Program only throws exception when variable is defined in “try”Do try/catch blocks hurt performance when exceptions are not thrown?Is it a good practice to use try-except-else in Python?Exception try and catch on split name into first and lastWhy is “except: pass” a bad programming practice?How to prevent a runtime error to stop the rest of the code in try/catch from executing?Python looping with try and exceptStruggling with try/catch InputMismatch Exception and retrieving data from an arrayPython ValueError exception - Name “a” is not defined errorrequests.get returns 400 bad url when given a variable containing a url, but not when given a string with the same urlJava JSoup Exception ignores try catch?
Pronouncing Dictionary.com's W.O.D "vade mecum" in English
Infinite past with a beginning?
Compute hash value according to multiplication method
Why is an old chain unsafe?
Download, install and reboot computer at night if needed
A function which translates a sentence to title-case
What are these boxed doors outside store fronts in New York?
How to make payment on the internet without leaving a money trail?
I see my dog run
Is it possible to make sharp wind that can cut stuff from afar?
Draw simple lines in Inkscape
Why doesn't Newton's third law mean a person bounces back to where they started when they hit the ground?
What do you call something that goes against the spirit of the law, but is legal when interpreting the law to the letter?
The use of multiple foreign keys on same column in SQL Server
How to type dʒ symbol (IPA) on Mac?
Can I make popcorn with any corn?
Is there really no realistic way for a skeleton monster to move around without magic?
Modification to Chariots for Heavy Cavalry Analogue for 4-armed race
How is it possible for user to changed after storage was encrypted? (on OS X, Android)
How is the claim "I am in New York only if I am in America" the same as "If I am in New York, then I am in America?
How is it possible to have an ability score that is less than 3?
What would happen to a modern skyscraper if it rains micro blackholes?
Patience, young "Padovan"
How old can references or sources in a thesis be?
Program only throws exception when variable is defined in “try”
Do try/catch blocks hurt performance when exceptions are not thrown?Is it a good practice to use try-except-else in Python?Exception try and catch on split name into first and lastWhy is “except: pass” a bad programming practice?How to prevent a runtime error to stop the rest of the code in try/catch from executing?Python looping with try and exceptStruggling with try/catch InputMismatch Exception and retrieving data from an arrayPython ValueError exception - Name “a” is not defined errorrequests.get returns 400 bad url when given a variable containing a url, but not when given a string with the same urlJava JSoup Exception ignores try catch?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
I am trying to write a program that will read a webpage. The following code is not valid:
String[] lines;
try
lines = loadStrings(url);
catch(IOException e)
return;
But this code is:
try
String[] lines = loadStrings(url);
catch(IOException e)
return;
I cannot use the code in the second example because I need to use the variable "lines" later in the program. Is there some other way to catch this exception (504 error)?
Example:
PFont f;
String webpage;
void setup()
size(400, 400);
f = createFont("Arial", 16);
void draw()
background(255);
textFont(f);
fill(0);
text("Press Return to Start", 25, 90);
void keyPressed()
if (key == 'n')
webpage = "processing.org";
loadData(webpage);
void loadData(String webpage)
String url = "www.processing.org";
try
String[] lines = loadStrings(url);
catch(IOException e)
return;
saveStrings("Stuff on Webpage.txt", lines);
This is where the program crashes:
java.io.IOException: Server returned HTTP response code: 504 for URL: http://www.processing.org/
In the event of this exception, I would like the program to exit out of the loadData function and continue with keyPressed.
html url try-catch processing
add a comment |
I am trying to write a program that will read a webpage. The following code is not valid:
String[] lines;
try
lines = loadStrings(url);
catch(IOException e)
return;
But this code is:
try
String[] lines = loadStrings(url);
catch(IOException e)
return;
I cannot use the code in the second example because I need to use the variable "lines" later in the program. Is there some other way to catch this exception (504 error)?
Example:
PFont f;
String webpage;
void setup()
size(400, 400);
f = createFont("Arial", 16);
void draw()
background(255);
textFont(f);
fill(0);
text("Press Return to Start", 25, 90);
void keyPressed()
if (key == 'n')
webpage = "processing.org";
loadData(webpage);
void loadData(String webpage)
String url = "www.processing.org";
try
String[] lines = loadStrings(url);
catch(IOException e)
return;
saveStrings("Stuff on Webpage.txt", lines);
This is where the program crashes:
java.io.IOException: Server returned HTTP response code: 504 for URL: http://www.processing.org/
In the event of this exception, I would like the program to exit out of the loadData function and continue with keyPressed.
html url try-catch processing
Can you please post a Minimal, Complete, and Verifiable example?
– Kevin Workman
Mar 22 at 0:57
I have updated it with some code.
– user11006750
Mar 23 at 2:13
add a comment |
I am trying to write a program that will read a webpage. The following code is not valid:
String[] lines;
try
lines = loadStrings(url);
catch(IOException e)
return;
But this code is:
try
String[] lines = loadStrings(url);
catch(IOException e)
return;
I cannot use the code in the second example because I need to use the variable "lines" later in the program. Is there some other way to catch this exception (504 error)?
Example:
PFont f;
String webpage;
void setup()
size(400, 400);
f = createFont("Arial", 16);
void draw()
background(255);
textFont(f);
fill(0);
text("Press Return to Start", 25, 90);
void keyPressed()
if (key == 'n')
webpage = "processing.org";
loadData(webpage);
void loadData(String webpage)
String url = "www.processing.org";
try
String[] lines = loadStrings(url);
catch(IOException e)
return;
saveStrings("Stuff on Webpage.txt", lines);
This is where the program crashes:
java.io.IOException: Server returned HTTP response code: 504 for URL: http://www.processing.org/
In the event of this exception, I would like the program to exit out of the loadData function and continue with keyPressed.
html url try-catch processing
I am trying to write a program that will read a webpage. The following code is not valid:
String[] lines;
try
lines = loadStrings(url);
catch(IOException e)
return;
But this code is:
try
String[] lines = loadStrings(url);
catch(IOException e)
return;
I cannot use the code in the second example because I need to use the variable "lines" later in the program. Is there some other way to catch this exception (504 error)?
Example:
PFont f;
String webpage;
void setup()
size(400, 400);
f = createFont("Arial", 16);
void draw()
background(255);
textFont(f);
fill(0);
text("Press Return to Start", 25, 90);
void keyPressed()
if (key == 'n')
webpage = "processing.org";
loadData(webpage);
void loadData(String webpage)
String url = "www.processing.org";
try
String[] lines = loadStrings(url);
catch(IOException e)
return;
saveStrings("Stuff on Webpage.txt", lines);
This is where the program crashes:
java.io.IOException: Server returned HTTP response code: 504 for URL: http://www.processing.org/
In the event of this exception, I would like the program to exit out of the loadData function and continue with keyPressed.
html url try-catch processing
html url try-catch processing
edited Mar 23 at 16:10
Kevin Workman
34.4k54273
34.4k54273
asked Mar 22 at 0:54
user11006750user11006750
83
83
Can you please post a Minimal, Complete, and Verifiable example?
– Kevin Workman
Mar 22 at 0:57
I have updated it with some code.
– user11006750
Mar 23 at 2:13
add a comment |
Can you please post a Minimal, Complete, and Verifiable example?
– Kevin Workman
Mar 22 at 0:57
I have updated it with some code.
– user11006750
Mar 23 at 2:13
Can you please post a Minimal, Complete, and Verifiable example?
– Kevin Workman
Mar 22 at 0:57
Can you please post a Minimal, Complete, and Verifiable example?
– Kevin Workman
Mar 22 at 0:57
I have updated it with some code.
– user11006750
Mar 23 at 2:13
I have updated it with some code.
– user11006750
Mar 23 at 2:13
add a comment |
1 Answer
1
active
oldest
votes
Let's look at this code:
String message;
if(random(1) < .5)
message = "hello";
println(message);
This code will generate an error that says message cannot be resolved to a variable
. This is because the compiler is smart enough to know that depending on what the if
statement does, message
might not contain a value. Processing / Java does not assign a default value to local variables.
We could fix this compiler error by assigning a default value ourselves:
String message = null;
if(random(1) < .5)
message = "hello";
println(message);
Now message
has a default value of null
, which matches what happens with sketch / class variables by default.
Another way we could fix this is by putting the println()
call inside the if
statement:
String message;
if(random(1) < .5)
message = "hello";
println(message);
We could simplify this a bit:
if(random(1) < .5)
String message = "hello";
println(message);
Now, back to your code:
try
String[] lines = loadStrings(url);
catch(IOException e)
return;
saveStrings("Stuff on Webpage.txt", lines);
This doesn't work because lines
is declared inside the try
block, so it goes out of scope outside of that block, and you can't use it afterwards.
String[] lines;
try
lines = loadStrings(url);
catch(IOException e)
return;
saveStrings("Stuff on Webpage.txt", lines);
This doesn't work, because the compiler is not smart enough to see that return
statement. It just knows that one branch of the code skips over the lines = loadStrings(url);
line, so it knows that lines
might be unassigned when you try to use it.
You can fix this by either assigning a default value yourself, or moving everything to be inside the same block, just like we did with our simplified example above.
add a comment |
Your Answer
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: "1"
;
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
);
);
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%2fstackoverflow.com%2fquestions%2f55291364%2fprogram-only-throws-exception-when-variable-is-defined-in-try%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
Let's look at this code:
String message;
if(random(1) < .5)
message = "hello";
println(message);
This code will generate an error that says message cannot be resolved to a variable
. This is because the compiler is smart enough to know that depending on what the if
statement does, message
might not contain a value. Processing / Java does not assign a default value to local variables.
We could fix this compiler error by assigning a default value ourselves:
String message = null;
if(random(1) < .5)
message = "hello";
println(message);
Now message
has a default value of null
, which matches what happens with sketch / class variables by default.
Another way we could fix this is by putting the println()
call inside the if
statement:
String message;
if(random(1) < .5)
message = "hello";
println(message);
We could simplify this a bit:
if(random(1) < .5)
String message = "hello";
println(message);
Now, back to your code:
try
String[] lines = loadStrings(url);
catch(IOException e)
return;
saveStrings("Stuff on Webpage.txt", lines);
This doesn't work because lines
is declared inside the try
block, so it goes out of scope outside of that block, and you can't use it afterwards.
String[] lines;
try
lines = loadStrings(url);
catch(IOException e)
return;
saveStrings("Stuff on Webpage.txt", lines);
This doesn't work, because the compiler is not smart enough to see that return
statement. It just knows that one branch of the code skips over the lines = loadStrings(url);
line, so it knows that lines
might be unassigned when you try to use it.
You can fix this by either assigning a default value yourself, or moving everything to be inside the same block, just like we did with our simplified example above.
add a comment |
Let's look at this code:
String message;
if(random(1) < .5)
message = "hello";
println(message);
This code will generate an error that says message cannot be resolved to a variable
. This is because the compiler is smart enough to know that depending on what the if
statement does, message
might not contain a value. Processing / Java does not assign a default value to local variables.
We could fix this compiler error by assigning a default value ourselves:
String message = null;
if(random(1) < .5)
message = "hello";
println(message);
Now message
has a default value of null
, which matches what happens with sketch / class variables by default.
Another way we could fix this is by putting the println()
call inside the if
statement:
String message;
if(random(1) < .5)
message = "hello";
println(message);
We could simplify this a bit:
if(random(1) < .5)
String message = "hello";
println(message);
Now, back to your code:
try
String[] lines = loadStrings(url);
catch(IOException e)
return;
saveStrings("Stuff on Webpage.txt", lines);
This doesn't work because lines
is declared inside the try
block, so it goes out of scope outside of that block, and you can't use it afterwards.
String[] lines;
try
lines = loadStrings(url);
catch(IOException e)
return;
saveStrings("Stuff on Webpage.txt", lines);
This doesn't work, because the compiler is not smart enough to see that return
statement. It just knows that one branch of the code skips over the lines = loadStrings(url);
line, so it knows that lines
might be unassigned when you try to use it.
You can fix this by either assigning a default value yourself, or moving everything to be inside the same block, just like we did with our simplified example above.
add a comment |
Let's look at this code:
String message;
if(random(1) < .5)
message = "hello";
println(message);
This code will generate an error that says message cannot be resolved to a variable
. This is because the compiler is smart enough to know that depending on what the if
statement does, message
might not contain a value. Processing / Java does not assign a default value to local variables.
We could fix this compiler error by assigning a default value ourselves:
String message = null;
if(random(1) < .5)
message = "hello";
println(message);
Now message
has a default value of null
, which matches what happens with sketch / class variables by default.
Another way we could fix this is by putting the println()
call inside the if
statement:
String message;
if(random(1) < .5)
message = "hello";
println(message);
We could simplify this a bit:
if(random(1) < .5)
String message = "hello";
println(message);
Now, back to your code:
try
String[] lines = loadStrings(url);
catch(IOException e)
return;
saveStrings("Stuff on Webpage.txt", lines);
This doesn't work because lines
is declared inside the try
block, so it goes out of scope outside of that block, and you can't use it afterwards.
String[] lines;
try
lines = loadStrings(url);
catch(IOException e)
return;
saveStrings("Stuff on Webpage.txt", lines);
This doesn't work, because the compiler is not smart enough to see that return
statement. It just knows that one branch of the code skips over the lines = loadStrings(url);
line, so it knows that lines
might be unassigned when you try to use it.
You can fix this by either assigning a default value yourself, or moving everything to be inside the same block, just like we did with our simplified example above.
Let's look at this code:
String message;
if(random(1) < .5)
message = "hello";
println(message);
This code will generate an error that says message cannot be resolved to a variable
. This is because the compiler is smart enough to know that depending on what the if
statement does, message
might not contain a value. Processing / Java does not assign a default value to local variables.
We could fix this compiler error by assigning a default value ourselves:
String message = null;
if(random(1) < .5)
message = "hello";
println(message);
Now message
has a default value of null
, which matches what happens with sketch / class variables by default.
Another way we could fix this is by putting the println()
call inside the if
statement:
String message;
if(random(1) < .5)
message = "hello";
println(message);
We could simplify this a bit:
if(random(1) < .5)
String message = "hello";
println(message);
Now, back to your code:
try
String[] lines = loadStrings(url);
catch(IOException e)
return;
saveStrings("Stuff on Webpage.txt", lines);
This doesn't work because lines
is declared inside the try
block, so it goes out of scope outside of that block, and you can't use it afterwards.
String[] lines;
try
lines = loadStrings(url);
catch(IOException e)
return;
saveStrings("Stuff on Webpage.txt", lines);
This doesn't work, because the compiler is not smart enough to see that return
statement. It just knows that one branch of the code skips over the lines = loadStrings(url);
line, so it knows that lines
might be unassigned when you try to use it.
You can fix this by either assigning a default value yourself, or moving everything to be inside the same block, just like we did with our simplified example above.
answered Mar 23 at 16:09
Kevin WorkmanKevin Workman
34.4k54273
34.4k54273
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- 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.
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%2fstackoverflow.com%2fquestions%2f55291364%2fprogram-only-throws-exception-when-variable-is-defined-in-try%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
Can you please post a Minimal, Complete, and Verifiable example?
– Kevin Workman
Mar 22 at 0:57
I have updated it with some code.
– user11006750
Mar 23 at 2:13