Java八进制到二进制的转换（无预定义方法）

Question

I'm trying to create a calculator that converts an octal value to binary.

temp = txt.getText();
temptemp = "";
if (prev == 3) {
    for (int i = 0; i < temp.length() - 1; i++) {
        if (temp.charAt(i) == '1') {
            temptemp = temptemp + "000";
        } else if (temp.charAt(i) == '2') {
            temptemp = temptemp + "001";
        } else if (temp.charAt(i) == '3') {
            temptemp = temptemp + "010";
        } else if (temp.charAt(i) == '4') {
            temptemp = temptemp + "011";
        } else if (temp.charAt(i) == '5') {
            temptemp = temptemp + "100";
        } else if (temp.charAt(i) == '6') {
            temptemp = temptemp + "101";
        } else if (temp.charAt(i) == '7') {
            temptemp = temptemp + "111";
        }
    }
    temp = temptemp;
    txt.setText(temp);

What's wrong with my looping statement? Please help. Thank you :)

EDIT:

I know now the problem. Thanks for all the comments and answers guys. I was off by 1 increment. I should've start at == 0 . Sorry and thank you :)

Answer 1

You've done several mistakes in your conversion table. (no 0 and no 7, wrong conversions, ommitting the last character by stopping the loop too eraly).

After you corrected those you should consider using a StringBuilder instead of just concatenating strings! And a switch statement might be a bit more readable than an if-else chain

StringBuilder strBuilder = new StringBuilder();
for (int i = 0; i < temp.length(); i++) {
    switch(temp.charAt(i)) {
    case '0':
        strBuilder.append("000");
        break;
    case '1':
        strBuilder.append("001");
        break;
    case '2':
        strBuilder.append("010");
        break;
    case '3':
        strBuilder.append("011");
        break;
    case '4':
        strBuilder.append("100");
        break;
    case '5':
        strBuilder.append("101");
        break;
    case '6':
        strBuilder.append("110");
        break;
    case '7':
        strBuilder.append("111");
        break;
    }
}
String temptemp = strBuilder.toString();

There's another possibility which is much shorter (and completely unreadable ;) ):

String[] convArray = { "000", "001", "010", "011", "100", "101", "110", "111" };
StringBuilder strBuild = new StringBuilder();
for (int i = 0; i < temp.length(); i++)
    strBuild.append(convArray[temp.charAt(i)-48]);
String temptemp = strBuild.toString();

Please not that this will only work if the String really only contains the numbers 0-7

Answer 2

Its easy to do it without considering all 8 branches

for (int i = 0; i < temp.length(); i++) {
  int d = Character.getNumericValue(temp.charAt(i));                           
  for (int k=2; k >= 0; k--)
    temptemp = temptemp + Character.forDigit((i >> k) & 1, 2);
}

Answer 3

You can skip having this big long list of explicit cases by doing some bitwise operations:

StringBuilder sb = new StringBuilder(3 * temp.length());
for (int i = 0; i < temp.length(); i++) {
  // + check that the character is actually an octal digit.
  int digit = Character.digit(temp.charAt(i), 10);
  for (int b = 2; b >= 0; --b) {
    sb.append(Character.forDigit((digit >> b) & 0x1, 10));
  }
}

Answer 4

private static String octalToBinary(int octal) {
        if (octal < 1 || octal > 7) {
            throw new IllegalArgumentException("Not an ocatl number");
        }

        String binary = new String();
        int myOctal = octal / 2;
        for (; myOctal >= 1;) {
            binary = octal % 2 + binary;
            octal = myOctal;
            myOctal = myOctal / 2;
        }
        binary = octal + binary;

        // if you want to make it of length 3
        if (binary.length() == 1) {
            binary = "00" + binary;
        }

        if (binary.length() == 2) {
            binary = "0" + binary;
        }
        return binary;
    }

Answer 5

For your case it's the right converts. You can check the converts here

for (int i = 0; i < temp.length() - 1; i++) {
   if (temp.charAt(i) == '0') {
         temptemp = temptemp + "000";
   } else if (temp.charAt(i) == '1') {
        temptemp = temptemp + "001";
   } else if (temp.charAt(i) == '2') {
        temptemp = temptemp + "010";
   } else if (temp.charAt(i) == '3') {
        temptemp = temptemp + "011";
    } else if (temp.charAt(i) == '4') {
        temptemp = temptemp + "100";
    } else if (temp.charAt(i) == '5') {
        temptemp = temptemp + "101";
    } else if (temp.charAt(i) == '6') {
         temptemp = temptemp + "110";
    } else if (temp.charAt(i) == '7') {
        temptemp = temptemp + "111";
   }
}

Instead, you can loop over the number, each time div the number by 2, inside the loop check his result by % 2 (or by using >> ) if it's 0 , you add to result the digit 0 , else 1