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
vector rust
add a comment |
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
vector rust
add a comment |
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
vector rust
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
vector rust
edited Nov 12 at 1:36
Shepmaster
146k11281413
146k11281413
asked Nov 12 at 0:14
WarSame
729
729
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1 Answer
1
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3
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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).
2
It's worth noting that this would normally idiomatically be written aslist.iter().max().unwrap()
– Shepmaster
Nov 12 at 1:37
add a comment |
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1 Answer
1
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).
2
It's worth noting that this would normally idiomatically be written aslist.iter().max().unwrap()
– Shepmaster
Nov 12 at 1:37
add a comment |
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).
2
It's worth noting that this would normally idiomatically be written aslist.iter().max().unwrap()
– Shepmaster
Nov 12 at 1:37
add a comment |
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).
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).
answered Nov 12 at 0:52
Francis Gagné
31.6k26478
31.6k26478
2
It's worth noting that this would normally idiomatically be written aslist.iter().max().unwrap()
– Shepmaster
Nov 12 at 1:37
add a comment |
2
It's worth noting that this would normally idiomatically be written aslist.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
add a comment |
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