Converting String binary to integer

Question

How can I do this without using multiplication, division or mod?

I come with the solution but it needs multiplication.

public StringBToInt(String b) {
    int value = 0;
    for(int z = 0; z < b.length(); z++) {
        value = value * 2 + (int)b.charAt(i) - 48;
    }
}

EDIT: SORRY! Only 3 java API are allowed. length(), charAt(), and equals()

This homework problem clearly requires addition. Can you use arrays or lists? Is the input string a fixed length? — Blake Yarbrough
– Blake Yarbrough, Commented Jan 27, 2016 at 16:45
Yes, it does allow addition and subtract. No you cannot use array or lists. only Integer and String with charAt(), length(), and equals. — Joedie 123
– Joedie 123, Commented Jan 27, 2016 at 16:48
What about input length? Is it fixed input length, is there variability within a range, or is it potentially infinitely long? — Blake Yarbrough
– Blake Yarbrough, Commented Jan 27, 2016 at 16:52

Matteo Di Napoli · Accepted Answer · 2016-01-27 17:09:36Z

2

Without multiplication, use bitwise shift operator:

public StringBToInt(String b) {
    int value = 0;
    for(int z = 0; z < b.length(); z++) {
        if(b.charAt(z) == '1'){
            shift = b.length()-z-1;
            value += (1 << shift);
        }
    }
}

edited Jan 27, 2016 at 17:09

answered Jan 27, 2016 at 16:55

Matteo Di Napoli

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1 Comment

Clashsoft Over a year ago

if(b.charAt(z).equals('1'){ is missing a closing parenthesis and doesn't work - the primitive type char doesn't have an equals method!

cjstehno · Accepted Answer · 2016-01-27 16:41:30Z

1

Use the Integer.valueOf(String, int) method:

Integer.valueOf('10101',2)

answered Jan 27, 2016 at 16:41

cjstehno

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Comments

Abdelhak · Accepted Answer · 2016-01-27 17:10:35Z

1

Try to use Integer.parseInt(..) like this:

  int value = Integer.parseInt(b, 2);

Ofcourse b is a binary String.

edited Jan 27, 2016 at 17:10

answered Jan 27, 2016 at 16:41

Abdelhak

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1 Comment

Clashsoft Over a year ago

b.toCharArray()[i] is the wasteful version of b.charAt(i)

codecubed · Accepted Answer · 2016-01-27 17:20:21Z

0

You can use the method Integer.parseInt to do this.

String binary = "101010"
int value = Integer.parseInt(binary, 2);

The '2' in Integer.parseInt means to parse the String in base 2.

answered Jan 27, 2016 at 17:20

codecubed

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