Validating user inputValidating 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
Russian cases: A few examples, I'm really confused
Bash replace string at multiple places in a file from command line
How to make healing in an exploration game interesting
Be in awe of my brilliance!
Validating user input
The use of "touch" and "touch on" in context
It's a yearly task, alright
Is a lawful good "antagonist" effective?
What is a good source for large tables on the properties of water?
Did CPM support custom hardware using device drivers?
How to deal with a cynical class?
Am I not good enough for you?
Happy pi day, everyone!
How do anti-virus programs start at Windows boot?
Could the Saturn V actually have launched astronauts around Venus?
Know when to turn notes upside-down(eighth notes, sixteen notes, etc.)
Does the statement `int val = (++i > ++j) ? ++i : ++j;` invoke undefined behavior?
RegionDifference for Cylinder and Cuboid
My adviser wants to be the first author
What does it mean to make a bootable LiveUSB?
Schematic conventions for different supply rails
Ban on all campaign finance?
Science-fiction short story where space navy wanted hospital ships and settlers had guns mounted everywhere
Why are there 40 737 Max planes in flight when they have been grounded as not airworthy?
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
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
);
);
, "mathjax-editing");
StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "196"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
PerfectContrast is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f215452%2fvalidating-user-input%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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.
Thanks for contributing an answer to Code Review Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodereview.stackexchange.com%2fquestions%2f215452%2fvalidating-user-input%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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