How do I get a Duration as a number of milliseconds in Rust

I have a time::Duration. How can I get the number of milliseconds represented by this duration as an integer? There used to be a num_milliseconds() function, but it is no longer available.

edited Apr 23 '16 at 20:57

Shepmaster

157k14316457

asked Apr 23 '16 at 20:32

Kevin Burke

20.6k45132236

In addition to the other answers, keep an eye on this RFC issue for any future RFCs regarding this.

– robinst
Jan 4 '17 at 1:34

use chrono exercism.io/submissions/1264089798d244e5b8278e588919901c

– rofrol
May 18 '18 at 14:04

See also How can I get the current time in milliseconds?

– Shepmaster
Dec 31 '18 at 14:28

add a comment |

I have a time::Duration. How can I get the number of milliseconds represented by this duration as an integer? There used to be a num_milliseconds() function, but it is no longer available.

edited Apr 23 '16 at 20:57

Shepmaster

157k14316457

asked Apr 23 '16 at 20:32

Kevin Burke

20.6k45132236

In addition to the other answers, keep an eye on this RFC issue for any future RFCs regarding this.

– robinst
Jan 4 '17 at 1:34

use chrono exercism.io/submissions/1264089798d244e5b8278e588919901c

– rofrol
May 18 '18 at 14:04

See also How can I get the current time in milliseconds?

– Shepmaster
Dec 31 '18 at 14:28

add a comment |

I have a time::Duration. How can I get the number of milliseconds represented by this duration as an integer? There used to be a num_milliseconds() function, but it is no longer available.

edited Apr 23 '16 at 20:57

Shepmaster

157k14316457

asked Apr 23 '16 at 20:32

Kevin Burke

20.6k45132236

I have a time::Duration. How can I get the number of milliseconds represented by this duration as an integer? There used to be a num_milliseconds() function, but it is no longer available.

rust

edited Apr 23 '16 at 20:57

Shepmaster

157k14316457

asked Apr 23 '16 at 20:32

Kevin Burke

20.6k45132236

edited Apr 23 '16 at 20:57

Shepmaster

157k14316457

asked Apr 23 '16 at 20:32

Kevin Burke

20.6k45132236

edited Apr 23 '16 at 20:57

Shepmaster

157k14316457

edited Apr 23 '16 at 20:57

Shepmaster

157k14316457

edited Apr 23 '16 at 20:57

Shepmaster

157k14316457

asked Apr 23 '16 at 20:32

Kevin Burke

20.6k45132236

asked Apr 23 '16 at 20:32

Kevin Burke

20.6k45132236

asked Apr 23 '16 at 20:32

Kevin Burke

20.6k45132236

In addition to the other answers, keep an eye on this RFC issue for any future RFCs regarding this.

– robinst
Jan 4 '17 at 1:34

use chrono exercism.io/submissions/1264089798d244e5b8278e588919901c

– rofrol
May 18 '18 at 14:04

See also How can I get the current time in milliseconds?

– Shepmaster
Dec 31 '18 at 14:28

add a comment |

In addition to the other answers, keep an eye on this RFC issue for any future RFCs regarding this.

– robinst
Jan 4 '17 at 1:34

use chrono exercism.io/submissions/1264089798d244e5b8278e588919901c

– rofrol
May 18 '18 at 14:04

See also How can I get the current time in milliseconds?

– Shepmaster
Dec 31 '18 at 14:28

In addition to the other answers, keep an eye on this RFC issue for any future RFCs regarding this.

– robinst
Jan 4 '17 at 1:34

use chrono exercism.io/submissions/1264089798d244e5b8278e588919901c

– rofrol
May 18 '18 at 14:04

See also How can I get the current time in milliseconds?

– Shepmaster
Dec 31 '18 at 14:28

add a comment |

3 Answers
3

active

oldest

votes

Here is the solution I came up with, which is to multiply the seconds by a billion, add it to the nanoseconds, then divide by 1e6.

let nanos = timeout_duration.subsec_nanos() as u64;

let ms = (1000*1000*1000 * timeout_duration.as_secs() + nanos)/(1000 * 1000);

answered Apr 23 '16 at 20:50

Kevin Burke

20.6k45132236

7

I'd rather multiply seconds by 1000, then add nanos/1000000. Less risk of overflow.

– llogiq
Apr 24 '16 at 10:22

subsec_nanos doesn't return the number of elapsed nanoseconds. It represents the precision of the Duration.

– w.brian
Mar 13 '17 at 1:15

2

@w.brian the example in the documentation doc.rust-lang.org/std/time/… supports the facts that the method returns the fractional part of the duration in nanoseconds, so the answer seems correct to me.

– poros
Jan 30 '18 at 18:32

add a comment |

Use time::Duration from the time crate on crates.io which provides a num_milliseconds() method.

edited Apr 23 '16 at 21:01

answered Apr 23 '16 at 20:58

aeveris

335

add a comment |

Since Rust 1.33.0, there is an as_millis() function:

use std::time::SystemTime;



fn main() {

    let now = SystemTime::now().duration_since(SystemTime::UNIX_EPOCH).expect("get millis error");

    println!("now millis: {}", now.as_millis());

}

Since Rust 1.27.0, there is a subsec_millis() function:

use std::time::SystemTime;



fn main() {

    let since_the_epoch = SystemTime::now().duration_since(SystemTime::UNIX_EPOCH).expect("get millis error");

    let seconds = since_the_epoch.as_secs();

    let subsec_millis = since_the_epoch.subsec_millis() as u64;

    println!("now millis: {}", seconds * 1000 + subsec_millis);

}

Since Rust 1.8, there is a subsec_nanos function:

let in_ms = since_the_epoch.as_secs() * 1000 +

            since_the_epoch.subsec_nanos() as u64 / 1_000_000;

See also:

How can I get the current time in milliseconds?

edited Jan 1 at 16:39

Shepmaster

157k14316457

answered Dec 31 '18 at 10:49

iceblue

8916

What does Rust 2018 have to do with the problem? Why are you using SystemTime instead of directly using a Duration?

– Shepmaster
Dec 31 '18 at 14:25

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

Here is the solution I came up with, which is to multiply the seconds by a billion, add it to the nanoseconds, then divide by 1e6.

let nanos = timeout_duration.subsec_nanos() as u64;

let ms = (1000*1000*1000 * timeout_duration.as_secs() + nanos)/(1000 * 1000);

answered Apr 23 '16 at 20:50

Kevin Burke

20.6k45132236

7

I'd rather multiply seconds by 1000, then add nanos/1000000. Less risk of overflow.

– llogiq
Apr 24 '16 at 10:22

subsec_nanos doesn't return the number of elapsed nanoseconds. It represents the precision of the Duration.

– w.brian
Mar 13 '17 at 1:15

2

@w.brian the example in the documentation doc.rust-lang.org/std/time/… supports the facts that the method returns the fractional part of the duration in nanoseconds, so the answer seems correct to me.

– poros
Jan 30 '18 at 18:32

add a comment |

Here is the solution I came up with, which is to multiply the seconds by a billion, add it to the nanoseconds, then divide by 1e6.

let nanos = timeout_duration.subsec_nanos() as u64;

let ms = (1000*1000*1000 * timeout_duration.as_secs() + nanos)/(1000 * 1000);

answered Apr 23 '16 at 20:50

Kevin Burke

20.6k45132236

7

I'd rather multiply seconds by 1000, then add nanos/1000000. Less risk of overflow.

– llogiq
Apr 24 '16 at 10:22

subsec_nanos doesn't return the number of elapsed nanoseconds. It represents the precision of the Duration.

– w.brian
Mar 13 '17 at 1:15

2

@w.brian the example in the documentation doc.rust-lang.org/std/time/… supports the facts that the method returns the fractional part of the duration in nanoseconds, so the answer seems correct to me.

– poros
Jan 30 '18 at 18:32

add a comment |

Here is the solution I came up with, which is to multiply the seconds by a billion, add it to the nanoseconds, then divide by 1e6.

let nanos = timeout_duration.subsec_nanos() as u64;

let ms = (1000*1000*1000 * timeout_duration.as_secs() + nanos)/(1000 * 1000);

answered Apr 23 '16 at 20:50

Kevin Burke

20.6k45132236

Here is the solution I came up with, which is to multiply the seconds by a billion, add it to the nanoseconds, then divide by 1e6.

let nanos = timeout_duration.subsec_nanos() as u64;

let ms = (1000*1000*1000 * timeout_duration.as_secs() + nanos)/(1000 * 1000);

answered Apr 23 '16 at 20:50

Kevin Burke

20.6k45132236

answered Apr 23 '16 at 20:50

Kevin Burke

20.6k45132236

answered Apr 23 '16 at 20:50

Kevin Burke

20.6k45132236

answered Apr 23 '16 at 20:50

Kevin Burke

20.6k45132236

7

I'd rather multiply seconds by 1000, then add nanos/1000000. Less risk of overflow.

– llogiq
Apr 24 '16 at 10:22

subsec_nanos doesn't return the number of elapsed nanoseconds. It represents the precision of the Duration.

– w.brian
Mar 13 '17 at 1:15

2

@w.brian the example in the documentation doc.rust-lang.org/std/time/… supports the facts that the method returns the fractional part of the duration in nanoseconds, so the answer seems correct to me.

– poros
Jan 30 '18 at 18:32

add a comment |

7

I'd rather multiply seconds by 1000, then add nanos/1000000. Less risk of overflow.

– llogiq
Apr 24 '16 at 10:22

subsec_nanos doesn't return the number of elapsed nanoseconds. It represents the precision of the Duration.

– w.brian
Mar 13 '17 at 1:15

2

@w.brian the example in the documentation doc.rust-lang.org/std/time/… supports the facts that the method returns the fractional part of the duration in nanoseconds, so the answer seems correct to me.

– poros
Jan 30 '18 at 18:32

I'd rather multiply seconds by 1000, then add nanos/1000000. Less risk of overflow.

– llogiq
Apr 24 '16 at 10:22

subsec_nanos doesn't return the number of elapsed nanoseconds. It represents the precision of the Duration.

– w.brian
Mar 13 '17 at 1:15

@w.brian the example in the documentation doc.rust-lang.org/std/time/… supports the facts that the method returns the fractional part of the duration in nanoseconds, so the answer seems correct to me.

– poros
Jan 30 '18 at 18:32

add a comment |

Use time::Duration from the time crate on crates.io which provides a num_milliseconds() method.

edited Apr 23 '16 at 21:01

answered Apr 23 '16 at 20:58

aeveris

335

add a comment |

Use time::Duration from the time crate on crates.io which provides a num_milliseconds() method.

edited Apr 23 '16 at 21:01

answered Apr 23 '16 at 20:58

aeveris

335

add a comment |

Use time::Duration from the time crate on crates.io which provides a num_milliseconds() method.

edited Apr 23 '16 at 21:01

answered Apr 23 '16 at 20:58

aeveris

335

Use time::Duration from the time crate on crates.io which provides a num_milliseconds() method.

edited Apr 23 '16 at 21:01

answered Apr 23 '16 at 20:58

aeveris

335

edited Apr 23 '16 at 21:01

answered Apr 23 '16 at 20:58

aeveris

335

answered Apr 23 '16 at 20:58

aeveris

335

answered Apr 23 '16 at 20:58

aeveris

335

add a comment |

Since Rust 1.33.0, there is an as_millis() function:

use std::time::SystemTime;



fn main() {

    let now = SystemTime::now().duration_since(SystemTime::UNIX_EPOCH).expect("get millis error");

    println!("now millis: {}", now.as_millis());

}

Since Rust 1.27.0, there is a subsec_millis() function:

use std::time::SystemTime;



fn main() {

    let since_the_epoch = SystemTime::now().duration_since(SystemTime::UNIX_EPOCH).expect("get millis error");

    let seconds = since_the_epoch.as_secs();

    let subsec_millis = since_the_epoch.subsec_millis() as u64;

    println!("now millis: {}", seconds * 1000 + subsec_millis);

}

Since Rust 1.8, there is a subsec_nanos function:

let in_ms = since_the_epoch.as_secs() * 1000 +

            since_the_epoch.subsec_nanos() as u64 / 1_000_000;

See also:

How can I get the current time in milliseconds?

edited Jan 1 at 16:39

Shepmaster

157k14316457

answered Dec 31 '18 at 10:49

iceblue

8916

What does Rust 2018 have to do with the problem? Why are you using SystemTime instead of directly using a Duration?

– Shepmaster
Dec 31 '18 at 14:25

add a comment |

Since Rust 1.33.0, there is an as_millis() function:

use std::time::SystemTime;



fn main() {

    let now = SystemTime::now().duration_since(SystemTime::UNIX_EPOCH).expect("get millis error");

    println!("now millis: {}", now.as_millis());

}

Since Rust 1.27.0, there is a subsec_millis() function:

use std::time::SystemTime;



fn main() {

    let since_the_epoch = SystemTime::now().duration_since(SystemTime::UNIX_EPOCH).expect("get millis error");

    let seconds = since_the_epoch.as_secs();

    let subsec_millis = since_the_epoch.subsec_millis() as u64;

    println!("now millis: {}", seconds * 1000 + subsec_millis);

}

Since Rust 1.8, there is a subsec_nanos function:

let in_ms = since_the_epoch.as_secs() * 1000 +

            since_the_epoch.subsec_nanos() as u64 / 1_000_000;

See also:

How can I get the current time in milliseconds?

edited Jan 1 at 16:39

Shepmaster

157k14316457

answered Dec 31 '18 at 10:49

iceblue

8916

What does Rust 2018 have to do with the problem? Why are you using SystemTime instead of directly using a Duration?

– Shepmaster
Dec 31 '18 at 14:25

add a comment |

Since Rust 1.33.0, there is an as_millis() function:

use std::time::SystemTime;



fn main() {

    let now = SystemTime::now().duration_since(SystemTime::UNIX_EPOCH).expect("get millis error");

    println!("now millis: {}", now.as_millis());

}

Since Rust 1.27.0, there is a subsec_millis() function:

use std::time::SystemTime;



fn main() {

    let since_the_epoch = SystemTime::now().duration_since(SystemTime::UNIX_EPOCH).expect("get millis error");

    let seconds = since_the_epoch.as_secs();

    let subsec_millis = since_the_epoch.subsec_millis() as u64;

    println!("now millis: {}", seconds * 1000 + subsec_millis);

}

Since Rust 1.8, there is a subsec_nanos function:

let in_ms = since_the_epoch.as_secs() * 1000 +

            since_the_epoch.subsec_nanos() as u64 / 1_000_000;

See also:

How can I get the current time in milliseconds?

edited Jan 1 at 16:39

Shepmaster

157k14316457

answered Dec 31 '18 at 10:49

iceblue

8916

Since Rust 1.33.0, there is an as_millis() function:

use std::time::SystemTime;



fn main() {

    let now = SystemTime::now().duration_since(SystemTime::UNIX_EPOCH).expect("get millis error");

    println!("now millis: {}", now.as_millis());

}

Since Rust 1.27.0, there is a subsec_millis() function:

use std::time::SystemTime;



fn main() {

    let since_the_epoch = SystemTime::now().duration_since(SystemTime::UNIX_EPOCH).expect("get millis error");

    let seconds = since_the_epoch.as_secs();

    let subsec_millis = since_the_epoch.subsec_millis() as u64;

    println!("now millis: {}", seconds * 1000 + subsec_millis);

}

Since Rust 1.8, there is a subsec_nanos function:

let in_ms = since_the_epoch.as_secs() * 1000 +

            since_the_epoch.subsec_nanos() as u64 / 1_000_000;

See also:

How can I get the current time in milliseconds?

edited Jan 1 at 16:39

Shepmaster

157k14316457

answered Dec 31 '18 at 10:49

iceblue

8916

edited Jan 1 at 16:39

Shepmaster

157k14316457

edited Jan 1 at 16:39

Shepmaster

157k14316457

edited Jan 1 at 16:39

Shepmaster

157k14316457

answered Dec 31 '18 at 10:49

iceblue

8916

answered Dec 31 '18 at 10:49

iceblue

8916

answered Dec 31 '18 at 10:49

iceblue

8916

What does Rust 2018 have to do with the problem? Why are you using SystemTime instead of directly using a Duration?

– Shepmaster
Dec 31 '18 at 14:25

add a comment |

What does Rust 2018 have to do with the problem? Why are you using SystemTime instead of directly using a Duration?

– Shepmaster
Dec 31 '18 at 14:25

What does Rust 2018 have to do with the problem? Why are you using SystemTime instead of directly using a Duration?

– Shepmaster
Dec 31 '18 at 14:25

add a comment |

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