This is an homework problem
Is there a way tor reverse a number in Java without using any loops? The only solution I can think of is reversing it using String and then casting it back to an integer.
If you want to reverse a number withour using any loop you can use Recursion method call. Following program is doing same
public static void reverseMethod(int number) {
if (number < 10) {
System.out.println(number);
return;
} else {
System.out.print(number % 10);
reverseMethod(number / 10);
}
}
public static void main(String args[]) {
int num = 4567;
reverseMethod(num);
}
Even if you were to reverse the number by casting it into a String, you would still need a loop if you want the program to work when having ints of different sizes. If I were to make a method to reverse a number but could not do it with loops, I would probably do it with recursion (which still uses loops indirectly). The code will look something like this:
class Main {
public static void main(String[] args) {
String input = "1234"; // or scanner to take in input can be implemented
System.out.println(Integer.parseInt(reverseInt(input)));
}
public static String reverseInt(String x) {
if (x.length() == 1) {
return x;
} else {
return x.substring(x.length() - 1) + reverseInt(x.substring(0, x.length() - 1));
}
}
}
Hope this helps!
By using reverse() of StringBuilder:
int number = 1234;
String str = String.valueOf(number);
StringBuilder builder = new StringBuilder(str);
builder.reverse();
number = Integer.parseInt(builder.toString());
System.out.println(number);
will print:
4321
if you want reverse method without loop and recursion then use this code
int a=12345;
int b,c,d,e,f;
b=a%10;
c=a%100/10;
d=a%1000/100;
e=a%10000/1000;
f=a%100000/10000;
System.out.println(b+","+c+","+d+","+e+","+f);
you can go like :
public int reverse(int x) {
String o = "";
if (x < 0) {
x *= -1;
String s = Integer.toString(x);
o += "-";
o += new StringBuilder(s).reverse().toString();
}
else {
String s = Integer.toString(x);
o += new StringBuilder(s).reverse().toString();
}
try {
int out = Integer.parseInt(o);
//System.out.println(s);
return out; }
catch (NumberFormatException e) {
return 0;
}
}
This is a solution using recursive method call
public class Tester{
public static int findReverse(int num, int temp){
if(num==0){
return temp;
}else if(num<10){
return temp*10 + num; //up to this is stopping condition
}else{
temp = temp*10 + num%10;
return findReverse(num/10, temp);
}
}
public static void main(String args[]){
int num = 120021;
int reverseNum = findReverse(num, 0);
System.out.println(reverseNum);
if(num == reverseNum)
System.out.println(num +" is a palindrome!");
else
System.out.println(num +" is not a palindrome!");
}
}
This will be fast.
static int reverseNum(int num) {
StringBuilder sb = new StringBuilder(String.valueOf(num));
sb.reverse();
return Integer.parseInt(sb.toString());
}
I submitted one code in code chef but it's giving wrong answer even if it's correct
can anybody help me to identify that please.
I have tried so many inputs and calculated manually and they are correct so why they gave me wrong answer.
so,anybody who can find the TEST Case which give incorrect output by this code ?.
Here is Problem definition.
import java.util.Scanner;
import java.lang.Math;
class Codechef {
static int get(int n,int i,int digit)
{
int p;
p=(int)Math.pow(10,i-1);
n=n/p;
return n%10;
}
static boolean check_pal(int n)
{
int digit;
digit=(int) (Math.log10(n)+1);
int a=0,b=0,i,j,p;
int sum=0;
for(i=1,j=digit-1 ; i<=digit ; i++,j-- )
{
a=get(n,i,digit);
sum+=a*Math.pow(10,j);
}
if(sum==n)
return true;
else
return false;
}
static int reverse(int n)
{
int digit;
digit=(int) (Math.log10(n)+1);
int a=0,b=0,i,j,p;
int sum=0;
for(i=1,j=digit-1 ; i<=digit ; i++,j-- )
{
a=get(n,i,digit);
sum+=a*Math.pow(10,j);
}
return n+sum;
}
public static void main(String[] args) {
try{
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
if(n<10 || n>999){
System.out.println("NONE");
return;
}
boolean c;
for(int i=1 ; i<=100 ; i++)
{
c=check_pal(n);
if(c==true)
{
System.out.println(n);
return;
}
n=reverse(n);
}
System.out.println("NONE");
}
catch(Exception e)
{
System.out.println("NONE");
}
}
}
Here is one more output.
for 99 it gives 99 and which is correct as it's palindrome.
For 89 (or 98 for that matter), your code returns "NONE", although you reach the answer 8813200023188 after only 24 steps.
Another case is that for 177 and 276 you should get 8836886388 instead of NONE
I didn't debug your code, I just wrote a program that does the same, and compared the output my program gave to the one your program gave. Since you just requested a testcase, that should suffice :) My gutfeeling is that you overflow... an int is not large enough to hold the answer in all cases.
Happy bughunting.
Edit (on Request) with my code.
I didn't change your code, except that I extracted your logic into a getResult(integer) methode so that I could bypass the scanning of the input and simply return a string as result. It prints out all the differences between our versions. I used BigInteger as the type to hold my results.
public class Main {
public static void main(String[] args) {
Main m = new Main();
for (int i=10; i < 1000; i++) {
String myResult = null;
String hisResult = null;
try {
myResult = m.getResultAsString(i);
} catch (Exception e){
System.out.println("Your code threw an exception for " + i);
}
try{
hisResult = Codechef.getResult(i);
} catch (Exception e){
System.out.println("His code threw an exception for " + i);
}
if (myResult != null && hisResult != null && ! myResult.equals(hisResult)) {
System.out.println("For " + i + " you have " + myResult + " but he has " + hisResult);
}
}
}
public String getResultAsString(int inputNumber) {
BigInteger res = getResultAsBigInteger(new BigInteger(""+inputNumber));
if (res != null) {
return res.toString();
} else {
return "NONE";
}
}
public BigInteger getResultAsBigInteger(BigInteger inputNumber) {
int numberOfSteps = 0;
BigInteger currentValue = inputNumber;
while (numberOfSteps < 101 && ! isPalindrome(currentValue)) {
numberOfSteps++;
currentValue = currentValue.add(reverseDigits(currentValue));
}
return numberOfSteps < 101 ? currentValue : null;
}
public boolean isPalindrome(BigInteger number) {
return number.equals(reverseDigits(number));
}
public BigInteger reverseDigits(BigInteger input) {
String inputString = input.toString();
String output = "";
for (int i = inputString.length() - 1; i >= 0; i--)
{
output += inputString.charAt(i);
}
return new BigInteger(output);
}
}
There is an overflow error in your code.
for input 89 it's not working as #Yves V. said
Suggestion is to use BigInteger class of lang.Match it will be useful to eliminate this overflow error.
This is an extension to below question in StackOverflow as I am not understanding the accepted answer.
How do you determine the type of data contained in a string?
If someone could please give a sample code for below case:
I have a String array as below:
String[] myArray = new String[4];
myArray[0] = "one";
myArray[1] = "2012-02-25";
myArray[2] = "12345.58";
myArray[3] = "1245";
I want something as below:
for(String s:myArray){
if(s is a number){
System.out.println("A number!");
}
else if(s is a float){
System.out.println("A float!");
}
else if(s is a date){
System.out.println("A date!");
}
else if(s is a text){
System.out.println("A text!");
}
}
But I don't know what will come inside the IF conditions to determine the type of data in given String.
Thanks for reading!
The simplest way to do it is to create the aformentioned methods an simply try to parse the string like (but you have to remember that you cannot tell 100% what datatype a certain pattern is, can be multiple times at once):
public static boolean isANumber(String s) {
try {
BigDecimal d = new BigDecimal(s);
return true;
} catch (Exception e) {
return false;
}
}
public static boolean isAFloat(String s) {
//same as number, unless you want is a number to
//check if it an integer or not
try {
BigDecimal d = new BigDecimal(s);
return true;
} catch (Exception e) {
return false;
}
}
public static boolean isADate(String s) {
SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd");
try {
sdf.parse(s);
return true;
} catch (Exception e) {
return false;
}
}
public static boolean isAText(String s) {
//everything could be considered a text
// unless you want to check that characters are within some ranges
return true;
}
public static void main(String[] args) {
String[] myArray = new String[4];
myArray[0] = "one";
myArray[1] = "2012-02-25";
myArray[2] = "12345.58";
myArray[3] = "1245";
for(String str : myArray){
if(str.matches("[\\d]+")){
System.out.println(str + " is integer");
}
else if(str.matches("[\\D]+")){
System.out.println(str + " is word");
}
else if(str.matches("[\\d.]+")){
System.out.println(str + " is double");
}
else if(str.matches("\\d{4}-\\d{2}-\\d{2}")){
System.out.println(str + " is date");
}
}
}
Output:
one is word
2012-02-25 is date
12345.58 is double
1245 is integer
One way to do it is by trying to parse the string.
Here is an example for one of the types;:
boolean isFloat(String x) {
try {
Float.parseFloat(x);
} catch (Throwable e) {
return false;
}
return true;
}
Note:
As an integer can be parsed a a float you need to test if it is an integer first.
So:
if (isInteger(s)) {
// Integer
else if (isFloat(s)) {
...
for(String s:myArray){
if(NumberUtils.isNumber(s)){ //From Apache commons
double d= Double.valueOf(s);
if (d==(int)d){
System.out.println("A number!");
}else{
System.out.println("A Float!");
}
}else if(isValidDate(s)){
System.out.println("A date!");
}else{
System.out.println("A text!");
}
}
Method for valid date check
public static boolean isDateValid(String date)
{
try {
DateFormat df = new SimpleDateFormat("yyyy-MM-dd");
df.parse(date);
return true;
} catch (ParseException e) {
return false;
}
}
Write function like :
private boolean isFloat(String str) {
try {
Float.parseFloat(str);
return true;
} catch (NumberFormatException) {
return false;
}
}
And check the type. Or you can just use apache commons NumberUtils
There's no way of defining an array of strings with a declaration of a specific datatype. If you need something like that, use collection like List, e.g. a list of date objects. For strings, you have to programatically determine it by analyzing the data or if you already know what type of data an array of strings or a string contains in your program since you are the one who wrote it, then handle accordingly.
You can create a custom object/class which can take the value and type of data that value represents.
I have a project in which we often use Integer.parseInt() to convert a String to an int. When something goes wrong (for example, the String is not a number but the letter a, or whatever) this method will throw an exception. However, if I have to handle exceptions in my code everywhere, this starts to look very ugly very quickly. I would like to put this in a method, however, I have no clue how to return a clean value in order to show that the conversion went wrong.
In C++ I could have created a method that accepted a pointer to an int and let the method itself return true or false. However, as far as I know, this is not possible in Java. I could also create an object that contains a true/false variable and the converted value, but this does not seem ideal either. The same thing goes for a global value, and this might give me some trouble with multithreading.
So is there a clean way to do this?
You could return an Integer instead of an int, returning null on parse failure.
It's a shame Java doesn't provide a way of doing this without there being an exception thrown internally though - you can hide the exception (by catching it and returning null), but it could still be a performance issue if you're parsing hundreds of thousands of bits of user-provided data.
EDIT: Code for such a method:
public static Integer tryParse(String text) {
try {
return Integer.parseInt(text);
} catch (NumberFormatException e) {
return null;
}
}
Note that I'm not sure off the top of my head what this will do if text is null. You should consider that - if it represents a bug (i.e. your code may well pass an invalid value, but should never pass null) then throwing an exception is appropriate; if it doesn't represent a bug then you should probably just return null as you would for any other invalid value.
Originally this answer used the new Integer(String) constructor; it now uses Integer.parseInt and a boxing operation; in this way small values will end up being boxed to cached Integer objects, making it more efficient in those situations.
What behaviour do you expect when it's not a number?
If, for example, you often have a default value to use when the input is not a number, then a method such as this could be useful:
public static int parseWithDefault(String number, int defaultVal) {
try {
return Integer.parseInt(number);
} catch (NumberFormatException e) {
return defaultVal;
}
}
Similar methods can be written for different default behaviour when the input can't be parsed.
In some cases you should handle parsing errors as fail-fast situations, but in others cases, such as application configuration, I prefer to handle missing input with default values using Apache Commons Lang 3 NumberUtils.
int port = NumberUtils.toInt(properties.getProperty("port"), 8080);
To avoid handling exceptions use a regular expression to make sure you have all digits first:
//Checking for Regular expression that matches digits
if(value.matches("\\d+")) {
Integer.parseInt(value);
}
There is Ints.tryParse() in Guava. It doesn't throw exception on non-numeric string, however it does throw exception on null string.
After reading the answers to the question I think encapsulating or wrapping the parseInt method is not necessary, maybe even not a good idea.
You could return 'null' as Jon suggested, but that's more or less replacing a try/catch construct by a null-check. There's just a slight difference on the behaviour if you 'forget' error handling: if you don't catch the exception, there's no assignment and the left hand side variable keeps it old value. If you don't test for null, you'll probably get hit by the JVM (NPE).
yawn's suggestion looks more elegant to me, because I do not like returning null to signal some errors or exceptional states. Now you have to check referential equality with a predefined object, that indicates a problem. But, as others argue, if again you 'forget' to check and a String is unparsable, the program continous with the wrapped int inside your 'ERROR' or 'NULL' object.
Nikolay's solution is even more object orientated and will work with parseXXX methods from other wrapper classes aswell. But in the end, he just replaced the NumberFormatException by an OperationNotSupported exception - again you need a try/catch to handle unparsable inputs.
So, its my conclusion to not encapsulate the plain parseInt method. I'd only encapsulate if I could add some (application depended) error handling as well.
May be you can use something like this:
public class Test {
public interface Option<T> {
T get();
T getOrElse(T def);
boolean hasValue();
}
final static class Some<T> implements Option<T> {
private final T value;
public Some(T value) {
this.value = value;
}
#Override
public T get() {
return value;
}
#Override
public T getOrElse(T def) {
return value;
}
#Override
public boolean hasValue() {
return true;
}
}
final static class None<T> implements Option<T> {
#Override
public T get() {
throw new UnsupportedOperationException();
}
#Override
public T getOrElse(T def) {
return def;
}
#Override
public boolean hasValue() {
return false;
}
}
public static Option<Integer> parseInt(String s) {
Option<Integer> result = new None<Integer>();
try {
Integer value = Integer.parseInt(s);
result = new Some<Integer>(value);
} catch (NumberFormatException e) {
}
return result;
}
}
You could also replicate the C++ behaviour that you want very simply
public static boolean parseInt(String str, int[] byRef) {
if(byRef==null) return false;
try {
byRef[0] = Integer.parseInt(prop);
return true;
} catch (NumberFormatException ex) {
return false;
}
}
You would use the method like so:
int[] byRef = new int[1];
boolean result = parseInt("123",byRef);
After that the variable result it's true if everything went allright and byRef[0] contains the parsed value.
Personally, I would stick to catching the exception.
I know that this is quite an old question, but I was looking for a modern solution to solve that issue.
I came up with the following solution:
public static OptionalInt tryParseInt(String string) {
try {
return OptionalInt.of(Integer.parseInt(string));
} catch (NumberFormatException e) {
return OptionalInt.empty();
}
}
Usage:
#Test
public void testTryParseIntPositive() {
// given
int expected = 5;
String value = "" + expected;
// when
OptionalInt optionalInt = tryParseInt(value);
// then
Assert.assertTrue(optionalInt.isPresent());
Assert.assertEquals(expected, optionalInt.getAsInt());
}
#Test
public void testTryParseIntNegative() {
// given
int expected = 5;
String value = "x" + expected;
// when
OptionalInt optionalInt = tryParseInt(value);
// then
Assert.assertTrue(optionalInt.isEmpty());
}
My Java is a little rusty, but let me see if I can point you in the right direction:
public class Converter {
public static Integer parseInt(String str) {
Integer n = null;
try {
n = new Integer(Integer.tryParse(str));
} catch (NumberFormatException ex) {
// leave n null, the string is invalid
}
return n;
}
}
If your return value is null, you have a bad value. Otherwise, you have a valid Integer.
The answer given by Jon Skeet is fine, but I don't like giving back a null Integer object. I find this confusing to use. Since Java 8 there is a better option (in my opinion), using the OptionalInt:
public static OptionalInt tryParse(String value) {
try {
return OptionalInt.of(Integer.parseInt(value));
} catch (NumberFormatException e) {
return OptionalInt.empty();
}
}
This makes it explicit that you have to handle the case where no value is available. I would prefer if this kind of function would be added to the java library in the future, but I don't know if that will ever happen.
What about forking the parseInt method?
It's easy, just copy-paste the contents to a new utility that returns Integer or Optional<Integer> and replace throws with returns. It seems there are no exceptions in the underlying code, but better check.
By skipping the whole exception handling stuff, you can save some time on invalid inputs. And the method is there since JDK 1.0, so it is not likely you will have to do much to keep it up-to-date.
If you're using Java 8 or up, you can use a library I just released: https://github.com/robtimus/try-parse. It has support for int, long and boolean that doesn't rely on catching exceptions. Unlike Guava's Ints.tryParse it returns OptionalInt / OptionalLong / Optional, much like in https://stackoverflow.com/a/38451745/1180351 but more efficient.
Maybe someone is looking for a more generic approach, since Java 8 there is the Package java.util.function that allows to define Supplier Functions. You could have a function that takes a supplier and a default value as follows:
public static <T> T tryGetOrDefault(Supplier<T> supplier, T defaultValue) {
try {
return supplier.get();
} catch (Exception e) {
return defaultValue;
}
}
With this function, you can execute any parsing method or even other methods that could throw an Exception while ensuring that no Exception can ever be thrown:
Integer i = tryGetOrDefault(() -> Integer.parseInt(stringValue), 0);
Long l = tryGetOrDefault(() -> Long.parseLong(stringValue), 0l);
Double d = tryGetOrDefault(() -> Double.parseDouble(stringValue), 0d);
I would suggest you consider a method like
IntegerUtilities.isValidInteger(String s)
which you then implement as you see fit. If you want the result carried back - perhaps because you use Integer.parseInt() anyway - you can use the array trick.
IntegerUtilities.isValidInteger(String s, int[] result)
where you set result[0] to the integer value found in the process.
This is somewhat similar to Nikolay's solution:
private static class Box<T> {
T me;
public Box() {}
public T get() { return me; }
public void set(T fromParse) { me = fromParse; }
}
private interface Parser<T> {
public void setExclusion(String regex);
public boolean isExcluded(String s);
public T parse(String s);
}
public static <T> boolean parser(Box<T> ref, Parser<T> p, String toParse) {
if (!p.isExcluded(toParse)) {
ref.set(p.parse(toParse));
return true;
} else return false;
}
public static void main(String args[]) {
Box<Integer> a = new Box<Integer>();
Parser<Integer> intParser = new Parser<Integer>() {
String myExclusion;
public void setExclusion(String regex) {
myExclusion = regex;
}
public boolean isExcluded(String s) {
return s.matches(myExclusion);
}
public Integer parse(String s) {
return new Integer(s);
}
};
intParser.setExclusion("\\D+");
if (parser(a,intParser,"123")) System.out.println(a.get());
if (!parser(a,intParser,"abc")) System.out.println("didn't parse "+a.get());
}
The main method demos the code. Another way to implement the Parser interface would obviously be to just set "\D+" from construction, and have the methods do nothing.
They way I handle this problem is recursively. For example when reading data from the console:
Java.util.Scanner keyboard = new Java.util.Scanner(System.in);
public int GetMyInt(){
int ret;
System.out.print("Give me an Int: ");
try{
ret = Integer.parseInt(keyboard.NextLine());
}
catch(Exception e){
System.out.println("\nThere was an error try again.\n");
ret = GetMyInt();
}
return ret;
}
To avoid an exception, you can use Java's Format.parseObject method. The code below is basically a simplified version of Apache Common's IntegerValidator class.
public static boolean tryParse(String s, int[] result)
{
NumberFormat format = NumberFormat.getIntegerInstance();
ParsePosition position = new ParsePosition(0);
Object parsedValue = format.parseObject(s, position);
if (position.getErrorIndex() > -1)
{
return false;
}
if (position.getIndex() < s.length())
{
return false;
}
result[0] = ((Long) parsedValue).intValue();
return true;
}
You can either use AtomicInteger or the int[] array trick depending upon your preference.
Here is my test that uses it -
int[] i = new int[1];
Assert.assertTrue(IntUtils.tryParse("123", i));
Assert.assertEquals(123, i[0]);
I was also having the same problem. This is a method I wrote to ask the user for an input and not accept the input unless its an integer. Please note that I am a beginner so if the code is not working as expected, blame my inexperience !
private int numberValue(String value, boolean val) throws IOException {
//prints the value passed by the code implementer
System.out.println(value);
//returns 0 is val is passed as false
Object num = 0;
while (val) {
num = br.readLine();
try {
Integer numVal = Integer.parseInt((String) num);
if (numVal instanceof Integer) {
val = false;
num = numVal;
}
} catch (Exception e) {
System.out.println("Error. Please input a valid number :-");
}
}
return ((Integer) num).intValue();
}
This is an answer to question 8391979, "Does java have a int.tryparse that doesn't throw an exception for bad data? [duplicate]" which is closed and linked to this question.
Edit 2016 08 17: Added ltrimZeroes methods and called them in tryParse(). Without leading zeroes in numberString may give false results (see comments in code). There is now also public static String ltrimZeroes(String numberString) method which works for positive and negative "numbers"(END Edit)
Below you find a rudimentary Wrapper (boxing) class for int with an highly speed optimized tryParse() method (similar as in C#) which parses the string itself and is a little bit faster than Integer.parseInt(String s) from Java:
public class IntBoxSimple {
// IntBoxSimple - Rudimentary class to implement a C#-like tryParse() method for int
// A full blown IntBox class implementation can be found in my Github project
// Copyright (c) 2016, Peter Sulzer, Fürth
// Program is published under the GNU General Public License (GPL) Version 1 or newer
protected int _n; // this "boxes" the int value
// BEGIN The following statements are only executed at the
// first instantiation of an IntBox (i. e. only once) or
// already compiled into the code at compile time:
public static final int MAX_INT_LEN =
String.valueOf(Integer.MAX_VALUE).length();
public static final int MIN_INT_LEN =
String.valueOf(Integer.MIN_VALUE).length();
public static final int MAX_INT_LASTDEC =
Integer.parseInt(String.valueOf(Integer.MAX_VALUE).substring(1));
public static final int MAX_INT_FIRSTDIGIT =
Integer.parseInt(String.valueOf(Integer.MAX_VALUE).substring(0, 1));
public static final int MIN_INT_LASTDEC =
-Integer.parseInt(String.valueOf(Integer.MIN_VALUE).substring(2));
public static final int MIN_INT_FIRSTDIGIT =
Integer.parseInt(String.valueOf(Integer.MIN_VALUE).substring(1,2));
// END The following statements...
// ltrimZeroes() methods added 2016 08 16 (are required by tryParse() methods)
public static String ltrimZeroes(String s) {
if (s.charAt(0) == '-')
return ltrimZeroesNegative(s);
else
return ltrimZeroesPositive(s);
}
protected static String ltrimZeroesNegative(String s) {
int i=1;
for ( ; s.charAt(i) == '0'; i++);
return ("-"+s.substring(i));
}
protected static String ltrimZeroesPositive(String s) {
int i=0;
for ( ; s.charAt(i) == '0'; i++);
return (s.substring(i));
}
public static boolean tryParse(String s,IntBoxSimple intBox) {
if (intBox == null)
// intBoxSimple=new IntBoxSimple(); // This doesn't work, as
// intBoxSimple itself is passed by value and cannot changed
// for the caller. I. e. "out"-arguments of C# cannot be simulated in Java.
return false; // so we simply return false
s=s.trim(); // leading and trailing whitespace is allowed for String s
int len=s.length();
int rslt=0, d, dfirst=0, i, j;
char c=s.charAt(0);
if (c == '-') {
if (len > MIN_INT_LEN) { // corrected (added) 2016 08 17
s = ltrimZeroesNegative(s);
len = s.length();
}
if (len >= MIN_INT_LEN) {
c = s.charAt(1);
if (!Character.isDigit(c))
return false;
dfirst = c-'0';
if (len > MIN_INT_LEN || dfirst > MIN_INT_FIRSTDIGIT)
return false;
}
for (i = len - 1, j = 1; i >= 2; --i, j *= 10) {
c = s.charAt(i);
if (!Character.isDigit(c))
return false;
rslt -= (c-'0')*j;
}
if (len < MIN_INT_LEN) {
c = s.charAt(i);
if (!Character.isDigit(c))
return false;
rslt -= (c-'0')*j;
} else {
if (dfirst >= MIN_INT_FIRSTDIGIT && rslt < MIN_INT_LASTDEC)
return false;
rslt -= dfirst * j;
}
} else {
if (len > MAX_INT_LEN) { // corrected (added) 2016 08 16
s = ltrimZeroesPositive(s);
len=s.length();
}
if (len >= MAX_INT_LEN) {
c = s.charAt(0);
if (!Character.isDigit(c))
return false;
dfirst = c-'0';
if (len > MAX_INT_LEN || dfirst > MAX_INT_FIRSTDIGIT)
return false;
}
for (i = len - 1, j = 1; i >= 1; --i, j *= 10) {
c = s.charAt(i);
if (!Character.isDigit(c))
return false;
rslt += (c-'0')*j;
}
if (len < MAX_INT_LEN) {
c = s.charAt(i);
if (!Character.isDigit(c))
return false;
rslt += (c-'0')*j;
}
if (dfirst >= MAX_INT_FIRSTDIGIT && rslt > MAX_INT_LASTDEC)
return false;
rslt += dfirst*j;
}
intBox._n=rslt;
return true;
}
// Get the value stored in an IntBoxSimple:
public int get_n() {
return _n;
}
public int v() { // alternative shorter version, v for "value"
return _n;
}
// Make objects of IntBoxSimple (needed as constructors are not public):
public static IntBoxSimple makeIntBoxSimple() {
return new IntBoxSimple();
}
public static IntBoxSimple makeIntBoxSimple(int integerNumber) {
return new IntBoxSimple(integerNumber);
}
// constructors are not public(!=:
protected IntBoxSimple() {} {
_n=0; // default value an IntBoxSimple holds
}
protected IntBoxSimple(int integerNumber) {
_n=integerNumber;
}
}
Test/example program for class IntBoxSimple:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class IntBoxSimpleTest {
public static void main (String args[]) {
IntBoxSimple ibs = IntBoxSimple.makeIntBoxSimple();
String in = null;
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
do {
System.out.printf(
"Enter an integer number in the range %d to %d:%n",
Integer.MIN_VALUE, Integer.MAX_VALUE);
try { in = br.readLine(); } catch (IOException ex) {}
} while(! IntBoxSimple.tryParse(in, ibs));
System.out.printf("The number you have entered was: %d%n", ibs.v());
}
}
Try with regular expression and default parameters argument
public static int parseIntWithDefault(String str, int defaultInt) {
return str.matches("-?\\d+") ? Integer.parseInt(str) : defaultInt;
}
int testId = parseIntWithDefault("1001", 0);
System.out.print(testId); // 1001
int testId = parseIntWithDefault("test1001", 0);
System.out.print(testId); // 1001
int testId = parseIntWithDefault("-1001", 0);
System.out.print(testId); // -1001
int testId = parseIntWithDefault("test", 0);
System.out.print(testId); // 0
if you're using apache.commons.lang3 then by using NumberUtils:
int testId = NumberUtils.toInt("test", 0);
System.out.print(testId); // 0
I would like to throw in another proposal that works if one specifically requests integers: Simply use long and use Long.MIN_VALUE for error cases. This is similar to the approach that is used for chars in Reader where Reader.read() returns an integer in the range of a char or -1 if the reader is empty.
For Float and Double, NaN can be used in a similar way.
public static long parseInteger(String s) {
try {
return Integer.parseInt(s);
} catch (NumberFormatException e) {
return Long.MIN_VALUE;
}
}
// ...
long l = parseInteger("ABC");
if (l == Long.MIN_VALUE) {
// ... error
} else {
int i = (int) l;
}
Considering existing answers, I've copy-pasted and enhanced source code of Integer.parseInt to do the job, and my solution
does not use potentially slow try-catch (unlike Lang 3 NumberUtils),
does not use regexps which can't catch too big numbers,
avoids boxing (unlike Guava's Ints.tryParse()),
does not require any allocations (unlike int[], Box, OptionalInt),
accepts any CharSequence or a part of it instead of a whole String,
can use any radix which Integer.parseInt can, i.e. [2,36],
does not depend on any libraries.
The only downside is that there's no difference between toIntOfDefault("-1", -1) and toIntOrDefault("oops", -1).
public static int toIntOrDefault(CharSequence s, int def) {
return toIntOrDefault0(s, 0, s.length(), 10, def);
}
public static int toIntOrDefault(CharSequence s, int def, int radix) {
radixCheck(radix);
return toIntOrDefault0(s, 0, s.length(), radix, def);
}
public static int toIntOrDefault(CharSequence s, int start, int endExclusive, int def) {
boundsCheck(start, endExclusive, s.length());
return toIntOrDefault0(s, start, endExclusive, 10, def);
}
public static int toIntOrDefault(CharSequence s, int start, int endExclusive, int radix, int def) {
radixCheck(radix);
boundsCheck(start, endExclusive, s.length());
return toIntOrDefault0(s, start, endExclusive, radix, def);
}
private static int toIntOrDefault0(CharSequence s, int start, int endExclusive, int radix, int def) {
if (start == endExclusive) return def; // empty
boolean negative = false;
int limit = -Integer.MAX_VALUE;
char firstChar = s.charAt(start);
if (firstChar < '0') { // Possible leading "+" or "-"
if (firstChar == '-') {
negative = true;
limit = Integer.MIN_VALUE;
} else if (firstChar != '+') {
return def;
}
start++;
// Cannot have lone "+" or "-"
if (start == endExclusive) return def;
}
int multmin = limit / radix;
int result = 0;
while (start < endExclusive) {
// Accumulating negatively avoids surprises near MAX_VALUE
int digit = Character.digit(s.charAt(start++), radix);
if (digit < 0 || result < multmin) return def;
result *= radix;
if (result < limit + digit) return def;
result -= digit;
}
return negative ? result : -result;
}
private static void radixCheck(int radix) {
if (radix < Character.MIN_RADIX || radix > Character.MAX_RADIX)
throw new NumberFormatException(
"radix=" + radix + " ∉ [" + Character.MIN_RADIX + "," + Character.MAX_RADIX + "]");
}
private static void boundsCheck(int start, int endExclusive, int len) {
if (start < 0 || start > len || start > endExclusive)
throw new IndexOutOfBoundsException("start=" + start + " ∉ [0, min(" + len + ", " + endExclusive + ")]");
if (endExclusive > len)
throw new IndexOutOfBoundsException("endExclusive=" + endExclusive + " > s.length=" + len);
}
I've been using a helper class that contains a static Queue of parsed values, and I find it to look quite clean. This would be the helper class could look like:
public static class Parsing {
// Could optimise with specific queues for primitive types
// and also using a circular queue, instead of LinkedList
private static final Queue<Number> QUEUE = new LinkedList<Number>();
public static boolean parseInt(String value) {
// Could implement custom integer parsing here, which does not throw
try {
QUEUE.offer(Integer.parseInt(value));
return true;
}
catch (Throwable ignored) {
return false;
}
}
public static int getInt() {
return QUEUE.remove().intValue(); // user's fault if this throws :)
}
}
And then in code, you use it like this:
public Vector3 parseVector(String content) {
if (Parsing.parseInt(content)) {
return new Vector3(Parsing.getInt());
}
else {
String[] parts = content.split(",");
if (Parsing.parseInt(parts[0]) && Parsing.parseInt(parts[1]) && Parsing.parseInt(parts[2])) {
// the queue ensures these are in the same order they are parsed
return new Vector3(Parsing.getInt(), Parsing.getInt(), Parsing.getInt());
}
else {
throw new RuntimeException("Invalid Vector3");
}
}
}
The only problem with this, is that if you use multiple calls like i did above, but maybe the last one fails, then you'd have to roll back or clear the queue
Edit: You could remove the above problem and include some thread safely, by making the class non-static and, maybe for slightly cleaner code, make the class implement AutoCloseable so that you could do something like this:
public Vector3 parseVector(String content) {
try (Parsing parser = Parsing.of()) {
if (parser.parseInt(content)) {
return new Vector3(parser.getInt());
}
else {
String[] parts = content.split(",");
if (parser.parseInt(parts[0]) && parser.parseInt(parts[1]) && parser.parseInt(parts[2])) {
// the queue ensures these are in the same order they are parsed
return new Vector3(parser.getInt(), parser.getInt(), parser.getInt());
}
else {
throw new RuntimeException("Invalid Vector3");
}
}
}
}
You can use a Null-Object like so:
public class Convert {
#SuppressWarnings({"UnnecessaryBoxing"})
public static final Integer NULL = new Integer(0);
public static Integer convert(String integer) {
try {
return Integer.valueOf(integer);
} catch (NumberFormatException e) {
return NULL;
}
}
public static void main(String[] args) {
Integer a = convert("123");
System.out.println("a.equals(123) = " + a.equals(123));
System.out.println("a == NULL " + (a == NULL));
Integer b = convert("onetwothree");
System.out.println("b.equals(123) = " + b.equals(123));
System.out.println("b == NULL " + (b == NULL));
Integer c = convert("0");
System.out.println("equals(0) = " + c.equals(0));
System.out.println("c == NULL " + (c == NULL));
}
}
The result of main in this example is:
a.equals(123) = true
a == NULL false
b.equals(123) = false
b == NULL true
c.equals(0) = true
c == NULL false
This way you can always test for failed conversion but still work with the results as Integer instances. You might also want to tweak the number NULL represents (≠ 0).
You could roll your own, but it's just as easy to use commons lang's StringUtils.isNumeric() method. It uses Character.isDigit() to iterate over each character in the String.
You shouldn't use Exceptions to validate your values.
For single character there is a simple solution:
Character.isDigit()
For longer values it's better to use some utils. NumberUtils provided by Apache would work perfectly here:
NumberUtils.isNumber()
Please check https://commons.apache.org/proper/commons-lang/javadocs/api-2.6/org/apache/commons/lang/math/NumberUtils.html