将String转换为char []-Java

Question

This is my function of converting binary to Gray Code.

public void convert(String bin)
{
    char[] b = bin.toCharArray();
    char[] g = new char[100];
    System.out.print(g[0]);
    for(int i=1;i<b.length;i++)
    {
        System.out.print(g[i] = (b[i-1] + b[i]) - 96);
        if(g[i] == '2')
        {
            System.out.print(0);
            i++;
            continue;
        }
        System.out.print(g[i] - 0);
    }
}

I have above function which works perfectly fine but I want to return the converted string from this function. So I come up with the code given below, which is not working fine and it only give me the single digit which I store in starting ie g[0] = b[0] as a result.

public String convert(String bin)
{
    char[] b = bin.toCharArray();
    char[] g = new char[100];
    g[0] = b[0];
    for(int i=1;i<b.length;i++)
    {
        g[i] = (char)((b[i-1] + b[i]) - 96);
        if(g[i] == '2')
        {
            g[i] = 0;
            i++;
            continue;
        }
        g[i] = g[i] - 0;
    }
    String gray = String.valueOf(g);
    return gray;
}

How can I do this so that it give me the result which I want.

Thanks

Answer 1

    public static String convert(String bin)
    {

        //System.out.println( "The Gray Equivalent Is: ");
        char[] b = bin.toCharArray();
        StringBuilder g = new StringBuilder(); // Use StringBuilder
        g.append(b[0]);
        //System.out.print(g[0]);
        for(int i=1;i<b.length;i++)
        {
            int val = (int)b[i-1] ^(int) b[i]; // use exclusive-or( ^ ) 
            if(val == '2')
            {
                //System.out.print(0);
                g.append(0);
                i++;
                continue;
        }

        //System.out.print(g[i] - 0);
        g.append(val);
    }

    String gray = String.valueOf(g);
    return gray;
}

Answer 2

I see what you want to achieve. But you are mistaking the integer values with the character values. Look:

An int is a integer numeric value, which can contain positive and negative numbers: -3, -2, -1, 0, 1, 2, 3...

A char is still a numeric value, but represented as letters: 'a'(97), 'b'(98), 'c'(99)...

I know you already know this, because you have been careful enough to compute the sum of two chars and normalize it substracting 2*'0' (=96). Good.

But you must notice that every number included in your code is implicitly an int . Now, realise that you are mixing ints and chars in several lines:

if(g[i] == '2')
g[i] = 0;
g[i] = g[i] - 0;

My suggestion: Follow an order:

1. Fist, normalize your data and store them into temporary int variables: int digit0=b[i - 1]-'0'; int digit1=b[i]-'0'; int digit0=b[i - 1]-'0'; int digit1=b[i]-'0';
2. Perform the calculations and store it into a temporary int variable: int result=digit0 + digit1; if (result==2) { result=0; } int result=digit0 + digit1; if (result==2) { result=0; }
3. Last, de-normalize the result and store it into the definitive output variable: g[i]=(char)(result + '0'); In the last place, you must also control the length of the arrays: If you know what is the length of the input array, you should pre-size the output array with the same length.