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Validating user input
Validating a userFinding potential thread safety issues and race conditions in my multithreading codeScrubbing user inputValidating input against rulesDriver license program which grades an individual's responsesValidating input values in C#Validating integer or string inputValidating user input in C# coming from XAML controlsValidating input variablesValidating proper input
$begingroup$
I've written the below code to validate user input but I feel it's quite excessive for what seems to be a simple operation. I want to only accept non-negative integers or doubles as an input. This block of code is required three times in my program so I will need to put it into a method or something but in the mean time is there a simpler way to validate user input?
Scanner input = new Scanner(System.in);
boolean validWidth = false;
double width = 0.0;;
do // do while runs until the input is validated
{
System.out.print("Enter width: ");
if (input.hasNextInt()) // we accept an int
{
width = input.nextInt();
if (width > 0) // we accept a non-negative
{
validWidth = true;
}
else // refuse a negative
{
System.out.println("Input error. Try again.");
validWidth = false;
}
}
else if (input.hasNextDouble())// accept a double
{
width = input.nextDouble();
if (width > 0) // we accept a non-negative
{
validWidth = true;
}
else // and refuse a negative
{
System.out.println("Input error. Try again.");
validWidth = false;
}
}
else
{
System.out.println("Input error. Try again.");
validInput = false;
input.next();
}
} while (!(validWidth));
java validation
New contributor
$endgroup$
add a comment |
$begingroup$
I've written the below code to validate user input but I feel it's quite excessive for what seems to be a simple operation. I want to only accept non-negative integers or doubles as an input. This block of code is required three times in my program so I will need to put it into a method or something but in the mean time is there a simpler way to validate user input?
Scanner input = new Scanner(System.in);
boolean validWidth = false;
double width = 0.0;;
do // do while runs until the input is validated
{
System.out.print("Enter width: ");
if (input.hasNextInt()) // we accept an int
{
width = input.nextInt();
if (width > 0) // we accept a non-negative
{
validWidth = true;
}
else // refuse a negative
{
System.out.println("Input error. Try again.");
validWidth = false;
}
}
else if (input.hasNextDouble())// accept a double
{
width = input.nextDouble();
if (width > 0) // we accept a non-negative
{
validWidth = true;
}
else // and refuse a negative
{
System.out.println("Input error. Try again.");
validWidth = false;
}
}
else
{
System.out.println("Input error. Try again.");
validInput = false;
input.next();
}
} while (!(validWidth));
java validation
New contributor
$endgroup$
$begingroup$
Welcome to Code Review. Unfortunately, your code is missing some important context, for exampleinput
's type, how it's initialized, and the declarations of the other variables. Keep in mind that It's fine to drop a lot of code here, as long as you properly explain it beforehand.
$endgroup$
– Zeta
4 hours ago
$begingroup$
Sorry, I have added those now.
$endgroup$
– PerfectContrast
4 hours ago
$begingroup$
Perfect, thanks. I hope that one of the Java devs will review your code soon.
$endgroup$
– Zeta
4 hours ago
add a comment |
$begingroup$
I've written the below code to validate user input but I feel it's quite excessive for what seems to be a simple operation. I want to only accept non-negative integers or doubles as an input. This block of code is required three times in my program so I will need to put it into a method or something but in the mean time is there a simpler way to validate user input?
Scanner input = new Scanner(System.in);
boolean validWidth = false;
double width = 0.0;;
do // do while runs until the input is validated
{
System.out.print("Enter width: ");
if (input.hasNextInt()) // we accept an int
{
width = input.nextInt();
if (width > 0) // we accept a non-negative
{
validWidth = true;
}
else // refuse a negative
{
System.out.println("Input error. Try again.");
validWidth = false;
}
}
else if (input.hasNextDouble())// accept a double
{
width = input.nextDouble();
if (width > 0) // we accept a non-negative
{
validWidth = true;
}
else // and refuse a negative
{
System.out.println("Input error. Try again.");
validWidth = false;
}
}
else
{
System.out.println("Input error. Try again.");
validInput = false;
input.next();
}
} while (!(validWidth));
java validation
New contributor
$endgroup$
I've written the below code to validate user input but I feel it's quite excessive for what seems to be a simple operation. I want to only accept non-negative integers or doubles as an input. This block of code is required three times in my program so I will need to put it into a method or something but in the mean time is there a simpler way to validate user input?
Scanner input = new Scanner(System.in);
boolean validWidth = false;
double width = 0.0;;
do // do while runs until the input is validated
{
System.out.print("Enter width: ");
if (input.hasNextInt()) // we accept an int
{
width = input.nextInt();
if (width > 0) // we accept a non-negative
{
validWidth = true;
}
else // refuse a negative
{
System.out.println("Input error. Try again.");
validWidth = false;
}
}
else if (input.hasNextDouble())// accept a double
{
width = input.nextDouble();
if (width > 0) // we accept a non-negative
{
validWidth = true;
}
else // and refuse a negative
{
System.out.println("Input error. Try again.");
validWidth = false;
}
}
else
{
System.out.println("Input error. Try again.");
validInput = false;
input.next();
}
} while (!(validWidth));
java validation
java validation
New contributor
New contributor
edited 4 hours ago
PerfectContrast
New contributor
asked 4 hours ago
PerfectContrastPerfectContrast
235
235
New contributor
New contributor
$begingroup$
Welcome to Code Review. Unfortunately, your code is missing some important context, for exampleinput
's type, how it's initialized, and the declarations of the other variables. Keep in mind that It's fine to drop a lot of code here, as long as you properly explain it beforehand.
$endgroup$
– Zeta
4 hours ago
$begingroup$
Sorry, I have added those now.
$endgroup$
– PerfectContrast
4 hours ago
$begingroup$
Perfect, thanks. I hope that one of the Java devs will review your code soon.
$endgroup$
– Zeta
4 hours ago
add a comment |
$begingroup$
Welcome to Code Review. Unfortunately, your code is missing some important context, for exampleinput
's type, how it's initialized, and the declarations of the other variables. Keep in mind that It's fine to drop a lot of code here, as long as you properly explain it beforehand.
$endgroup$
– Zeta
4 hours ago
$begingroup$
Sorry, I have added those now.
$endgroup$
– PerfectContrast
4 hours ago
$begingroup$
Perfect, thanks. I hope that one of the Java devs will review your code soon.
$endgroup$
– Zeta
4 hours ago
$begingroup$
Welcome to Code Review. Unfortunately, your code is missing some important context, for example
input
's type, how it's initialized, and the declarations of the other variables. Keep in mind that It's fine to drop a lot of code here, as long as you properly explain it beforehand.$endgroup$
– Zeta
4 hours ago
$begingroup$
Welcome to Code Review. Unfortunately, your code is missing some important context, for example
input
's type, how it's initialized, and the declarations of the other variables. Keep in mind that It's fine to drop a lot of code here, as long as you properly explain it beforehand.$endgroup$
– Zeta
4 hours ago
$begingroup$
Sorry, I have added those now.
$endgroup$
– PerfectContrast
4 hours ago
$begingroup$
Sorry, I have added those now.
$endgroup$
– PerfectContrast
4 hours ago
$begingroup$
Perfect, thanks. I hope that one of the Java devs will review your code soon.
$endgroup$
– Zeta
4 hours ago
$begingroup$
Perfect, thanks. I hope that one of the Java devs will review your code soon.
$endgroup$
– Zeta
4 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are distinguishing between hasNextInt()
and hasNextDouble()
in the input stream, but treating them identically afterwards. hasNextDouble()
will return true
if the next token is an integer, because integers can be successfully parsed as a double too.
So your loop could be simplified into:
double width = 0.0;
for(;;) {
System.out.print("Enter width: ");
if (input.hasNextDouble()) {
width = input.nextDouble();
if (width > 0)
break;
}
System.out.println("Input error. Try again.");
input.nextLine();
}
A few notes:
The unnecessary
validWidth
variable has been removed. Abreak
statement exits the infinitefor(;;)
loop, skipping over the "cleanup, try again" code..nextLine()
is used to cleanup after invalid input. This is important, because if the user enters"one fish two fish red fish blue fish"
instead of say-12
, your current approach will print out"Input error. Try again"
8 times. Using.nextLine()
discards everything up to and including the new line character. Which brings us to zero and negative numbers. If the user enters a valid integer/double, but not a positive one, your code didn't skip any tokens, where as my code (as mentioned above) will skip the remaining input up to and including the new line character. Assuming the next character was a new line (as in the user entered-12
and pressed the return key) the following.hasNextDouble()
will skip over the new line (and any other blank space) looking for the next token, so the effect is approximately the same.But my code performs differently if the user enters
red 5
. You original code will print"Invalid input. Try again."
, and then immediately consume the5
as valid input, where as my code will discard the remainder of the line, and wait for the user to enter another line.
Since you are using the code multiple places, putting it into a method is prudent. Here is one using a DoublePredicate
functional interface in order to validate the input according to the caller's requirements:
double getDouble(Scanner input, String prompt, DoublePredicate validate) {
double value = 0.0;
for(;;) {
System.out.print(prompt);
if (input.hasNextDouble()) {
value = input.nextDouble();
if (validate.test(value))
return value;
}
System.out.println("Input error. Try again.");
input.nextLine();
}
}
Which you could use with a lambda function like:
double width = getDouble(input, "Enter width: ", x -> x > 0);
$endgroup$
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are distinguishing between hasNextInt()
and hasNextDouble()
in the input stream, but treating them identically afterwards. hasNextDouble()
will return true
if the next token is an integer, because integers can be successfully parsed as a double too.
So your loop could be simplified into:
double width = 0.0;
for(;;) {
System.out.print("Enter width: ");
if (input.hasNextDouble()) {
width = input.nextDouble();
if (width > 0)
break;
}
System.out.println("Input error. Try again.");
input.nextLine();
}
A few notes:
The unnecessary
validWidth
variable has been removed. Abreak
statement exits the infinitefor(;;)
loop, skipping over the "cleanup, try again" code..nextLine()
is used to cleanup after invalid input. This is important, because if the user enters"one fish two fish red fish blue fish"
instead of say-12
, your current approach will print out"Input error. Try again"
8 times. Using.nextLine()
discards everything up to and including the new line character. Which brings us to zero and negative numbers. If the user enters a valid integer/double, but not a positive one, your code didn't skip any tokens, where as my code (as mentioned above) will skip the remaining input up to and including the new line character. Assuming the next character was a new line (as in the user entered-12
and pressed the return key) the following.hasNextDouble()
will skip over the new line (and any other blank space) looking for the next token, so the effect is approximately the same.But my code performs differently if the user enters
red 5
. You original code will print"Invalid input. Try again."
, and then immediately consume the5
as valid input, where as my code will discard the remainder of the line, and wait for the user to enter another line.
Since you are using the code multiple places, putting it into a method is prudent. Here is one using a DoublePredicate
functional interface in order to validate the input according to the caller's requirements:
double getDouble(Scanner input, String prompt, DoublePredicate validate) {
double value = 0.0;
for(;;) {
System.out.print(prompt);
if (input.hasNextDouble()) {
value = input.nextDouble();
if (validate.test(value))
return value;
}
System.out.println("Input error. Try again.");
input.nextLine();
}
}
Which you could use with a lambda function like:
double width = getDouble(input, "Enter width: ", x -> x > 0);
$endgroup$
add a comment |
$begingroup$
You are distinguishing between hasNextInt()
and hasNextDouble()
in the input stream, but treating them identically afterwards. hasNextDouble()
will return true
if the next token is an integer, because integers can be successfully parsed as a double too.
So your loop could be simplified into:
double width = 0.0;
for(;;) {
System.out.print("Enter width: ");
if (input.hasNextDouble()) {
width = input.nextDouble();
if (width > 0)
break;
}
System.out.println("Input error. Try again.");
input.nextLine();
}
A few notes:
The unnecessary
validWidth
variable has been removed. Abreak
statement exits the infinitefor(;;)
loop, skipping over the "cleanup, try again" code..nextLine()
is used to cleanup after invalid input. This is important, because if the user enters"one fish two fish red fish blue fish"
instead of say-12
, your current approach will print out"Input error. Try again"
8 times. Using.nextLine()
discards everything up to and including the new line character. Which brings us to zero and negative numbers. If the user enters a valid integer/double, but not a positive one, your code didn't skip any tokens, where as my code (as mentioned above) will skip the remaining input up to and including the new line character. Assuming the next character was a new line (as in the user entered-12
and pressed the return key) the following.hasNextDouble()
will skip over the new line (and any other blank space) looking for the next token, so the effect is approximately the same.But my code performs differently if the user enters
red 5
. You original code will print"Invalid input. Try again."
, and then immediately consume the5
as valid input, where as my code will discard the remainder of the line, and wait for the user to enter another line.
Since you are using the code multiple places, putting it into a method is prudent. Here is one using a DoublePredicate
functional interface in order to validate the input according to the caller's requirements:
double getDouble(Scanner input, String prompt, DoublePredicate validate) {
double value = 0.0;
for(;;) {
System.out.print(prompt);
if (input.hasNextDouble()) {
value = input.nextDouble();
if (validate.test(value))
return value;
}
System.out.println("Input error. Try again.");
input.nextLine();
}
}
Which you could use with a lambda function like:
double width = getDouble(input, "Enter width: ", x -> x > 0);
$endgroup$
add a comment |
$begingroup$
You are distinguishing between hasNextInt()
and hasNextDouble()
in the input stream, but treating them identically afterwards. hasNextDouble()
will return true
if the next token is an integer, because integers can be successfully parsed as a double too.
So your loop could be simplified into:
double width = 0.0;
for(;;) {
System.out.print("Enter width: ");
if (input.hasNextDouble()) {
width = input.nextDouble();
if (width > 0)
break;
}
System.out.println("Input error. Try again.");
input.nextLine();
}
A few notes:
The unnecessary
validWidth
variable has been removed. Abreak
statement exits the infinitefor(;;)
loop, skipping over the "cleanup, try again" code..nextLine()
is used to cleanup after invalid input. This is important, because if the user enters"one fish two fish red fish blue fish"
instead of say-12
, your current approach will print out"Input error. Try again"
8 times. Using.nextLine()
discards everything up to and including the new line character. Which brings us to zero and negative numbers. If the user enters a valid integer/double, but not a positive one, your code didn't skip any tokens, where as my code (as mentioned above) will skip the remaining input up to and including the new line character. Assuming the next character was a new line (as in the user entered-12
and pressed the return key) the following.hasNextDouble()
will skip over the new line (and any other blank space) looking for the next token, so the effect is approximately the same.But my code performs differently if the user enters
red 5
. You original code will print"Invalid input. Try again."
, and then immediately consume the5
as valid input, where as my code will discard the remainder of the line, and wait for the user to enter another line.
Since you are using the code multiple places, putting it into a method is prudent. Here is one using a DoublePredicate
functional interface in order to validate the input according to the caller's requirements:
double getDouble(Scanner input, String prompt, DoublePredicate validate) {
double value = 0.0;
for(;;) {
System.out.print(prompt);
if (input.hasNextDouble()) {
value = input.nextDouble();
if (validate.test(value))
return value;
}
System.out.println("Input error. Try again.");
input.nextLine();
}
}
Which you could use with a lambda function like:
double width = getDouble(input, "Enter width: ", x -> x > 0);
$endgroup$
You are distinguishing between hasNextInt()
and hasNextDouble()
in the input stream, but treating them identically afterwards. hasNextDouble()
will return true
if the next token is an integer, because integers can be successfully parsed as a double too.
So your loop could be simplified into:
double width = 0.0;
for(;;) {
System.out.print("Enter width: ");
if (input.hasNextDouble()) {
width = input.nextDouble();
if (width > 0)
break;
}
System.out.println("Input error. Try again.");
input.nextLine();
}
A few notes:
The unnecessary
validWidth
variable has been removed. Abreak
statement exits the infinitefor(;;)
loop, skipping over the "cleanup, try again" code..nextLine()
is used to cleanup after invalid input. This is important, because if the user enters"one fish two fish red fish blue fish"
instead of say-12
, your current approach will print out"Input error. Try again"
8 times. Using.nextLine()
discards everything up to and including the new line character. Which brings us to zero and negative numbers. If the user enters a valid integer/double, but not a positive one, your code didn't skip any tokens, where as my code (as mentioned above) will skip the remaining input up to and including the new line character. Assuming the next character was a new line (as in the user entered-12
and pressed the return key) the following.hasNextDouble()
will skip over the new line (and any other blank space) looking for the next token, so the effect is approximately the same.But my code performs differently if the user enters
red 5
. You original code will print"Invalid input. Try again."
, and then immediately consume the5
as valid input, where as my code will discard the remainder of the line, and wait for the user to enter another line.
Since you are using the code multiple places, putting it into a method is prudent. Here is one using a DoublePredicate
functional interface in order to validate the input according to the caller's requirements:
double getDouble(Scanner input, String prompt, DoublePredicate validate) {
double value = 0.0;
for(;;) {
System.out.print(prompt);
if (input.hasNextDouble()) {
value = input.nextDouble();
if (validate.test(value))
return value;
}
System.out.println("Input error. Try again.");
input.nextLine();
}
}
Which you could use with a lambda function like:
double width = getDouble(input, "Enter width: ", x -> x > 0);
answered 2 hours ago
AJNeufeldAJNeufeld
6,1601520
6,1601520
add a comment |
add a comment |
PerfectContrast is a new contributor. Be nice, and check out our Code of Conduct.
PerfectContrast is a new contributor. Be nice, and check out our Code of Conduct.
PerfectContrast is a new contributor. Be nice, and check out our Code of Conduct.
PerfectContrast is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Welcome to Code Review. Unfortunately, your code is missing some important context, for example
input
's type, how it's initialized, and the declarations of the other variables. Keep in mind that It's fine to drop a lot of code here, as long as you properly explain it beforehand.$endgroup$
– Zeta
4 hours ago
$begingroup$
Sorry, I have added those now.
$endgroup$
– PerfectContrast
4 hours ago
$begingroup$
Perfect, thanks. I hope that one of the Java devs will review your code soon.
$endgroup$
– Zeta
4 hours ago