I am wondering if the first way to throw an IOException is the better way to do this,
Like this , but I can't trigger an IOException to test it.
public static String inputStr (String prompt) throws IOException
{
String inputStr = "";
System.out.print(prompt);
inputStr = stdin.readLine();
if (inputStr.length() == 0)
{
System.out.println("Error! Enter valid field!");
inputStr(prompt);
}
return inputStr;
}
Or within the method.
public static String inputStr (String prompt)
{
String inputStr = "";
System.out.print(prompt);
try
{
inputStr = stdin.readLine();
if (inputStr.length() == 0)
{
System.out.println("Error! Enter valid field!");
inputStr(prompt);
}
} catch (IOException e)
{
System.out.println("Error! Enter valid field!");
inputStr(prompt);
}
return inputStr;
}
Is it necessary to include a customised message in case one is thrown? Re this question Java - What throws an IOException I am unable to test for what will happen.
Not sure if this should be codereview, as I am unsure if it actually will work.
If stdin.readLine() throws an IOException, trying to re-read again from stdin won't succeed: you'll get another IOException, making your second snippet go into an infinite recursive loop, ending with a StackOverflowError.
You can trigger an IOException to test. Just added this in your if statement:
throw new IOException("Testing Failure");
As far as how you deal with the exception it really depends on if you want the current method to deal with it or if you just want to pass it to whatever called the method.
Related
I want to make a test that reads from a file some data and passes that data to a function. That function calls other methods and some of them throw some exceptions. I'm interested in how can I check whether or not calling the method with the parameters from the file triggered an IOException somewhere along. I know that the code snippet provided will stop the execution because I've used assert. How should I write if I want to check if an IOException was thrown and if it was, to get the error message, without stopping the execution of the test? Thanks!
void test() throws IOException {
Service service = helperFunction();
File articles = new File("file.txt");
Scanner scanner = new Scanner(articles);
while(scanner.hasNextLine()) {
String line = scanner.nextLine();
line = line.replaceAll("[^\\d]", " ");
line = line.trim();
line = line.replaceAll(" +", " ");
String[] numberOnTheLine = line.split(" ");
List<Integer> list = Arrays.stream(numberOnTheLine).map(Integer::valueOf).collect(Collectors.toList());
Article article = new Article(Long.valueOf(list.get(0)),
new HashSet<>(List.of(new Version(list.get(1)))));
List<List<Article>> listOfArticles = Collections.singletonList(List.of(article));
Assertions.assertThrows(IOException.class,
() -> service.etlArticles(listOfArticles.stream().flatMap(List::stream).collect(Collectors.toList())));
}
}
Simple; a try/catch statement will take care of it.
Replace this:
service.etlArticles(listOfArticles.stream().flatMap(List::stream).collect(Collectors.toList())));
With:
try {
service.etlArticles(listOfArticles.stream().flatMap(List::stream).collect(Collectors.toList())));
} catch (IOException e) {
// Code jumps to here if an IOException occurs during the execution of anything in the try block
}
You are free to e.g. do some logging and then just Assert.fail, if you want.
assertThrows is quite simple, all it does is this:
try {
runThatCode();
} catch (Throwable e) {
if (e instanceof TypeThatShouldBeThrown) {
// Great, that means the code is working as designed, so, just...
return;
}
// If we get here, an exception was thrown, but it wasn't the right type.
// Let's just throw it, the test framework will register it as a fail.
throw e;
}
// If we get here, the exception was NOT thrown, and that's bad, so..
Assert.fail("Expected exception " + expected + " but didn't see it.");
}
Now that you know how it works, you can write it yourself and thus add or change or log or whatever you want to do during this process at the right place. However given you know it's IOException, instead of an instanceof check you can just catch (IOException e), simpler.
I have a serial device connected to my PI that I read data from it...
everything is good , but sometime the cable move or the serial device is unplug.
then the
line = r.readLine();
get stuck
I have try to overcome this problem by:
BufferedReader r = new BufferedReader (newnputStreamReader(p.getInputStream()));
try
{
line = r.readLine();
if (line.length() == 0)
{
logfile.append("problem");
logfile.close();
System.out.println("Problem");
TimeUnit.SECONDS.sleep(5);
break;
}
}
catch (IOException e)
{
logfile.append(line);
logfile.close();
}
but it doens't do nothing (because I geuss he is still waiting for data)
not even throw an exception
how can I make him say I have a problem ?
maybe to use a timer or something like this ?
if no data for 5 seconds ?
Thanks ,
Your assumption in this case is right. The readLine() method of the BufferedReader has an internal while-loop which will retrieve all bytes from the underlying input stream and will only break if the arriving character is either \n or \r.
Think of it like:
while(lastChar != '\n' || lastChar != '\r')
{
//read from stream
}
However once entered the method will not return. The only exceptions are the occurence of those two special characters or if the InputStream is closed (in which case null ist returned).
The trick is to no enter until there is something to read from the InputStream:
public static void main( String[] args ) throws IOException
{
boolean awaitInput = true;
while(awaitInput)
{
if(System.in.available() != 0)
{
awaitInput = false;
// read logic
}
}
}
This is only one of many possible solutions and I used System.in as an example since it is a InputStream like any other as well. But there is also a method called BufferedReader#ready which returns true if there is something to read:
public static void main( String[] args ) throws IOException
{
BufferedReader br = new BufferedReader( new InputStreamReader(System.in) );
boolean awaitInput = true;
while(awaitInput)
{
if(br.ready())
{
awaitInput = false;
String line = br.readLine();
// read logic
}
}
}
At last if you want a timeout you can easily do it yourself like this:
public static void main( String[] args ) throws IOException
{
BufferedReader br = new BufferedReader( new InputStreamReader(System.in) );
boolean awaitInput = true;
long timeout = System.currentTimeMillis() + 5_000;
// ^^^^^ 5_000ms = 5 sec
while(awaitInput && System.currentTimeMillis() < timeout)
{
if(br.ready())
{
awaitInput = false;
String line = br.readLine();
// read logic
}
}
}
You can use CompletableFuture to read concurrently and be able to use a timeout.
// wrap the readLine into a concurrent call
CompletableFuture<String> lineFuture = CompletableFuture.supplyAsync(() -> r.readLine());
try {
// do the call, but with a timeout
String readLine = lineFuture.get(5, TimeUnit.SECONDS);
// do stuff with the line you read
} catch (TimeoutException e) {
// plug pulled?
}
I am checking if number the user entered is Zeckendorf and I want to display an exception if it is not, how do i do that? Also how do I convert the Zeckondorf to its decimal equivalent?
import java.util.Scanner;
public class IZeckendorfNumberConvertor {
static String number;
int zeckonderEquivalent;
static Scanner scanner = new Scanner(System.in);
public static void main(String[] args) {
convertToZeckonderNumber();
isTrueZeckonderNumber();
}
private static boolean isTrueZeckonderNumber() {
System.out.println("Enter a Zeckonder Number:");
number = scanner.nextLine();
if (number.equals("10100"))
{
return true; }
else if (number.equals("010011") || number.equals("010100"))
{
return false; }
return false;
}
private static void convertToZeckonderNumber() {
}}
I advise you not to display an exception (i.e. trace and such) as it is very user Unfriendly.
You can use the throw syntax to throw a proper exception :
throw new Exception("Given number is not a Zeckendorf number");
but be sure to catch it and display a nice and clean message :
try {
// Your input code goes here
} catch (Exception e) {
System.out.println(e.getMessage());
}
Another easier option will be to just check the return value of the method and print the results accordingly.
I will recommend to use the latter solution as exceptions are used when something bad and unexpected happens in your program and you want to handle it gracefully. In your case the behavior is expected (i.e. user giving a wrong number) so checking the return value will be much clean and easier.
Use try catch block for catch an exception
try {
} catch (Exception e) {
e.printStackTrace();
}
Also use throw for throw a new exception
Assuming to really do want to display the exception, and not a more user friendly message, the first step is probably to get the exception as a string. Then you can do what you like with that string (echo to console, place in a javax.swing.JTextArea, email it, etc).
If you just want the message from the exception, then getMessage() will do:
try { ... }
catch(FooException e) {
String msg = e.getMessage();
}
However, if you want the whole exception, stack trace and all, you'll want something like this:
public static String stackTrace(Exception e) {
StringWriter w = new StringWriter();
e.printStackTrace(new PrintWriter(w));
return w.toString();
}
// ...
try { ... }
catch(FooException e) {
String st = stackTrace(e);
}
If you just want to echo the full stack trace to the console, there is the printStackTrace(), no-arg method:
try { ... }
catch(FooException e) {
e.printStackTrace();
}
If you want to take more control of the presentation of the stack trace you can get the details with:
try { ... }
catch(FooException e) {
StackTraceElement[] stes = e.getStackTrace();
// Do something pretty with 'stes' here...
}
You can just print a error message to the user saying that the input is wrong using a simple if.
if(yourCondition){
// Happy scenario
// Go shead
}else{
// Error Scenario
System.out.println("Error. Invalid Input.");
// If you persist to throw an exception, then you can do something like this
// throw new Exception("Exception Explanation"); // I've commented this, but you can uncomment it if needed
// But the advice is to display an error message, than throw a exception.
}
And regarding the conversion, you can convert binary to decimal like this
int i = Integer.parseInt(binaryString, 2); // 2 indicates the radix here, since you want to convert from binary.
With this code snippet you can convert the String into an integer :
int numberAsInt;
try {
numberAsInt = Integer.parseInt(number);
} catch (NumberFormatException e) {
//Will throw an Exception
}
If you want to create your own Exception class, you can do it like shown here or just throw a RuntimeException with
throw new RuntimeException("Your Message");
My opinion, you can try some thing like following
public static void main(String[] args) {
if(!isTrueZeckonderNumber()){
// your message should goes here
System.out.println("Your message");
}
}
If you really want to throws an exception do following
private static boolean isTrueZeckonderNumber() throws Exception{
System.out.println("Enter a Zeckonder Number:");
number = scanner.nextLine();
if (number.equals("10100")) {
return true;
} else{
throw new Exception("your message");
}
}
What do you mean you want to display an exception?
I would suggest just giving the user feedback instead, as exceptions are used more commonly for EXCEPTIONAL actions that are not supposed to happen.
However if you do want to, you can print a message explaining what happened.
try {
} catch (Exception e) {
System.out.println(e.getMessage());
}
Original Question
For the following small code I'm getting the error...
import java.io.*;
class test
{
public static void main(String args[]) throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int i;
System.out.println("Enter no of processes ");
int no_of_process=Integer.parseInt(br.readLine());
int process[]=new int[no_of_process];
System.out.println("Enter the values");
for(i=0;i<no_of_process;i++)
process[i]=Integer.parseInt(br.readLine());
for(i=0;i<no_of_process;i++)
System.out.println(process[i]);
}
}
Input:
Enter no of processes
5
Enter the values
1
2
Exception in thread "main" java.lang.NumberFormatException: For input string: ""
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Integer.parseInt(Integer.java:470)
at java.lang.Integer.parseInt(Integer.java:499)
at test.main(test.java:17)
Process completed.
I think I have written the code properly and proper integer input is also given. How do I get rid of the above error without using any explicit exception handling statements?
Further Question:
Thanks guys for your answers...it is working. But there is one new question in my mind.
I tried the following modification in the code and executed it. To my surprise, input was accepted properly without any run time error.
for(i=0;i<no_of_process;i++)
{
System.out.println(Write anything or even keep it blank);
process[i]=Integer.parseInt(br.readLine());
}
By adding just one Print statement before (or even after) the input statement in the for loop, my program worked correctly and no exception was thrown while giving input.
Can you guys explain the reason behind this?
Again if I remove the Print statement from there, the same error gets repeated. I am really confused about the reason behind this. Please help.
Without any error handling statements? check to see if br.readLine() is returning "" before you attempt to parse it, like so:
String line = br.readLine();
if(!String.isEmpty(line))
{
//Do stuff
}
else
{
//Do other stuff
}
How to get rid of above error without using any explicit exception handling statements?
for (i = 0; i < no_of_process; i++) {
String input;
do {
input = br.readLine();
if (isInteger(input)) {
process[i]=Integer.parseInt(input);
} else {
//error handling here
System.err.println("you entered an invalid input");
}
} while(isInteger(input));
}
And isIntegerlooks like this:
public static boolean isInteger(String s)
{
try {
Integer.parseInt(s);
}
catch (Exception e) {
return false;
}
return true;
}
I think I have written properly and proper integer input is also given.
I think not ;) i think you pressed the return with out typing anything
Try this:
import java.io.*;
public class test
{
public static void main(String args[]) throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int i;
System.out.println("Enter no of processes ");
try{
int no_of_process=Integer.parseInt(br.readLine());
int process[]=new int[no_of_process];
System.out.println("Enter the values");
for(i=0;i<no_of_process;i++)
process[i]=Integer.parseInt(br.readLine());
for(i=0;i<no_of_process;i++)
System.out.println(process[i]);
}
catch(NumberFormatException n)
{
System.out.println(n.getMessage());
}
}
}
I think it's easier to just show the code and the output I'm getting than trying to explain it :)
This is from my main method:
//prompt user for filename
System.out.println("Please enter the text file name. (Example: file.txt):");
String filename = ""; //will be used to hold filename
//loop until user enters valid file name
valid = false;
while(!valid)
{
filename = in.next();
try
{
reader.checkIfValid(filename);
valid = true; //file exists and contains text
}
catch (Exception e)
{
System.out.println(e + "\nPlease try again.");
}
}
And this is the reader.checkIfValid method:
public void checkIfValid(String filename) throws InvalidFileException, FileNotFoundException
{
try
{
in = new Scanner(new File(filename));
if (!in.hasNextLine()) // can't read first line
throw new InvalidFileException("File contains no readable text.");
}
finally
{
in.close();
}
}
This is the output I get when a nonexistent file is entered:
Please enter the text file name. (Example: file.txt):
doesNotExist.txt
java.lang.NullPointerException
Please try again.
Why is the System.out.println(e) getting a NullPointerException? When I enter an empty file or a file with text, it works just fine. The empty file prints the InvalidFileException (a custom exception) message.
When I put a try-catch statement around the "in = new Scanner(new File(filename));", and have the catch block display the exception, I do get the FileNotFoundException printed out, followed by the NullPointerException (I'm not entirely sure why the catch block in the main method would be activated if the exception was already caught in the checkIfValid method...).
I've spent a while on this and I'm completely clueless as to what's wrong. Any help would be appreciated. Thanks!
edited: I think the null pointer comes from the call to reader, it is poor practise to catch all exceptions as you no longer know where they came from!
Maybe the checkIfValid method should just check if the filename is valid?
public boolean checkIfValid(String filename) {
try {
File file = new File(filename);
return file.exists();
} catch (FileNotFoundException) {
System.out.println("Invalid filename ["+filename+"] "+e);
}
}
Then the code calling it could look like;
filename = in.next();
valid = reader.checkIfValid(filename);
if (valid)
List<String> fileContents = readFromFile(filename);
Then contain all the file reading logic in it's own method like this;
public List<String> readFromFile(filename) {
List<String> fileContents = new ArrayList<String>();
try {
in = new Scanner(new File(filename));
while (in.hasNextLine()) {
fileContents.add(in.nextLine);
}
} catch (IOException e){
//do something with the exception
} finally {
in.close();
}
return fileContents;
}
My mistake was something only I could've seen. I was catching all the exceptions so I wasn't able to see where it was coming from. Thank you for helping!