.
Put spaces between ) and { and before / after keywords.
Use curly brackets around the "then" and "else" parts of an if statement, even if they are not strictly necessary. (This is to avoid a problem where indentation mistakes cause you to misread the code.)

Assuming the string starts with the MSB, there is a bug (the OP didn't tell, but it's the most common I think). Also, the code is a bit faster (I get 20%) when replacing the `if` with `no |= 1 & value.charAt(i);`. But then of course it's not using only `+-*/`. — , Dec 18 '18 at 16:53
1) Corrected. Thanks. 2) *"Then of course it's not using only +-*/"* - Correct! Furthermore, I do not think it is helpful for a beginner to delve into micro-optimizations. — Stephen C, Dec 19 '18 at 01:05

Java converting a string binary to integer without using maths pow

3 Answers3