Slice a string containing Unicode chars

Clash Royale CLAN TAG#URR8PPP

up vote
13
down vote

favorite

I have a piece of text with characters of different bytelength.

let text = "Hello привет";

I need to take a slice of the string given start (included) and end (excluded) character indices. I tried this

let slice = &text[start..end];

and got the following error

thread 'main' panicked at 'byte index 7 is not a char boundary; it is inside 'п' (bytes 6..8) of `Hello привет`'

I suppose it happens since Cyrillic letters are multi-byte and the [..] notation takes chars using byte indices. What can I use if I want to slice using character indices, like I do in Python:

slice = text[start:end] ?

I know I can use the chars() iterator and manually walk through the desired substring, but is there a more concise way?

string unicode rust slice

share|improve this question

edited Aug 23 at 19:16

Matthieu M.

194k26257487

asked Aug 23 at 9:52

Sasha Tsukanov

414112

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I think chars() is the way to go here: text.chars().take(end).skip(start)
  – Tim Diekmann
  Aug 23 at 10:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TimDiekmann how do I convert the Take<Chars> to &str then if the API needs it?
  – Sasha Tsukanov
  Aug 23 at 10:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should call collect(). See this question stackoverflow.com/questions/37157926/…
  – ozkriff
  Aug 23 at 10:18
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ozkriff collect() will result in String, not in &str. This is why I didn't marked this as duplicate to your linked question.
  – Tim Diekmann
  Aug 23 at 10:31
  

  
  
  
  

add a comment | 

up vote
13
down vote

favorite

I have a piece of text with characters of different bytelength.

let text = "Hello привет";

I need to take a slice of the string given start (included) and end (excluded) character indices. I tried this

let slice = &text[start..end];

and got the following error

thread 'main' panicked at 'byte index 7 is not a char boundary; it is inside 'п' (bytes 6..8) of `Hello привет`'

I suppose it happens since Cyrillic letters are multi-byte and the [..] notation takes chars using byte indices. What can I use if I want to slice using character indices, like I do in Python:

slice = text[start:end] ?

I know I can use the chars() iterator and manually walk through the desired substring, but is there a more concise way?

string unicode rust slice

share|improve this question

edited Aug 23 at 19:16

Matthieu M.

194k26257487

asked Aug 23 at 9:52

Sasha Tsukanov

414112

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I think chars() is the way to go here: text.chars().take(end).skip(start)
  – Tim Diekmann
  Aug 23 at 10:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TimDiekmann how do I convert the Take<Chars> to &str then if the API needs it?
  – Sasha Tsukanov
  Aug 23 at 10:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should call collect(). See this question stackoverflow.com/questions/37157926/…
  – ozkriff
  Aug 23 at 10:18
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ozkriff collect() will result in String, not in &str. This is why I didn't marked this as duplicate to your linked question.
  – Tim Diekmann
  Aug 23 at 10:31
  

  
  
  
  

add a comment | 

up vote
13
down vote

favorite

up vote
13
down vote

favorite

I have a piece of text with characters of different bytelength.

let text = "Hello привет";

I need to take a slice of the string given start (included) and end (excluded) character indices. I tried this

let slice = &text[start..end];

and got the following error

thread 'main' panicked at 'byte index 7 is not a char boundary; it is inside 'п' (bytes 6..8) of `Hello привет`'

I suppose it happens since Cyrillic letters are multi-byte and the [..] notation takes chars using byte indices. What can I use if I want to slice using character indices, like I do in Python:

slice = text[start:end] ?

I know I can use the chars() iterator and manually walk through the desired substring, but is there a more concise way?

string unicode rust slice

share|improve this question

edited Aug 23 at 19:16

Matthieu M.

194k26257487

asked Aug 23 at 9:52

Sasha Tsukanov

414112

I have a piece of text with characters of different bytelength.

let text = "Hello привет";

I need to take a slice of the string given start (included) and end (excluded) character indices. I tried this

let slice = &text[start..end];

and got the following error

thread 'main' panicked at 'byte index 7 is not a char boundary; it is inside 'п' (bytes 6..8) of `Hello привет`'

I suppose it happens since Cyrillic letters are multi-byte and the [..] notation takes chars using byte indices. What can I use if I want to slice using character indices, like I do in Python:

slice = text[start:end] ?

I know I can use the chars() iterator and manually walk through the desired substring, but is there a more concise way?

string unicode rust slice

share|improve this question

edited Aug 23 at 19:16

Matthieu M.

194k26257487

asked Aug 23 at 9:52

Sasha Tsukanov

414112

share|improve this question

share|improve this question

edited Aug 23 at 19:16

Matthieu M.

194k26257487

edited Aug 23 at 19:16

Matthieu M.

194k26257487

edited Aug 23 at 19:16

Matthieu M.

194k26257487

194k26257487

asked Aug 23 at 9:52

Sasha Tsukanov

414112

asked Aug 23 at 9:52

Sasha Tsukanov

414112

asked Aug 23 at 9:52

Sasha Tsukanov

414112

414112

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I think chars() is the way to go here: text.chars().take(end).skip(start)
  – Tim Diekmann
  Aug 23 at 10:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TimDiekmann how do I convert the Take<Chars> to &str then if the API needs it?
  – Sasha Tsukanov
  Aug 23 at 10:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should call collect(). See this question stackoverflow.com/questions/37157926/…
  – ozkriff
  Aug 23 at 10:18
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ozkriff collect() will result in String, not in &str. This is why I didn't marked this as duplicate to your linked question.
  – Tim Diekmann
  Aug 23 at 10:31
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  I think chars() is the way to go here: text.chars().take(end).skip(start)
  – Tim Diekmann
  Aug 23 at 10:10
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @TimDiekmann how do I convert the Take<Chars> to &str then if the API needs it?
  – Sasha Tsukanov
  Aug 23 at 10:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  You should call collect(). See this question stackoverflow.com/questions/37157926/…
  – ozkriff
  Aug 23 at 10:18
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ozkriff collect() will result in String, not in &str. This is why I didn't marked this as duplicate to your linked question.
  – Tim Diekmann
  Aug 23 at 10:31
  

  
  
  
  

2

2

I think chars() is the way to go here: text.chars().take(end).skip(start)
– Tim Diekmann
Aug 23 at 10:10

I think chars() is the way to go here: text.chars().take(end).skip(start)
– Tim Diekmann
Aug 23 at 10:10

@TimDiekmann how do I convert the Take<Chars> to &str then if the API needs it?
– Sasha Tsukanov
Aug 23 at 10:17

@TimDiekmann how do I convert the Take<Chars> to &str then if the API needs it?
– Sasha Tsukanov
Aug 23 at 10:17

You should call collect(). See this question stackoverflow.com/questions/37157926/…
– ozkriff
Aug 23 at 10:18

You should call collect(). See this question stackoverflow.com/questions/37157926/…
– ozkriff
Aug 23 at 10:18

1

1

@ozkriff collect() will result in String, not in &str. This is why I didn't marked this as duplicate to your linked question.
– Tim Diekmann
Aug 23 at 10:31

@ozkriff collect() will result in String, not in &str. This is why I didn't marked this as duplicate to your linked question.
– Tim Diekmann
Aug 23 at 10:31

add a comment | 

2 Answers
2

active

oldest

votes

up vote
20
down vote



accepted

Possible solutions to codepoint slicing

I know I can use the chars() iterator and manually walk through the desired substring, but is there a more concise way?

If you know the exact byte indices, you can slice a string:

let text = "Hello привет";
println!("", &text[2..10]);

This prints "llo пр". So the problem is to find out the exact byte position. You can do that fairly easily with the char_indices() iterator (alternatively you could use chars() with char::len_utf8()):

let text = "Hello привет";
let end = text.char_indices().map(|(i, _)| i).nth(8).unwrap();
println!("", &text[2..idx]);

As another alternative, you can first collect the string into Vec<char>. Then, indexing is simple, but to print it as a string, you have to collect it again or write your own function to do it.

let text = "Hello привет";
let text_vec = text.chars().collect::<Vec<_>>();
println!("", text_vec[2..8].iter().cloned().collect::<String>());

Why is this not easier?

As you can see, neither of these solutions is all that great. This is intentional, for two reasons:

As str is a simply UTF8 buffer, indexing by unicode codepoints is an O(n) operation. Usually, people expect the operator to be a O(1) operation. Rust makes this runtime complexity explicit and doesn't try to hide it. In both solutions above you can clearly see that it's not O(1).

But the more important reason:

Unicode codepoints are generally not a useful unit

What Python does (and what you think you want) is not all that useful. It all comes down to the complexity of language and thus the complexity of unicode. Python slices Unicode codepoints. This is what a Rust char represents. It's 32 bit big (a few fewer bits would suffice, but we round up to a power of 2).

But what you actually want to do is slice user perceived characters. But this is an explicitly loosely defined term. Different cultures and languages regard different things as "one character". The closest approximation is a "grapheme cluster". Such a cluster can consist of one or more unicode codepoints. Consider this Python 3 code:

>>> s = "Jürgen"
>>> s[0:2]
'Ju'

Surprising, right? This is because the string above is:

• 
  0x004A LATIN CAPITAL LETTER J


• 
  0x0075 LATIN SMALL LETTER U


• 
  0x0308 COMBINING DIAERESIS


• ...

This is an example of a combining character that is rendered as part of the previous character. Python slicing does the "wrong" thing here.

Another example:

>>> s = "fire"
>>> s[0:2]
'fir'

Also not what you'd expect. This time, fi is actually the ligature fi, which is one codepoint.

There are far more examples where Unicode behaves in a surprising way. See the links at the bottom for more information and examples.

So if you want to work with international strings that should be able to work everywhere, don't do codepoint slicing! If you really need to semantically view the string as a series of characters, use grapheme clusters. To do that, the crate unicode-segmentation is very useful.

---

Further resources on this topic:

• Blogpost "Let's stop ascribing meaning to unicode codepoints"


• Blogpost "Breaking our Latin-1 assumptions


• http://utf8everywhere.org/

share|improve this answer

edited Aug 23 at 10:37

answered Aug 23 at 10:23

Lukas Kalbertodt

22.4k249100

• 
  
  
  

  
  

  
  
  
  
  
  

  
  To make let end = text.char_indices().map(|(i, _)| i).nth(8).unwrap(); work when we want to slice till the last codepoint in the string (say, with index 11) by using, say, 12 as excluded bound we need more work. One can add something like let end = if end_codepoint_idx == text.chars().count() text.len() else i).nth(end_codepoint_idx).unwrap();;
  – Sasha Tsukanov
  Aug 23 at 13:58
  

  
  
  
  

add a comment | 

up vote
5
down vote

An UTF-8 encoded string may contain characters, which consists of multiple bytes. In your case, п starts at index 6 (inclusive) and ends at position 8 (exclusive) so indexing 7 is not the start of the character. This is why your error occurred.

You may use str::char_indices for solving this (remember, that getting to a position in UTF-8 is O(n)):

fn get_utf8_slice(string: &str, start: usize, end: usize) -> Option<&str> 
 assert!(end >= start);
 string.char_indices().nth(start).and_then(

playground

You may use str::chars() if you are fine with getting a String:

let string: String = text.chars().take(end).skip(start).collect();

share|improve this answer

edited Aug 23 at 10:37

answered Aug 23 at 10:26

Tim Diekmann

2,57481631

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
20
down vote



accepted

Possible solutions to codepoint slicing

I know I can use the chars() iterator and manually walk through the desired substring, but is there a more concise way?

If you know the exact byte indices, you can slice a string:

let text = "Hello привет";
println!("", &text[2..10]);

This prints "llo пр". So the problem is to find out the exact byte position. You can do that fairly easily with the char_indices() iterator (alternatively you could use chars() with char::len_utf8()):

let text = "Hello привет";
let end = text.char_indices().map(|(i, _)| i).nth(8).unwrap();
println!("", &text[2..idx]);

As another alternative, you can first collect the string into Vec<char>. Then, indexing is simple, but to print it as a string, you have to collect it again or write your own function to do it.

let text = "Hello привет";
let text_vec = text.chars().collect::<Vec<_>>();
println!("", text_vec[2..8].iter().cloned().collect::<String>());

Why is this not easier?

As you can see, neither of these solutions is all that great. This is intentional, for two reasons:

As str is a simply UTF8 buffer, indexing by unicode codepoints is an O(n) operation. Usually, people expect the operator to be a O(1) operation. Rust makes this runtime complexity explicit and doesn't try to hide it. In both solutions above you can clearly see that it's not O(1).

But the more important reason:

Unicode codepoints are generally not a useful unit

What Python does (and what you think you want) is not all that useful. It all comes down to the complexity of language and thus the complexity of unicode. Python slices Unicode codepoints. This is what a Rust char represents. It's 32 bit big (a few fewer bits would suffice, but we round up to a power of 2).

But what you actually want to do is slice user perceived characters. But this is an explicitly loosely defined term. Different cultures and languages regard different things as "one character". The closest approximation is a "grapheme cluster". Such a cluster can consist of one or more unicode codepoints. Consider this Python 3 code:

>>> s = "Jürgen"
>>> s[0:2]
'Ju'

Surprising, right? This is because the string above is:

• 
  0x004A LATIN CAPITAL LETTER J


• 
  0x0075 LATIN SMALL LETTER U


• 
  0x0308 COMBINING DIAERESIS


• ...

This is an example of a combining character that is rendered as part of the previous character. Python slicing does the "wrong" thing here.

Another example:

>>> s = "fire"
>>> s[0:2]
'fir'

Also not what you'd expect. This time, fi is actually the ligature fi, which is one codepoint.

There are far more examples where Unicode behaves in a surprising way. See the links at the bottom for more information and examples.

So if you want to work with international strings that should be able to work everywhere, don't do codepoint slicing! If you really need to semantically view the string as a series of characters, use grapheme clusters. To do that, the crate unicode-segmentation is very useful.

---

Further resources on this topic:

• Blogpost "Let's stop ascribing meaning to unicode codepoints"


• Blogpost "Breaking our Latin-1 assumptions


• http://utf8everywhere.org/

share|improve this answer

edited Aug 23 at 10:37

answered Aug 23 at 10:23

Lukas Kalbertodt

22.4k249100

• 
  
  
  

  
  

  
  
  
  
  
  

  
  To make let end = text.char_indices().map(|(i, _)| i).nth(8).unwrap(); work when we want to slice till the last codepoint in the string (say, with index 11) by using, say, 12 as excluded bound we need more work. One can add something like let end = if end_codepoint_idx == text.chars().count() text.len() else i).nth(end_codepoint_idx).unwrap();;
  – Sasha Tsukanov
  Aug 23 at 13:58
  

  
  
  
  

add a comment | 

up vote
20
down vote



accepted

Possible solutions to codepoint slicing

I know I can use the chars() iterator and manually walk through the desired substring, but is there a more concise way?

If you know the exact byte indices, you can slice a string:

let text = "Hello привет";
println!("", &text[2..10]);

This prints "llo пр". So the problem is to find out the exact byte position. You can do that fairly easily with the char_indices() iterator (alternatively you could use chars() with char::len_utf8()):

let text = "Hello привет";
let end = text.char_indices().map(|(i, _)| i).nth(8).unwrap();
println!("", &text[2..idx]);

As another alternative, you can first collect the string into Vec<char>. Then, indexing is simple, but to print it as a string, you have to collect it again or write your own function to do it.

let text = "Hello привет";
let text_vec = text.chars().collect::<Vec<_>>();
println!("", text_vec[2..8].iter().cloned().collect::<String>());

Why is this not easier?

As you can see, neither of these solutions is all that great. This is intentional, for two reasons:

As str is a simply UTF8 buffer, indexing by unicode codepoints is an O(n) operation. Usually, people expect the operator to be a O(1) operation. Rust makes this runtime complexity explicit and doesn't try to hide it. In both solutions above you can clearly see that it's not O(1).

But the more important reason:

Unicode codepoints are generally not a useful unit

What Python does (and what you think you want) is not all that useful. It all comes down to the complexity of language and thus the complexity of unicode. Python slices Unicode codepoints. This is what a Rust char represents. It's 32 bit big (a few fewer bits would suffice, but we round up to a power of 2).

But what you actually want to do is slice user perceived characters. But this is an explicitly loosely defined term. Different cultures and languages regard different things as "one character". The closest approximation is a "grapheme cluster". Such a cluster can consist of one or more unicode codepoints. Consider this Python 3 code:

>>> s = "Jürgen"
>>> s[0:2]
'Ju'

Surprising, right? This is because the string above is:

• 
  0x004A LATIN CAPITAL LETTER J


• 
  0x0075 LATIN SMALL LETTER U


• 
  0x0308 COMBINING DIAERESIS


• ...

This is an example of a combining character that is rendered as part of the previous character. Python slicing does the "wrong" thing here.

Another example:

>>> s = "fire"
>>> s[0:2]
'fir'

Also not what you'd expect. This time, fi is actually the ligature fi, which is one codepoint.

There are far more examples where Unicode behaves in a surprising way. See the links at the bottom for more information and examples.

So if you want to work with international strings that should be able to work everywhere, don't do codepoint slicing! If you really need to semantically view the string as a series of characters, use grapheme clusters. To do that, the crate unicode-segmentation is very useful.

---

Further resources on this topic:

• Blogpost "Let's stop ascribing meaning to unicode codepoints"


• Blogpost "Breaking our Latin-1 assumptions


• http://utf8everywhere.org/

share|improve this answer

edited Aug 23 at 10:37

answered Aug 23 at 10:23

Lukas Kalbertodt

22.4k249100

• 
  
  
  

  
  

  
  
  
  
  
  

  
  To make let end = text.char_indices().map(|(i, _)| i).nth(8).unwrap(); work when we want to slice till the last codepoint in the string (say, with index 11) by using, say, 12 as excluded bound we need more work. One can add something like let end = if end_codepoint_idx == text.chars().count() text.len() else i).nth(end_codepoint_idx).unwrap();;
  – Sasha Tsukanov
  Aug 23 at 13:58
  

  
  
  
  

add a comment | 

up vote
20
down vote



accepted

up vote
20
down vote



accepted

Possible solutions to codepoint slicing

I know I can use the chars() iterator and manually walk through the desired substring, but is there a more concise way?

If you know the exact byte indices, you can slice a string:

let text = "Hello привет";
println!("", &text[2..10]);

This prints "llo пр". So the problem is to find out the exact byte position. You can do that fairly easily with the char_indices() iterator (alternatively you could use chars() with char::len_utf8()):

let text = "Hello привет";
let end = text.char_indices().map(|(i, _)| i).nth(8).unwrap();
println!("", &text[2..idx]);

As another alternative, you can first collect the string into Vec<char>. Then, indexing is simple, but to print it as a string, you have to collect it again or write your own function to do it.

let text = "Hello привет";
let text_vec = text.chars().collect::<Vec<_>>();
println!("", text_vec[2..8].iter().cloned().collect::<String>());

Why is this not easier?

As you can see, neither of these solutions is all that great. This is intentional, for two reasons:

As str is a simply UTF8 buffer, indexing by unicode codepoints is an O(n) operation. Usually, people expect the operator to be a O(1) operation. Rust makes this runtime complexity explicit and doesn't try to hide it. In both solutions above you can clearly see that it's not O(1).

But the more important reason:

Unicode codepoints are generally not a useful unit

What Python does (and what you think you want) is not all that useful. It all comes down to the complexity of language and thus the complexity of unicode. Python slices Unicode codepoints. This is what a Rust char represents. It's 32 bit big (a few fewer bits would suffice, but we round up to a power of 2).

But what you actually want to do is slice user perceived characters. But this is an explicitly loosely defined term. Different cultures and languages regard different things as "one character". The closest approximation is a "grapheme cluster". Such a cluster can consist of one or more unicode codepoints. Consider this Python 3 code:

>>> s = "Jürgen"
>>> s[0:2]
'Ju'

Surprising, right? This is because the string above is:

• 
  0x004A LATIN CAPITAL LETTER J


• 
  0x0075 LATIN SMALL LETTER U


• 
  0x0308 COMBINING DIAERESIS


• ...

This is an example of a combining character that is rendered as part of the previous character. Python slicing does the "wrong" thing here.

Another example:

>>> s = "fire"
>>> s[0:2]
'fir'

Also not what you'd expect. This time, fi is actually the ligature fi, which is one codepoint.

There are far more examples where Unicode behaves in a surprising way. See the links at the bottom for more information and examples.

So if you want to work with international strings that should be able to work everywhere, don't do codepoint slicing! If you really need to semantically view the string as a series of characters, use grapheme clusters. To do that, the crate unicode-segmentation is very useful.

---

Further resources on this topic:

• Blogpost "Let's stop ascribing meaning to unicode codepoints"


• Blogpost "Breaking our Latin-1 assumptions


• http://utf8everywhere.org/

share|improve this answer

edited Aug 23 at 10:37

answered Aug 23 at 10:23

Lukas Kalbertodt

22.4k249100

Possible solutions to codepoint slicing

I know I can use the chars() iterator and manually walk through the desired substring, but is there a more concise way?

If you know the exact byte indices, you can slice a string:

let text = "Hello привет";
println!("", &text[2..10]);

This prints "llo пр". So the problem is to find out the exact byte position. You can do that fairly easily with the char_indices() iterator (alternatively you could use chars() with char::len_utf8()):

let text = "Hello привет";
let end = text.char_indices().map(|(i, _)| i).nth(8).unwrap();
println!("", &text[2..idx]);

As another alternative, you can first collect the string into Vec<char>. Then, indexing is simple, but to print it as a string, you have to collect it again or write your own function to do it.

let text = "Hello привет";
let text_vec = text.chars().collect::<Vec<_>>();
println!("", text_vec[2..8].iter().cloned().collect::<String>());

Why is this not easier?

As you can see, neither of these solutions is all that great. This is intentional, for two reasons:

As str is a simply UTF8 buffer, indexing by unicode codepoints is an O(n) operation. Usually, people expect the operator to be a O(1) operation. Rust makes this runtime complexity explicit and doesn't try to hide it. In both solutions above you can clearly see that it's not O(1).

But the more important reason:

Unicode codepoints are generally not a useful unit

What Python does (and what you think you want) is not all that useful. It all comes down to the complexity of language and thus the complexity of unicode. Python slices Unicode codepoints. This is what a Rust char represents. It's 32 bit big (a few fewer bits would suffice, but we round up to a power of 2).

But what you actually want to do is slice user perceived characters. But this is an explicitly loosely defined term. Different cultures and languages regard different things as "one character". The closest approximation is a "grapheme cluster". Such a cluster can consist of one or more unicode codepoints. Consider this Python 3 code:

>>> s = "Jürgen"
>>> s[0:2]
'Ju'

Surprising, right? This is because the string above is:

• 
  0x004A LATIN CAPITAL LETTER J


• 
  0x0075 LATIN SMALL LETTER U


• 
  0x0308 COMBINING DIAERESIS


• ...

This is an example of a combining character that is rendered as part of the previous character. Python slicing does the "wrong" thing here.

Another example:

>>> s = "fire"
>>> s[0:2]
'fir'

Also not what you'd expect. This time, fi is actually the ligature fi, which is one codepoint.

There are far more examples where Unicode behaves in a surprising way. See the links at the bottom for more information and examples.

So if you want to work with international strings that should be able to work everywhere, don't do codepoint slicing! If you really need to semantically view the string as a series of characters, use grapheme clusters. To do that, the crate unicode-segmentation is very useful.

---

Further resources on this topic:

• Blogpost "Let's stop ascribing meaning to unicode codepoints"


• Blogpost "Breaking our Latin-1 assumptions


• http://utf8everywhere.org/

share|improve this answer

edited Aug 23 at 10:37

answered Aug 23 at 10:23

Lukas Kalbertodt

22.4k249100

share|improve this answer

share|improve this answer

edited Aug 23 at 10:37

edited Aug 23 at 10:37

edited Aug 23 at 10:37

answered Aug 23 at 10:23

Lukas Kalbertodt

22.4k249100

answered Aug 23 at 10:23

Lukas Kalbertodt

22.4k249100

answered Aug 23 at 10:23

Lukas Kalbertodt

22.4k249100

22.4k249100

• 
  
  
  

  
  

  
  
  
  
  
  

  
  To make let end = text.char_indices().map(|(i, _)| i).nth(8).unwrap(); work when we want to slice till the last codepoint in the string (say, with index 11) by using, say, 12 as excluded bound we need more work. One can add something like let end = if end_codepoint_idx == text.chars().count() text.len() else i).nth(end_codepoint_idx).unwrap();;
  – Sasha Tsukanov
  Aug 23 at 13:58
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  To make let end = text.char_indices().map(|(i, _)| i).nth(8).unwrap(); work when we want to slice till the last codepoint in the string (say, with index 11) by using, say, 12 as excluded bound we need more work. One can add something like let end = if end_codepoint_idx == text.chars().count() text.len() else i).nth(end_codepoint_idx).unwrap();;
  – Sasha Tsukanov
  Aug 23 at 13:58
  

  
  
  
  

To make let end = text.char_indices().map(|(i, _)| i).nth(8).unwrap(); work when we want to slice till the last codepoint in the string (say, with index 11) by using, say, 12 as excluded bound we need more work. One can add something like let end = if end_codepoint_idx == text.chars().count() text.len() else i).nth(end_codepoint_idx).unwrap();;
– Sasha Tsukanov
Aug 23 at 13:58

To make let end = text.char_indices().map(|(i, _)| i).nth(8).unwrap(); work when we want to slice till the last codepoint in the string (say, with index 11) by using, say, 12 as excluded bound we need more work. One can add something like let end = if end_codepoint_idx == text.chars().count() text.len() else i).nth(end_codepoint_idx).unwrap();;
– Sasha Tsukanov
Aug 23 at 13:58

add a comment | 

up vote
5
down vote

An UTF-8 encoded string may contain characters, which consists of multiple bytes. In your case, п starts at index 6 (inclusive) and ends at position 8 (exclusive) so indexing 7 is not the start of the character. This is why your error occurred.

You may use str::char_indices for solving this (remember, that getting to a position in UTF-8 is O(n)):

fn get_utf8_slice(string: &str, start: usize, end: usize) -> Option<&str> 
 assert!(end >= start);
 string.char_indices().nth(start).and_then(

playground

You may use str::chars() if you are fine with getting a String:

let string: String = text.chars().take(end).skip(start).collect();

share|improve this answer

edited Aug 23 at 10:37

answered Aug 23 at 10:26

Tim Diekmann

2,57481631

add a comment | 

up vote
5
down vote

An UTF-8 encoded string may contain characters, which consists of multiple bytes. In your case, п starts at index 6 (inclusive) and ends at position 8 (exclusive) so indexing 7 is not the start of the character. This is why your error occurred.

You may use str::char_indices for solving this (remember, that getting to a position in UTF-8 is O(n)):

fn get_utf8_slice(string: &str, start: usize, end: usize) -> Option<&str> 
 assert!(end >= start);
 string.char_indices().nth(start).and_then(

playground

You may use str::chars() if you are fine with getting a String:

let string: String = text.chars().take(end).skip(start).collect();

share|improve this answer

edited Aug 23 at 10:37

answered Aug 23 at 10:26

Tim Diekmann

2,57481631

add a comment | 

up vote
5
down vote

up vote
5
down vote

An UTF-8 encoded string may contain characters, which consists of multiple bytes. In your case, п starts at index 6 (inclusive) and ends at position 8 (exclusive) so indexing 7 is not the start of the character. This is why your error occurred.

You may use str::char_indices for solving this (remember, that getting to a position in UTF-8 is O(n)):

fn get_utf8_slice(string: &str, start: usize, end: usize) -> Option<&str> 
 assert!(end >= start);
 string.char_indices().nth(start).and_then(

playground

You may use str::chars() if you are fine with getting a String:

let string: String = text.chars().take(end).skip(start).collect();

share|improve this answer

edited Aug 23 at 10:37

answered Aug 23 at 10:26

Tim Diekmann

2,57481631

An UTF-8 encoded string may contain characters, which consists of multiple bytes. In your case, п starts at index 6 (inclusive) and ends at position 8 (exclusive) so indexing 7 is not the start of the character. This is why your error occurred.

You may use str::char_indices for solving this (remember, that getting to a position in UTF-8 is O(n)):

fn get_utf8_slice(string: &str, start: usize, end: usize) -> Option<&str> 
 assert!(end >= start);
 string.char_indices().nth(start).and_then(

playground

You may use str::chars() if you are fine with getting a String:

let string: String = text.chars().take(end).skip(start).collect();

share|improve this answer

edited Aug 23 at 10:37

answered Aug 23 at 10:26

Tim Diekmann

2,57481631

share|improve this answer

share|improve this answer

edited Aug 23 at 10:37

edited Aug 23 at 10:37

edited Aug 23 at 10:37

answered Aug 23 at 10:26

Tim Diekmann

2,57481631

answered Aug 23 at 10:26

Tim Diekmann

2,57481631

answered Aug 23 at 10:26

Tim Diekmann

2,57481631

2,57481631

add a comment | 

add a comment | 

 

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