How to run tasks at a scheduled rate that DON'T wait for the task before it?
1
Currently, I have some code that I need to run every (for example) 33 milliseconds. However, the operation that I am calling requires around 270ms. Is there a way to schedule my tasks so that they run regardless of the task before them?
I have tried implementing a ScheduledExecutorService variable and running the task at a "ScheduledFixedRate" but that currently waits for the task before it.
Runnable imageCapture = new Runnable() {
public void run() {
// code that takes approximately 270ms
}
};
executor = Executors.newScheduledThreadPool(4);
executor.scheduleAtFixedRate(imageCapture, 0, 33, TimeUnit.MILLISECONDS);
java concurrency scheduledexecutorservice
asked Dec 31 '18 at 12:37
TylerTyler
133
add a comment |
1
Currently, I have some code that I need to run every (for example) 33 milliseconds. However, the operation that I am calling requires around 270ms. Is there a way to schedule my tasks so that they run regardless of the task before them?
I have tried implementing a ScheduledExecutorService variable and running the task at a "ScheduledFixedRate" but that currently waits for the task before it.
Runnable imageCapture = new Runnable() {
public void run() {
// code that takes approximately 270ms
}
};
executor = Executors.newScheduledThreadPool(4);
executor.scheduleAtFixedRate(imageCapture, 0, 33, TimeUnit.MILLISECONDS);
java concurrency scheduledexecutorservice
asked Dec 31 '18 at 12:37
TylerTyler
133
If you want it to run that frequently, you'll need more threads.
– Andy Turner
Dec 31 '18 at 12:40
Is there like a formula to use, I kind of just guessed on the number. Do you think adding more threads will solve my issue? @AndyTurner
– Tyler
Dec 31 '18 at 12:45
1
Multiply the time for one action to complete by the frequency. Equivalently, divide the time for one action by the time between two actions starting. 270/33=8 and a bit. So you'd need at least 9 threads.
– Andy Turner
Dec 31 '18 at 12:54
But that won't work with the code you have anyway, since the javadoc says "If any execution of this task takes longer than its period, then subsequent executions may start late, but will not concurrently execute (emphasis mine).
– JB Nizet
Dec 31 '18 at 13:05
@AndyTurner not only 9 threads, but also 9 processors (cores).
– Alexei Kaigorodov
Dec 31 '18 at 19:47
add a comment |
1
1
1
Currently, I have some code that I need to run every (for example) 33 milliseconds. However, the operation that I am calling requires around 270ms. Is there a way to schedule my tasks so that they run regardless of the task before them?
I have tried implementing a ScheduledExecutorService variable and running the task at a "ScheduledFixedRate" but that currently waits for the task before it.
Runnable imageCapture = new Runnable() {
public void run() {
// code that takes approximately 270ms
}
};
executor = Executors.newScheduledThreadPool(4);
executor.scheduleAtFixedRate(imageCapture, 0, 33, TimeUnit.MILLISECONDS);
java concurrency scheduledexecutorservice
asked Dec 31 '18 at 12:37
TylerTyler
133
Currently, I have some code that I need to run every (for example) 33 milliseconds. However, the operation that I am calling requires around 270ms. Is there a way to schedule my tasks so that they run regardless of the task before them?
I have tried implementing a ScheduledExecutorService variable and running the task at a "ScheduledFixedRate" but that currently waits for the task before it.
Runnable imageCapture = new Runnable() {
public void run() {
// code that takes approximately 270ms
}
};
executor = Executors.newScheduledThreadPool(4);
executor.scheduleAtFixedRate(imageCapture, 0, 33, TimeUnit.MILLISECONDS);
java concurrency scheduledexecutorservice
java concurrency scheduledexecutorservice
asked Dec 31 '18 at 12:37
TylerTyler
133
asked Dec 31 '18 at 12:37
TylerTyler
133
asked Dec 31 '18 at 12:37
TylerTyler
133
asked Dec 31 '18 at 12:37
TylerTyler
133
asked Dec 31 '18 at 12:37
TylerTyler
133
133
If you want it to run that frequently, you'll need more threads.
– Andy Turner
Dec 31 '18 at 12:40
Is there like a formula to use, I kind of just guessed on the number. Do you think adding more threads will solve my issue? @AndyTurner
– Tyler
Dec 31 '18 at 12:45
1
Multiply the time for one action to complete by the frequency. Equivalently, divide the time for one action by the time between two actions starting. 270/33=8 and a bit. So you'd need at least 9 threads.
– Andy Turner
Dec 31 '18 at 12:54
But that won't work with the code you have anyway, since the javadoc says "If any execution of this task takes longer than its period, then subsequent executions may start late, but will not concurrently execute (emphasis mine).
– JB Nizet
Dec 31 '18 at 13:05
@AndyTurner not only 9 threads, but also 9 processors (cores).
– Alexei Kaigorodov
Dec 31 '18 at 19:47
add a comment |
If you want it to run that frequently, you'll need more threads.
– Andy Turner
Dec 31 '18 at 12:40
Is there like a formula to use, I kind of just guessed on the number. Do you think adding more threads will solve my issue? @AndyTurner
– Tyler
Dec 31 '18 at 12:45
1
Multiply the time for one action to complete by the frequency. Equivalently, divide the time for one action by the time between two actions starting. 270/33=8 and a bit. So you'd need at least 9 threads.
– Andy Turner
Dec 31 '18 at 12:54
But that won't work with the code you have anyway, since the javadoc says "If any execution of this task takes longer than its period, then subsequent executions may start late, but will not concurrently execute (emphasis mine).
– JB Nizet
Dec 31 '18 at 13:05
@AndyTurner not only 9 threads, but also 9 processors (cores).
– Alexei Kaigorodov
Dec 31 '18 at 19:47
If you want it to run that frequently, you'll need more threads.
– Andy Turner
Dec 31 '18 at 12:40
If you want it to run that frequently, you'll need more threads.
– Andy Turner
Dec 31 '18 at 12:40
Is there like a formula to use, I kind of just guessed on the number. Do you think adding more threads will solve my issue? @AndyTurner
– Tyler
Dec 31 '18 at 12:45
Is there like a formula to use, I kind of just guessed on the number. Do you think adding more threads will solve my issue? @AndyTurner
– Tyler
Dec 31 '18 at 12:45
1
1
Multiply the time for one action to complete by the frequency. Equivalently, divide the time for one action by the time between two actions starting. 270/33=8 and a bit. So you'd need at least 9 threads.
– Andy Turner
Dec 31 '18 at 12:54
Multiply the time for one action to complete by the frequency. Equivalently, divide the time for one action by the time between two actions starting. 270/33=8 and a bit. So you'd need at least 9 threads.
– Andy Turner
Dec 31 '18 at 12:54
But that won't work with the code you have anyway, since the javadoc says "If any execution of this task takes longer than its period, then subsequent executions may start late, but will not concurrently execute (emphasis mine).
– JB Nizet
Dec 31 '18 at 13:05
But that won't work with the code you have anyway, since the javadoc says "If any execution of this task takes longer than its period, then subsequent executions may start late, but will not concurrently execute (emphasis mine).
– JB Nizet
Dec 31 '18 at 13:05
@AndyTurner not only 9 threads, but also 9 processors (cores).
– Alexei Kaigorodov
Dec 31 '18 at 19:47
@AndyTurner not only 9 threads, but also 9 processors (cores).
– Alexei Kaigorodov
Dec 31 '18 at 19:47
add a comment |
1 Answer
1
active
oldest
votes
0
Split the task in two: one makes actual computations and another is executed periodically and starts the first one:
executor = Executors.newScheduledThreadPool(4);
Runnable imageCapture = new Runnable() {
public void run() {
// code that takes approximately 270ms
}
};
Runnable launcher = new Runnable() {
public void run() {
executor.execute(imageCapture);
}
};
executor.scheduleAtFixedRate(launcher, 0, 33, TimeUnit.MILLISECONDS);
answered Dec 31 '18 at 15:14
Alexei KaigorodovAlexei Kaigorodov
10.1k11029
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
});
}
});
draft saved
draft discarded
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%2f53987629%2fhow-to-run-tasks-at-a-scheduled-rate-that-dont-wait-for-the-task-before-it%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
0
Split the task in two: one makes actual computations and another is executed periodically and starts the first one:
executor = Executors.newScheduledThreadPool(4);
Runnable imageCapture = new Runnable() {
public void run() {
// code that takes approximately 270ms
}
};
Runnable launcher = new Runnable() {
public void run() {
executor.execute(imageCapture);
}
};
executor.scheduleAtFixedRate(launcher, 0, 33, TimeUnit.MILLISECONDS);
answered Dec 31 '18 at 15:14
Alexei KaigorodovAlexei Kaigorodov
10.1k11029
add a comment |
0
Split the task in two: one makes actual computations and another is executed periodically and starts the first one:
executor = Executors.newScheduledThreadPool(4);
Runnable imageCapture = new Runnable() {
public void run() {
// code that takes approximately 270ms
}
};
Runnable launcher = new Runnable() {
public void run() {
executor.execute(imageCapture);
}
};
executor.scheduleAtFixedRate(launcher, 0, 33, TimeUnit.MILLISECONDS);
answered Dec 31 '18 at 15:14
Alexei KaigorodovAlexei Kaigorodov
10.1k11029
add a comment |
0
0
0
Split the task in two: one makes actual computations and another is executed periodically and starts the first one:
executor = Executors.newScheduledThreadPool(4);
Runnable imageCapture = new Runnable() {
public void run() {
// code that takes approximately 270ms
}
};
Runnable launcher = new Runnable() {
public void run() {
executor.execute(imageCapture);
}
};
executor.scheduleAtFixedRate(launcher, 0, 33, TimeUnit.MILLISECONDS);
answered Dec 31 '18 at 15:14
Alexei KaigorodovAlexei Kaigorodov
10.1k11029
Split the task in two: one makes actual computations and another is executed periodically and starts the first one:
executor = Executors.newScheduledThreadPool(4);
Runnable imageCapture = new Runnable() {
public void run() {
// code that takes approximately 270ms
}
};
Runnable launcher = new Runnable() {
public void run() {
executor.execute(imageCapture);
}
};
executor.scheduleAtFixedRate(launcher, 0, 33, TimeUnit.MILLISECONDS);
answered Dec 31 '18 at 15:14
Alexei KaigorodovAlexei Kaigorodov
10.1k11029
edited Dec 31 '18 at 19:44
answered Dec 31 '18 at 15:14
Alexei KaigorodovAlexei Kaigorodov
10.1k11029
answered Dec 31 '18 at 15:14
Alexei KaigorodovAlexei Kaigorodov
10.1k11029
answered Dec 31 '18 at 15:14
Alexei KaigorodovAlexei Kaigorodov
10.1k11029
10.1k11029
add a comment |
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f53987629%2fhow-to-run-tasks-at-a-scheduled-rate-that-dont-wait-for-the-task-before-it%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
If you want it to run that frequently, you'll need more threads.
– Andy Turner
Dec 31 '18 at 12:40
Is there like a formula to use, I kind of just guessed on the number. Do you think adding more threads will solve my issue? @AndyTurner
– Tyler
Dec 31 '18 at 12:45
1
Multiply the time for one action to complete by the frequency. Equivalently, divide the time for one action by the time between two actions starting. 270/33=8 and a bit. So you'd need at least 9 threads.
– Andy Turner
Dec 31 '18 at 12:54
But that won't work with the code you have anyway, since the javadoc says "If any execution of this task takes longer than its period, then subsequent executions may start late, but will not concurrently execute (emphasis mine).
– JB Nizet
Dec 31 '18 at 13:05
@AndyTurner not only 9 threads, but also 9 processors (cores).
– Alexei Kaigorodov
Dec 31 '18 at 19:47