nextInt() method of Scanner class does not ask me for input again in while loop?
In my main method is this code:
int hours = getHours();
Here is the get hours() code:
public static int getHours()
int hours = 0;
boolean hoursNotOk = true;
do
try
hours = console.nextInt();
hoursNotOk = false;
catch(Exception e)
System.out.print(e);
finally
if(hoursNotOk)
System.out.print(", please re-enter the hours again:");
else
System.out.print("**hours input accepted**");
while(hoursNotOk);
return hours;
For the first time console.nextInt() ask me for input, so lets say I put in a "two" in the console, it will throw an exception and loop through the try block again but this time it did not ask me for input and keeps printing out from the catch and finally block, why is this happening?
java exception-handling try-catch do-while
add a comment |
In my main method is this code:
int hours = getHours();
Here is the get hours() code:
public static int getHours()
int hours = 0;
boolean hoursNotOk = true;
do
try
hours = console.nextInt();
hoursNotOk = false;
catch(Exception e)
System.out.print(e);
finally
if(hoursNotOk)
System.out.print(", please re-enter the hours again:");
else
System.out.print("**hours input accepted**");
while(hoursNotOk);
return hours;
For the first time console.nextInt() ask me for input, so lets say I put in a "two" in the console, it will throw an exception and loop through the try block again but this time it did not ask me for input and keeps printing out from the catch and finally block, why is this happening?
java exception-handling try-catch do-while
Well, that's because you are not asking for input anywhere but thefinallyblock.
– Guy
Nov 11 at 12:19
add a comment |
In my main method is this code:
int hours = getHours();
Here is the get hours() code:
public static int getHours()
int hours = 0;
boolean hoursNotOk = true;
do
try
hours = console.nextInt();
hoursNotOk = false;
catch(Exception e)
System.out.print(e);
finally
if(hoursNotOk)
System.out.print(", please re-enter the hours again:");
else
System.out.print("**hours input accepted**");
while(hoursNotOk);
return hours;
For the first time console.nextInt() ask me for input, so lets say I put in a "two" in the console, it will throw an exception and loop through the try block again but this time it did not ask me for input and keeps printing out from the catch and finally block, why is this happening?
java exception-handling try-catch do-while
In my main method is this code:
int hours = getHours();
Here is the get hours() code:
public static int getHours()
int hours = 0;
boolean hoursNotOk = true;
do
try
hours = console.nextInt();
hoursNotOk = false;
catch(Exception e)
System.out.print(e);
finally
if(hoursNotOk)
System.out.print(", please re-enter the hours again:");
else
System.out.print("**hours input accepted**");
while(hoursNotOk);
return hours;
For the first time console.nextInt() ask me for input, so lets say I put in a "two" in the console, it will throw an exception and loop through the try block again but this time it did not ask me for input and keeps printing out from the catch and finally block, why is this happening?
java exception-handling try-catch do-while
java exception-handling try-catch do-while
asked Nov 11 at 12:05
Nigel Ng
436
436
Well, that's because you are not asking for input anywhere but thefinallyblock.
– Guy
Nov 11 at 12:19
add a comment |
Well, that's because you are not asking for input anywhere but thefinallyblock.
– Guy
Nov 11 at 12:19
Well, that's because you are not asking for input anywhere but the
finally block.– Guy
Nov 11 at 12:19
Well, that's because you are not asking for input anywhere but the
finally block.– Guy
Nov 11 at 12:19
add a comment |
2 Answers
2
active
oldest
votes
Because nextInt() only reads the number, and not the n appended after you hit return, you need to clear that before you can read a number again, in this example I do nextLine() in the catch block. here's more indepth explanation
Working example:
public static int getHours()
int hours = 0;
boolean hoursNotOk = true;
do
try
System.out.println("Here");
hours = console.nextInt();
hoursNotOk = false;
catch (Exception e)
e.printStackTrace();
console.nextLine();
finally
if (hoursNotOk)
System.out.println(", please re-enter the hours again:");
else
System.out.println("**hours input accepted**");
while (hoursNotOk);
return hours;
If I just put this in the main method while(true) someVar = console.nextInt(); why would it not need console.nextLine()?
– Nigel Ng
Nov 11 at 12:59
@NigelNg If I try that it either works when I enter a number or throws an exception when I enter a character.
– Mark
Nov 11 at 15:34
add a comment |
A simpler approach is to test whether you can read an int before throwing an exception. In any case, you need to discard the current word or line before trying again.
public static int getHours()
while (true)
if (console.hasNextInt())
System.out.print("**hours input accepted**");
return console.nextInt();
console.nextLine(); // discard the line and try again
System.out.print(", please re-enter the hours again:");
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Because nextInt() only reads the number, and not the n appended after you hit return, you need to clear that before you can read a number again, in this example I do nextLine() in the catch block. here's more indepth explanation
Working example:
public static int getHours()
int hours = 0;
boolean hoursNotOk = true;
do
try
System.out.println("Here");
hours = console.nextInt();
hoursNotOk = false;
catch (Exception e)
e.printStackTrace();
console.nextLine();
finally
if (hoursNotOk)
System.out.println(", please re-enter the hours again:");
else
System.out.println("**hours input accepted**");
while (hoursNotOk);
return hours;
If I just put this in the main method while(true) someVar = console.nextInt(); why would it not need console.nextLine()?
– Nigel Ng
Nov 11 at 12:59
@NigelNg If I try that it either works when I enter a number or throws an exception when I enter a character.
– Mark
Nov 11 at 15:34
add a comment |
Because nextInt() only reads the number, and not the n appended after you hit return, you need to clear that before you can read a number again, in this example I do nextLine() in the catch block. here's more indepth explanation
Working example:
public static int getHours()
int hours = 0;
boolean hoursNotOk = true;
do
try
System.out.println("Here");
hours = console.nextInt();
hoursNotOk = false;
catch (Exception e)
e.printStackTrace();
console.nextLine();
finally
if (hoursNotOk)
System.out.println(", please re-enter the hours again:");
else
System.out.println("**hours input accepted**");
while (hoursNotOk);
return hours;
If I just put this in the main method while(true) someVar = console.nextInt(); why would it not need console.nextLine()?
– Nigel Ng
Nov 11 at 12:59
@NigelNg If I try that it either works when I enter a number or throws an exception when I enter a character.
– Mark
Nov 11 at 15:34
add a comment |
Because nextInt() only reads the number, and not the n appended after you hit return, you need to clear that before you can read a number again, in this example I do nextLine() in the catch block. here's more indepth explanation
Working example:
public static int getHours()
int hours = 0;
boolean hoursNotOk = true;
do
try
System.out.println("Here");
hours = console.nextInt();
hoursNotOk = false;
catch (Exception e)
e.printStackTrace();
console.nextLine();
finally
if (hoursNotOk)
System.out.println(", please re-enter the hours again:");
else
System.out.println("**hours input accepted**");
while (hoursNotOk);
return hours;
Because nextInt() only reads the number, and not the n appended after you hit return, you need to clear that before you can read a number again, in this example I do nextLine() in the catch block. here's more indepth explanation
Working example:
public static int getHours()
int hours = 0;
boolean hoursNotOk = true;
do
try
System.out.println("Here");
hours = console.nextInt();
hoursNotOk = false;
catch (Exception e)
e.printStackTrace();
console.nextLine();
finally
if (hoursNotOk)
System.out.println(", please re-enter the hours again:");
else
System.out.println("**hours input accepted**");
while (hoursNotOk);
return hours;
edited Nov 11 at 12:23
answered Nov 11 at 12:16
Mark
3,3671926
3,3671926
If I just put this in the main method while(true) someVar = console.nextInt(); why would it not need console.nextLine()?
– Nigel Ng
Nov 11 at 12:59
@NigelNg If I try that it either works when I enter a number or throws an exception when I enter a character.
– Mark
Nov 11 at 15:34
add a comment |
If I just put this in the main method while(true) someVar = console.nextInt(); why would it not need console.nextLine()?
– Nigel Ng
Nov 11 at 12:59
@NigelNg If I try that it either works when I enter a number or throws an exception when I enter a character.
– Mark
Nov 11 at 15:34
If I just put this in the main method while(true) someVar = console.nextInt(); why would it not need console.nextLine()?
– Nigel Ng
Nov 11 at 12:59
If I just put this in the main method while(true) someVar = console.nextInt(); why would it not need console.nextLine()?
– Nigel Ng
Nov 11 at 12:59
@NigelNg If I try that it either works when I enter a number or throws an exception when I enter a character.
– Mark
Nov 11 at 15:34
@NigelNg If I try that it either works when I enter a number or throws an exception when I enter a character.
– Mark
Nov 11 at 15:34
add a comment |
A simpler approach is to test whether you can read an int before throwing an exception. In any case, you need to discard the current word or line before trying again.
public static int getHours()
while (true)
if (console.hasNextInt())
System.out.print("**hours input accepted**");
return console.nextInt();
console.nextLine(); // discard the line and try again
System.out.print(", please re-enter the hours again:");
add a comment |
A simpler approach is to test whether you can read an int before throwing an exception. In any case, you need to discard the current word or line before trying again.
public static int getHours()
while (true)
if (console.hasNextInt())
System.out.print("**hours input accepted**");
return console.nextInt();
console.nextLine(); // discard the line and try again
System.out.print(", please re-enter the hours again:");
add a comment |
A simpler approach is to test whether you can read an int before throwing an exception. In any case, you need to discard the current word or line before trying again.
public static int getHours()
while (true)
if (console.hasNextInt())
System.out.print("**hours input accepted**");
return console.nextInt();
console.nextLine(); // discard the line and try again
System.out.print(", please re-enter the hours again:");
A simpler approach is to test whether you can read an int before throwing an exception. In any case, you need to discard the current word or line before trying again.
public static int getHours()
while (true)
if (console.hasNextInt())
System.out.print("**hours input accepted**");
return console.nextInt();
console.nextLine(); // discard the line and try again
System.out.print(", please re-enter the hours again:");
answered Nov 11 at 12:48
Peter Lawrey
440k55558957
440k55558957
add a comment |
add a comment |
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Well, that's because you are not asking for input anywhere but the
finallyblock.– Guy
Nov 11 at 12:19