Java：设置字符串中字符位置的值时，我收到意外的类型错误（必填：变量；找到：值）

Question

public static void octIn(String[] input, String octString)
    {
        String strString, biString;
        String[] str = new String[2];
        //Octal to Binary
        biString = Integer.toBinaryString(Integer.parseInt(octString, 8));
        //Binary to String
        String[] bi = splitBi(biString);
        for (int i = 0; i < 3; i++)
        {
            for (int x = 0; x < 3; x++)
            {
                if (bi[i].charAt(x) == 0)
                    str[i].charAt(x) = '-';
                else 
                {
                    switch(x)
                    {
                        case 0:
                            str[i].charAt(x) = 'r';
                            break;
                        case 1:
                            str[i].charAt(x) = 'w';
                            break;
                        case 2:
                            str[i].charAt(x) = 'x';
                            break;
                    }
                }
            }
        }
    }

I am wondering why the following error is appearing, as well as how to fix it:

ACSL1.java:36: error: unexpected type
                bi[i].charAt(x) = '1';
                            ^
  required: variable
  found:    value

note: others are appearing, but I thought it would be redundant to show those as well.

Answer 1

str[i].charAt(x) is a char value, not a variable. You can't assign to it. It's similar to writing 'a' = '1'; .

Anyway, a String is immutable, so you can't modify its characters. If you want str[i] to have a new String value, you'll have to create a new String (for example, by appending characters to a StringBuilder) and assign the new String to str[i] .

    for (int i = 0; i < 3; i++)
    {
        StringBuilder sb = new StringBuilder();
        for (int x = 0; x < 3; x++)
        {
            if (bi[i].charAt(x) == 0)
                sb.append('-');
            else 
            {
                switch(x)
                {
                    case 0:
                        sb.append('r');
                        break;
                    case 1:
                        sb.append('w');
                        break;
                    case 2:
                        sb.append('x');
                        break;
                }
            }
        }
        str[i] = sb.toString();
    }