How do I convert the “largest value in a Vec” example in the Rust book to not use the Copy trait?









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I'm trying to accomplish an exercise "left to the reader" in the 2018 Rust book. The example they have, 10-15, uses the Copy trait. However, they recommend implementing the same without Copy and I've been really struggling with it.



Without Copy, I cannot use largest = list[0]. The compiler recommends using a reference instead. I do so, making largest into a &T. The compiler then complains that the largest used in the comparison is a &T, not T, so I change it to *largest to dereference the pointer. This goes fine, but then stumbles on largest = item, with complaints about T instead of &T. I switch to largest = &item. Then I get an error I cannot deal with:



error[E0597]: `item` does not live long enough
--> src/main.rs:6:24
|
6 | largest = &item;
| ^^^^ borrowed value does not live long enough
7 | }
8 | }
| - borrowed value only lives until here
|
note: borrowed value must be valid for the anonymous lifetime #1 defined on the function body at 1:1...


I do not understand how to lengthen the life of this value. It lives and dies in the list.iter(). How can I extend it while still only using references?



Here is my code for reference:



fn largest<T: PartialOrd>(list: &[T]) -> &T 
let mut largest = &list[0];

for &item in list.iter()
if item > *largest
largest = &item;



largest










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    up vote
    0
    down vote

    favorite












    I'm trying to accomplish an exercise "left to the reader" in the 2018 Rust book. The example they have, 10-15, uses the Copy trait. However, they recommend implementing the same without Copy and I've been really struggling with it.



    Without Copy, I cannot use largest = list[0]. The compiler recommends using a reference instead. I do so, making largest into a &T. The compiler then complains that the largest used in the comparison is a &T, not T, so I change it to *largest to dereference the pointer. This goes fine, but then stumbles on largest = item, with complaints about T instead of &T. I switch to largest = &item. Then I get an error I cannot deal with:



    error[E0597]: `item` does not live long enough
    --> src/main.rs:6:24
    |
    6 | largest = &item;
    | ^^^^ borrowed value does not live long enough
    7 | }
    8 | }
    | - borrowed value only lives until here
    |
    note: borrowed value must be valid for the anonymous lifetime #1 defined on the function body at 1:1...


    I do not understand how to lengthen the life of this value. It lives and dies in the list.iter(). How can I extend it while still only using references?



    Here is my code for reference:



    fn largest<T: PartialOrd>(list: &[T]) -> &T 
    let mut largest = &list[0];

    for &item in list.iter()
    if item > *largest
    largest = &item;



    largest










    share|improve this question

























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I'm trying to accomplish an exercise "left to the reader" in the 2018 Rust book. The example they have, 10-15, uses the Copy trait. However, they recommend implementing the same without Copy and I've been really struggling with it.



      Without Copy, I cannot use largest = list[0]. The compiler recommends using a reference instead. I do so, making largest into a &T. The compiler then complains that the largest used in the comparison is a &T, not T, so I change it to *largest to dereference the pointer. This goes fine, but then stumbles on largest = item, with complaints about T instead of &T. I switch to largest = &item. Then I get an error I cannot deal with:



      error[E0597]: `item` does not live long enough
      --> src/main.rs:6:24
      |
      6 | largest = &item;
      | ^^^^ borrowed value does not live long enough
      7 | }
      8 | }
      | - borrowed value only lives until here
      |
      note: borrowed value must be valid for the anonymous lifetime #1 defined on the function body at 1:1...


      I do not understand how to lengthen the life of this value. It lives and dies in the list.iter(). How can I extend it while still only using references?



      Here is my code for reference:



      fn largest<T: PartialOrd>(list: &[T]) -> &T 
      let mut largest = &list[0];

      for &item in list.iter()
      if item > *largest
      largest = &item;



      largest










      share|improve this question















      I'm trying to accomplish an exercise "left to the reader" in the 2018 Rust book. The example they have, 10-15, uses the Copy trait. However, they recommend implementing the same without Copy and I've been really struggling with it.



      Without Copy, I cannot use largest = list[0]. The compiler recommends using a reference instead. I do so, making largest into a &T. The compiler then complains that the largest used in the comparison is a &T, not T, so I change it to *largest to dereference the pointer. This goes fine, but then stumbles on largest = item, with complaints about T instead of &T. I switch to largest = &item. Then I get an error I cannot deal with:



      error[E0597]: `item` does not live long enough
      --> src/main.rs:6:24
      |
      6 | largest = &item;
      | ^^^^ borrowed value does not live long enough
      7 | }
      8 | }
      | - borrowed value only lives until here
      |
      note: borrowed value must be valid for the anonymous lifetime #1 defined on the function body at 1:1...


      I do not understand how to lengthen the life of this value. It lives and dies in the list.iter(). How can I extend it while still only using references?



      Here is my code for reference:



      fn largest<T: PartialOrd>(list: &[T]) -> &T 
      let mut largest = &list[0];

      for &item in list.iter()
      if item > *largest
      largest = &item;



      largest







      vector rust






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      edited Nov 12 at 1:36









      Shepmaster

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      146k11281413










      asked Nov 12 at 0:14









      WarSame

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          1 Answer
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          accepted










          When you write for &item, this destructures each reference returned by the iterator, making the type of item T. You don't want to destructure these references, you want to keep them! Otherwise, when you take a reference to item, you are taking a reference to a local variable, which you can't return because local variables don't live long enough.



          fn largest<T: PartialOrd>(list: &[T]) -> &T 
          let mut largest = &list[0];

          for item in list.iter()
          if item > largest
          largest = item;



          largest



          Note also how we can compare references directly, because references to types implementing PartialOrd also implement PartialOrd, deferring the comparison to their referents (i.e. it's not a pointer comparison, unlike for raw pointers).






          share|improve this answer
















          • 2




            It's worth noting that this would normally idiomatically be written as list.iter().max().unwrap()
            – Shepmaster
            Nov 12 at 1:37










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          1 Answer
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          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          3
          down vote



          accepted










          When you write for &item, this destructures each reference returned by the iterator, making the type of item T. You don't want to destructure these references, you want to keep them! Otherwise, when you take a reference to item, you are taking a reference to a local variable, which you can't return because local variables don't live long enough.



          fn largest<T: PartialOrd>(list: &[T]) -> &T 
          let mut largest = &list[0];

          for item in list.iter()
          if item > largest
          largest = item;



          largest



          Note also how we can compare references directly, because references to types implementing PartialOrd also implement PartialOrd, deferring the comparison to their referents (i.e. it's not a pointer comparison, unlike for raw pointers).






          share|improve this answer
















          • 2




            It's worth noting that this would normally idiomatically be written as list.iter().max().unwrap()
            – Shepmaster
            Nov 12 at 1:37














          up vote
          3
          down vote



          accepted










          When you write for &item, this destructures each reference returned by the iterator, making the type of item T. You don't want to destructure these references, you want to keep them! Otherwise, when you take a reference to item, you are taking a reference to a local variable, which you can't return because local variables don't live long enough.



          fn largest<T: PartialOrd>(list: &[T]) -> &T 
          let mut largest = &list[0];

          for item in list.iter()
          if item > largest
          largest = item;



          largest



          Note also how we can compare references directly, because references to types implementing PartialOrd also implement PartialOrd, deferring the comparison to their referents (i.e. it's not a pointer comparison, unlike for raw pointers).






          share|improve this answer
















          • 2




            It's worth noting that this would normally idiomatically be written as list.iter().max().unwrap()
            – Shepmaster
            Nov 12 at 1:37












          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          When you write for &item, this destructures each reference returned by the iterator, making the type of item T. You don't want to destructure these references, you want to keep them! Otherwise, when you take a reference to item, you are taking a reference to a local variable, which you can't return because local variables don't live long enough.



          fn largest<T: PartialOrd>(list: &[T]) -> &T 
          let mut largest = &list[0];

          for item in list.iter()
          if item > largest
          largest = item;



          largest



          Note also how we can compare references directly, because references to types implementing PartialOrd also implement PartialOrd, deferring the comparison to their referents (i.e. it's not a pointer comparison, unlike for raw pointers).






          share|improve this answer












          When you write for &item, this destructures each reference returned by the iterator, making the type of item T. You don't want to destructure these references, you want to keep them! Otherwise, when you take a reference to item, you are taking a reference to a local variable, which you can't return because local variables don't live long enough.



          fn largest<T: PartialOrd>(list: &[T]) -> &T 
          let mut largest = &list[0];

          for item in list.iter()
          if item > largest
          largest = item;



          largest



          Note also how we can compare references directly, because references to types implementing PartialOrd also implement PartialOrd, deferring the comparison to their referents (i.e. it's not a pointer comparison, unlike for raw pointers).







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 12 at 0:52









          Francis Gagné

          31.6k26478




          31.6k26478







          • 2




            It's worth noting that this would normally idiomatically be written as list.iter().max().unwrap()
            – Shepmaster
            Nov 12 at 1:37












          • 2




            It's worth noting that this would normally idiomatically be written as list.iter().max().unwrap()
            – Shepmaster
            Nov 12 at 1:37







          2




          2




          It's worth noting that this would normally idiomatically be written as list.iter().max().unwrap()
          – Shepmaster
          Nov 12 at 1:37




          It's worth noting that this would normally idiomatically be written as list.iter().max().unwrap()
          – Shepmaster
          Nov 12 at 1:37

















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