Try/Catch exception help in Java
1
The program I am making has a console-based text menu to select an option between 1 and 2, I need it to catch for any input that isnt a number, and isnt a number between 1 and 2. This is what I have
Scanner scan = new Scanner(System.in);
int number = 0;
try {
System.out.println("Enter a number ");
System.out.println("1.");
System.out.println("2.");
number = scan.nextInt();
}
catch (ArithmeticException e) {
System.out.println("Arithmetic Exception");
}
catch (Exception e) {
if (!(number == 1) || !(number == 2)) {
System.out.println("Exception");
}
}
Any insight on where I have gone wrong is appreciated!
java
asked Nov 14 '18 at 19:59
Paul JonesPaul Jones
266
|
show 1 more comment
1
The program I am making has a console-based text menu to select an option between 1 and 2, I need it to catch for any input that isnt a number, and isnt a number between 1 and 2. This is what I have
Scanner scan = new Scanner(System.in);
int number = 0;
try {
System.out.println("Enter a number ");
System.out.println("1.");
System.out.println("2.");
number = scan.nextInt();
}
catch (ArithmeticException e) {
System.out.println("Arithmetic Exception");
}
catch (Exception e) {
if (!(number == 1) || !(number == 2)) {
System.out.println("Exception");
}
}
Any insight on where I have gone wrong is appreciated!
java
asked Nov 14 '18 at 19:59
Paul JonesPaul Jones
266
2
You say you’ve gone wrong - where does the actual behaviour differ from the intended?
– MTCoster
Nov 14 '18 at 20:00
1
What's the problem? You didn't specify.
– NiVeR
Nov 14 '18 at 20:01
2
Putting in the number3, for example, isn't a system exception but a program-defined boundary. Use an if-statement for that inside the try-block.
– Drew Kennedy
Nov 14 '18 at 20:01
2
You don't get an ArithmeticException from reading an int from a Scanner, as you are not doing any arithmetic. And a number is always either not equal to 1 or not equal to 2.
– Andy Turner
Nov 14 '18 at 20:02
It won't throw an exception for you, you need to do the logic and throw the exception.
– SPlatten
Nov 14 '18 at 20:03
|
show 1 more comment
1
1
1
The program I am making has a console-based text menu to select an option between 1 and 2, I need it to catch for any input that isnt a number, and isnt a number between 1 and 2. This is what I have
Scanner scan = new Scanner(System.in);
int number = 0;
try {
System.out.println("Enter a number ");
System.out.println("1.");
System.out.println("2.");
number = scan.nextInt();
}
catch (ArithmeticException e) {
System.out.println("Arithmetic Exception");
}
catch (Exception e) {
if (!(number == 1) || !(number == 2)) {
System.out.println("Exception");
}
}
Any insight on where I have gone wrong is appreciated!
java
asked Nov 14 '18 at 19:59
Paul JonesPaul Jones
266
The program I am making has a console-based text menu to select an option between 1 and 2, I need it to catch for any input that isnt a number, and isnt a number between 1 and 2. This is what I have
Scanner scan = new Scanner(System.in);
int number = 0;
try {
System.out.println("Enter a number ");
System.out.println("1.");
System.out.println("2.");
number = scan.nextInt();
}
catch (ArithmeticException e) {
System.out.println("Arithmetic Exception");
}
catch (Exception e) {
if (!(number == 1) || !(number == 2)) {
System.out.println("Exception");
}
}
Any insight on where I have gone wrong is appreciated!
java
java
asked Nov 14 '18 at 19:59
Paul JonesPaul Jones
266
asked Nov 14 '18 at 19:59
Paul JonesPaul Jones
266
asked Nov 14 '18 at 19:59
Paul JonesPaul Jones
266
asked Nov 14 '18 at 19:59
Paul JonesPaul Jones
266
asked Nov 14 '18 at 19:59
Paul JonesPaul Jones
266
266
2
You say you’ve gone wrong - where does the actual behaviour differ from the intended?
– MTCoster
Nov 14 '18 at 20:00
1
What's the problem? You didn't specify.
– NiVeR
Nov 14 '18 at 20:01
2
Putting in the number3, for example, isn't a system exception but a program-defined boundary. Use an if-statement for that inside the try-block.
– Drew Kennedy
Nov 14 '18 at 20:01
2
You don't get an ArithmeticException from reading an int from a Scanner, as you are not doing any arithmetic. And a number is always either not equal to 1 or not equal to 2.
– Andy Turner
Nov 14 '18 at 20:02
It won't throw an exception for you, you need to do the logic and throw the exception.
– SPlatten
Nov 14 '18 at 20:03
|
show 1 more comment
2
You say you’ve gone wrong - where does the actual behaviour differ from the intended?
– MTCoster
Nov 14 '18 at 20:00
1
What's the problem? You didn't specify.
– NiVeR
Nov 14 '18 at 20:01
2
Putting in the number3, for example, isn't a system exception but a program-defined boundary. Use an if-statement for that inside the try-block.
– Drew Kennedy
Nov 14 '18 at 20:01
2
You don't get an ArithmeticException from reading an int from a Scanner, as you are not doing any arithmetic. And a number is always either not equal to 1 or not equal to 2.
– Andy Turner
Nov 14 '18 at 20:02
It won't throw an exception for you, you need to do the logic and throw the exception.
– SPlatten
Nov 14 '18 at 20:03
2
2
You say you’ve gone wrong - where does the actual behaviour differ from the intended?
– MTCoster
Nov 14 '18 at 20:00
You say you’ve gone wrong - where does the actual behaviour differ from the intended?
– MTCoster
Nov 14 '18 at 20:00
1
1
What's the problem? You didn't specify.
– NiVeR
Nov 14 '18 at 20:01
What's the problem? You didn't specify.
– NiVeR
Nov 14 '18 at 20:01
2
2
Putting in the number
3, for example, isn't a system exception but a program-defined boundary. Use an if-statement for that inside the try-block.– Drew Kennedy
Nov 14 '18 at 20:01
Putting in the number
3, for example, isn't a system exception but a program-defined boundary. Use an if-statement for that inside the try-block.– Drew Kennedy
Nov 14 '18 at 20:01
2
2
You don't get an ArithmeticException from reading an int from a Scanner, as you are not doing any arithmetic. And a number is always either not equal to 1 or not equal to 2.
– Andy Turner
Nov 14 '18 at 20:02
You don't get an ArithmeticException from reading an int from a Scanner, as you are not doing any arithmetic. And a number is always either not equal to 1 or not equal to 2.
– Andy Turner
Nov 14 '18 at 20:02
It won't throw an exception for you, you need to do the logic and throw the exception.
– SPlatten
Nov 14 '18 at 20:03
It won't throw an exception for you, you need to do the logic and throw the exception.
– SPlatten
Nov 14 '18 at 20:03
|
show 1 more comment
3 Answers
3
active
oldest
votes
1
If you want to throw an exception, this is easy but take care with the condition:
Scanner scan = new Scanner(System.in);
int number = 0;
try {
System.out.println("Enter a number ");
System.out.println("1.");
System.out.println("2.");
number = scan.nextInt();
if ((number != 1) && (number != 2)){
throw new Exception();
}
}
catch (InputMismatchException e) {
System.out.println("This is not a number");
}
catch (Exception e) {
System.out.println("Inside here because the number is not 1 or 2");
}
The condition
(number != 1) && (number != 2)
is true if number is not (1 or 2)
answered Nov 14 '18 at 20:15
forpasforpas
13.6k3624
add a comment |
0
Scanner scan = new Scanner(System.in);
int number = 0;
try {
System.out.println("Enter a number ");
System.out.println("1.");
System.out.println("2.");
number = scan.nextInt();
if (number != 1 && number != 2){
throw new Exception();
}
}
catch (ArithmeticException e) {
System.out.println("Arithmetic Exception");
}
catch (Exception e) {
System.out.println("Inside here because the number is not 1 or 2 ");
}
answered Nov 14 '18 at 20:11
Akeme UAkeme U
113
add a comment |
0
As a suggestion, you can use
if ((number != 1) || (number != 2))
instead of
if (!(number == 1) || !(number == 2))
edited Nov 15 '18 at 5:53
Rahul Sharma
2,3191819
answered Nov 14 '18 at 20:09
kamilyrbkamilyrb
187
1
Sure, but why not just useif (true), that's equivalent and simpler again.
– Andy Turner
Nov 14 '18 at 20:15
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
If you want to throw an exception, this is easy but take care with the condition:
Scanner scan = new Scanner(System.in);
int number = 0;
try {
System.out.println("Enter a number ");
System.out.println("1.");
System.out.println("2.");
number = scan.nextInt();
if ((number != 1) && (number != 2)){
throw new Exception();
}
}
catch (InputMismatchException e) {
System.out.println("This is not a number");
}
catch (Exception e) {
System.out.println("Inside here because the number is not 1 or 2");
}
The condition
(number != 1) && (number != 2)
is true if number is not (1 or 2)
answered Nov 14 '18 at 20:15
forpasforpas
13.6k3624
add a comment |
1
If you want to throw an exception, this is easy but take care with the condition:
Scanner scan = new Scanner(System.in);
int number = 0;
try {
System.out.println("Enter a number ");
System.out.println("1.");
System.out.println("2.");
number = scan.nextInt();
if ((number != 1) && (number != 2)){
throw new Exception();
}
}
catch (InputMismatchException e) {
System.out.println("This is not a number");
}
catch (Exception e) {
System.out.println("Inside here because the number is not 1 or 2");
}
The condition
(number != 1) && (number != 2)
is true if number is not (1 or 2)
answered Nov 14 '18 at 20:15
forpasforpas
13.6k3624
add a comment |
1
1
1
If you want to throw an exception, this is easy but take care with the condition:
Scanner scan = new Scanner(System.in);
int number = 0;
try {
System.out.println("Enter a number ");
System.out.println("1.");
System.out.println("2.");
number = scan.nextInt();
if ((number != 1) && (number != 2)){
throw new Exception();
}
}
catch (InputMismatchException e) {
System.out.println("This is not a number");
}
catch (Exception e) {
System.out.println("Inside here because the number is not 1 or 2");
}
The condition
(number != 1) && (number != 2)
is true if number is not (1 or 2)
answered Nov 14 '18 at 20:15
forpasforpas
13.6k3624
If you want to throw an exception, this is easy but take care with the condition:
Scanner scan = new Scanner(System.in);
int number = 0;
try {
System.out.println("Enter a number ");
System.out.println("1.");
System.out.println("2.");
number = scan.nextInt();
if ((number != 1) && (number != 2)){
throw new Exception();
}
}
catch (InputMismatchException e) {
System.out.println("This is not a number");
}
catch (Exception e) {
System.out.println("Inside here because the number is not 1 or 2");
}
The condition
(number != 1) && (number != 2)
is true if number is not (1 or 2)
answered Nov 14 '18 at 20:15
forpasforpas
13.6k3624
edited Nov 14 '18 at 20:26
answered Nov 14 '18 at 20:15
forpasforpas
13.6k3624
answered Nov 14 '18 at 20:15
forpasforpas
13.6k3624
answered Nov 14 '18 at 20:15
forpasforpas
13.6k3624
13.6k3624
add a comment |
add a comment |
0
Scanner scan = new Scanner(System.in);
int number = 0;
try {
System.out.println("Enter a number ");
System.out.println("1.");
System.out.println("2.");
number = scan.nextInt();
if (number != 1 && number != 2){
throw new Exception();
}
}
catch (ArithmeticException e) {
System.out.println("Arithmetic Exception");
}
catch (Exception e) {
System.out.println("Inside here because the number is not 1 or 2 ");
}
answered Nov 14 '18 at 20:11
Akeme UAkeme U
113
add a comment |
0
Scanner scan = new Scanner(System.in);
int number = 0;
try {
System.out.println("Enter a number ");
System.out.println("1.");
System.out.println("2.");
number = scan.nextInt();
if (number != 1 && number != 2){
throw new Exception();
}
}
catch (ArithmeticException e) {
System.out.println("Arithmetic Exception");
}
catch (Exception e) {
System.out.println("Inside here because the number is not 1 or 2 ");
}
answered Nov 14 '18 at 20:11
Akeme UAkeme U
113
add a comment |
0
0
0
Scanner scan = new Scanner(System.in);
int number = 0;
try {
System.out.println("Enter a number ");
System.out.println("1.");
System.out.println("2.");
number = scan.nextInt();
if (number != 1 && number != 2){
throw new Exception();
}
}
catch (ArithmeticException e) {
System.out.println("Arithmetic Exception");
}
catch (Exception e) {
System.out.println("Inside here because the number is not 1 or 2 ");
}
answered Nov 14 '18 at 20:11
Akeme UAkeme U
113
Scanner scan = new Scanner(System.in);
int number = 0;
try {
System.out.println("Enter a number ");
System.out.println("1.");
System.out.println("2.");
number = scan.nextInt();
if (number != 1 && number != 2){
throw new Exception();
}
}
catch (ArithmeticException e) {
System.out.println("Arithmetic Exception");
}
catch (Exception e) {
System.out.println("Inside here because the number is not 1 or 2 ");
}
answered Nov 14 '18 at 20:11
Akeme UAkeme U
113
edited Nov 14 '18 at 23:45
answered Nov 14 '18 at 20:11
Akeme UAkeme U
113
answered Nov 14 '18 at 20:11
Akeme UAkeme U
113
answered Nov 14 '18 at 20:11
Akeme UAkeme U
113
113
add a comment |
add a comment |
0
As a suggestion, you can use
if ((number != 1) || (number != 2))
instead of
if (!(number == 1) || !(number == 2))
edited Nov 15 '18 at 5:53
Rahul Sharma
2,3191819
answered Nov 14 '18 at 20:09
kamilyrbkamilyrb
187
1
Sure, but why not just useif (true), that's equivalent and simpler again.
– Andy Turner
Nov 14 '18 at 20:15
add a comment |
0
As a suggestion, you can use
if ((number != 1) || (number != 2))
instead of
if (!(number == 1) || !(number == 2))
edited Nov 15 '18 at 5:53
Rahul Sharma
2,3191819
answered Nov 14 '18 at 20:09
kamilyrbkamilyrb
187
1
Sure, but why not just useif (true), that's equivalent and simpler again.
– Andy Turner
Nov 14 '18 at 20:15
add a comment |
0
0
0
As a suggestion, you can use
if ((number != 1) || (number != 2))
instead of
if (!(number == 1) || !(number == 2))
edited Nov 15 '18 at 5:53
Rahul Sharma
2,3191819
answered Nov 14 '18 at 20:09
kamilyrbkamilyrb
187
As a suggestion, you can use
if ((number != 1) || (number != 2))
instead of
if (!(number == 1) || !(number == 2))
edited Nov 15 '18 at 5:53
Rahul Sharma
2,3191819
answered Nov 14 '18 at 20:09
kamilyrbkamilyrb
187
edited Nov 15 '18 at 5:53
Rahul Sharma
2,3191819
edited Nov 15 '18 at 5:53
Rahul Sharma
2,3191819
edited Nov 15 '18 at 5:53
Rahul Sharma
2,3191819
2,3191819
answered Nov 14 '18 at 20:09
kamilyrbkamilyrb
187
answered Nov 14 '18 at 20:09
kamilyrbkamilyrb
187
answered Nov 14 '18 at 20:09
kamilyrbkamilyrb
187
187
1
Sure, but why not just useif (true), that's equivalent and simpler again.
– Andy Turner
Nov 14 '18 at 20:15
add a comment |
1
Sure, but why not just useif (true), that's equivalent and simpler again.
– Andy Turner
Nov 14 '18 at 20:15
1
1
Sure, but why not just use
if (true), that's equivalent and simpler again.– Andy Turner
Nov 14 '18 at 20:15
Sure, but why not just use
if (true), that's equivalent and simpler again.– Andy Turner
Nov 14 '18 at 20:15
add a comment |
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2
You say you’ve gone wrong - where does the actual behaviour differ from the intended?
– MTCoster
Nov 14 '18 at 20:00
1
What's the problem? You didn't specify.
– NiVeR
Nov 14 '18 at 20:01
2
Putting in the number
3, for example, isn't a system exception but a program-defined boundary. Use an if-statement for that inside the try-block.– Drew Kennedy
Nov 14 '18 at 20:01
2
You don't get an ArithmeticException from reading an int from a Scanner, as you are not doing any arithmetic. And a number is always either not equal to 1 or not equal to 2.
– Andy Turner
Nov 14 '18 at 20:02
It won't throw an exception for you, you need to do the logic and throw the exception.
– SPlatten
Nov 14 '18 at 20:03