I have the following code for a recursive function to convert binary to int:
public static int binaryToInt( String b ) {
if(b.length() < 2) {
return 0;
}
return b.charAt(0) * (int) Math.pow(2, b.length()) + binaryToInt(b.substring(1));
}
I am not getting the correct values for example: "101" I get 584. I thought my logic is correct but can someone please point out where i am going wrong?
There are quite a few problems
b.charAt(0) * ... - You are multiplying the ASCII value of the character and an int. Change it to b.charAt(0) - '0' to get the actual integer.
Math.pow(2, b.length()) - It has to Math.pow(2, b.length() - 1). Working for a few samples with a pen and paper will explain this.
if(b.length() == 0)
First: I changed your base criteria to allow all of the bits in calculation:
if(b.length() <= 0) {
return 0;
}
Second: b.charAt(0) return the ASCII value not the integer, so make it integer by using: (b.charAt(0) - '0')
Third: The power of each position will be length-1, so changed as following:
Math.pow(2, b.length()-1)
Final solution:
Please check the final solution:
public static int binaryToInt( String b ) {
if(b.length() <= 0) {
return 0;
}
return (b.charAt(0) - '0') * (int) Math.pow(2, b.length()-1) + binaryToInt(b.substring(1));
}
The code you have has the following errors:
b.charAt(0)- returns the ASCII value of the character that means result is calculated as ASCCI value (i.e if b.charAt(0) returns 1 then its corresponding ASCII value is 49).
if (b.length() < 2)- You are not checking the value of string what if the value is 1 still it returns 0.
the Math.pow(2, b.length()) to be modified as Math.pow(2, b.length()-1)
Here is the code:
public static int binaryToInt( String b ) {
if(b.length() < 2) {
if(b.charAt(0)=='1')
return 1;
return 0;
}
return Character.getNumericValue(b.charAt(0)) * (int) Math.pow(2, b.length()-1) + binaryToInt(b.substring(1));
}
