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content/courses/prog-intro/homeworks/sum/SumLongOctal/_index.md
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content/courses/prog-intro/homeworks/sum/SumLongOctal/_index.md
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---
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title: SumLongOctal
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weight: 6
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---
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# Task
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- The input data is in 64 bit decimal numbers
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- Octal numbers has suffix `"o"`
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- Class should be named `SumLongOctal`
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## Solution
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We need to change our `parseNumber` method yet again. Let's take a look at it from `SumHex` modification
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```java
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static long parseNumber(StringBuilder number) {
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if (!number.isEmpty()) {
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String numberString = number.toString();
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if (
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numberString.startsWith("0x") || numberString.startsWith("0X")
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) {
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return Long.parseLong(numberString.substring(2), 16);
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} else {
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return Integer.parseInt(numberString);
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}
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} else {
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return 0;
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}
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}
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```
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Let's adapt our new logic on a concept level
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```java
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static long parseNumber(StringBuilder number) {
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if (!number.isEmpty()) {
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String numberString = number.toString();
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int numberStringLength = numberString.length();
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if (
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/* number is octal */
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) {
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return Long.parseLong(numberString.substring(0, numberStringLength - 1), 8); // 16 -> 8
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} else {
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return Integer.parseInt(numberString);
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}
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} else {
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return 0;
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}
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}
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```
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As you can see is that we need to only change method of checking if the number is octal. It's actually fairly easy. With hexadecimal numbers we checked if the number ***started with*** `"0x"` or `"0X"`. Now we need to check if the number ***ends with*** `"o"` or `"O"`. To do that we will use [`String.endsWith(String suffix)`](https://docs.oracle.com/javase/8/docs/api/java/lang/String.html#endsWith-java.lang.String-) method that checks if the string ends with `prefix`.
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And also we change `16` to `8` here
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```java
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return Long.parseLong(numberString.substring(0, numberStringLength - 1), 8);
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```
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Because we are converting from octal, not hexadecimal anymore.
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> [!NOTE]
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> Note, that we alse changed how we crop out the last character of our `numberString`. First, we introduced new variable `numberStringLength`, that represents a length of `numberString` (or number of characters in the string). Then, we take a substring using `substring(0, numberStringLength - 1)`. We start at first index (`0`) and go all the way to the second from the end of the string. We use `- 1` and not `- 2` because last index is exclusive (meaning it doesn't count).
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So the code for `parseNumber` method will be
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```java
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static long parseNumber(StringBuilder number) {
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if (!number.isEmpty()) {
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String numberString = number.toString();
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int numberStringLength = numberString.length();
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if (
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numberString.endsWith("o") || numberString.endsWith("O")
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) {
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return Long.parseLong(numberString.substring(0, numberStringLength - 1), 8); // 16 -> 8
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} else {
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return Integer.parseInt(numberString);
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}
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} else {
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return 0;
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}
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}
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```
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So the solution would be
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```java
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// SumLongOctal.java
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public class SumLongOctal {
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public static void main(String[] args) {
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int sum = 0;
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for (String argument : args) {
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StringBuilder number = new StringBuilder();
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for (char c : argument.toCharArray()) {
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if (!Character.isWhitespace(c)) {
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number.append(c);
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} else {
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sum += parseNumber(number);
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number = new StringBuilder();
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}
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}
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sum += parseNumber(number);
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}
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System.out.println(sum);
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}
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static long parseNumber(StringBuilder number) {
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if (!number.isEmpty()) {
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String numberString = number.toString();
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int numberStringLength = numberString.length();
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if (numberString.endsWith("o") || numberString.endsWith("O")) {
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return Long.parseLong(numberString.substring(0, numberStringLength - ), 8);
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} else {
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return Integer.parseInt(numberString);
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}
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} else {
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return 0;
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}
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}
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}
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```
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Let's try to test it with test like this
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```bash
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$ java SumLongOctal " +12345678901234567 "
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Exception in thread "main" java.lang.NumberFormatException: For input string: "+12345678901234567"
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at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:67)
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at java.base/java.lang.Integer.parseInt(Integer.java:662)
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at java.base/java.lang.Integer.parseInt(Integer.java:778)
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at sum.SumLongOctal.parseNumber(SumLongOctal.java:32)
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at sum.SumLongOctal.main(SumLongOctal.java:15)
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```
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As you can see we got a `NumberFormatException`. Here it's not an octal number and it doesn't fit into `int` data type, so let's change `int` to `long` which is bigger.
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```java
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// SumLongOctal.java
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public class SumLongOctal {
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public static void main(String[] args) {
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long sum = 0; // int -> long
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for (String argument : args) {
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StringBuilder number = new StringBuilder();
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for (char c : argument.toCharArray()) {
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if (!Character.isWhitespace(c)) {
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number.append(c);
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} else {
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sum += parseNumber(number);
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number = new StringBuilder();
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}
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}
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sum += parseNumber(number);
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}
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System.out.println(sum);
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}
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static long parseNumber(StringBuilder number) {
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if (!number.isEmpty()) {
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String numberString = number.toString();
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int numberStringLength = numberString.length();
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if (numberString.endsWith("o") || numberString.endsWith("O")) {
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return Long.parseLong(numberString.substring(0, numberStringLength - 1), 8);
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} else {
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return Long.parseLong(numberString); // Integer.parseInt(...) -> Long.parseLong(...)
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}
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} else {
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return 0;
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}
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}
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}
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```
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> [!NOTE]
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> Notice we changed data type of `sum` to `long` in order to store 64 bit numbers.
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Let's test some more
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```bash
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$ java SumLongOctal.java "1777777777777777777776o"
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Exception in thread "main" java.lang.NumberFormatException: For input string: "1777777777777777777776" under radix 8
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at java.base/java.lang.NumberFormatException.forInputString(NumberFormatException.java:67)
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at java.base/java.lang.Long.parseLong(Long.java:709)
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at SumLongOctal.parseNumber(SumLongOctal.java:29)
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at SumLongOctal.main(SumLongOctal.java:18)
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```
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We can see that this number doesn't fit into `long` data type. So we need to use something called [`BigInteger`](https://docs.oracle.com/javase/8/docs/api/java/math/BigInteger.html) and more specifically we will be using its [`BigInteger.longValue()`](https://docs.oracle.com/javase/8/docs/api/java/math/BigInteger.html#longValue--) method.
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First we need to ***import*** it using
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```java
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import java.math.BigInteger;
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```
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and then we will replace
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```java
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return Long.parseLong(numberString.substring(0, numberStringLength - 1), 8);
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```
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with
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```java
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return new BigInteger(number.substring(0, numberStringLength - 1), 8).longValue();
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```
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So the final code would be
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```java
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// SumLongOctal.java
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import java.math.BigInteger;
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public class SumLongOctal {
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public static void main(String[] args) {
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long sum = 0;
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for (String argument : args) {
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StringBuilder number = new StringBuilder();
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for (char c : argument.toCharArray()) {
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if (!Character.isWhitespace(c)) {
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number.append(c);
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} else {
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sum += parseNumber(number);
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number = new StringBuilder();
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}
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}
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sum += parseNumber(number);
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}
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System.out.println(sum);
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}
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static long parseNumber(StringBuilder number) {
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if (!number.isEmpty()) {
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String numberString = number.toString();
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int numberStringLength = numberString.length();
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if (numberString.endsWith("o") || numberString.endsWith("O")) {
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return new BigInteger(
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number.substring(0, numberStringLength - 1),
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8
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).longValue();
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} else {
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return Long.parseLong(numberString);
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}
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} else {
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return 0;
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}
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}
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}
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```
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