Does manually breaking early from a for loop make sense when testing if a linked list contains an item?

While following a basic tutorial, I'm faced with this function:

use std::collections::LinkedList;

// ...

pub fn contains(&self, x: i32, y: i32) -> bool 
 let mut ch = 0;
 let list: &LinkedList<Block> = &self.body;

 for block in list 
 if block.x == x && block.y == y 
 return true;
 
 ch += 1;
 if ch == list.len() - 1 
 break;
 
 

 return false;

It felt obvious that I could get rid of the whole if ch == list.len() - 1 part and write it like so:

pub fn contains(&self, x: i32, y: i32) -> bool 
 for block in &self.body 
 if block.x == x && block.y == y 
 return true;
 
 
 return false;

It seems to work fine, but maybe there is something that I've missed? Is it just an unnecessary overhead that an author of the tutorial made by mistake?

edited Nov 13 '18 at 2:50

Shepmaster

149k12287422

asked Nov 12 '18 at 22:35

streletss

2,080421

1

Is this the LinkedList from the standard library? If not, perhaps it is a circularly linked list that iterates infinitely by linking the tail to the head.
– Benjamin Lindley
Nov 12 '18 at 23:12

Yes, it is. use std::collections::LinkedList;
– streletss
Nov 12 '18 at 23:14

4

even better self.body.iter().any(|block| block.x == x && block.y == y)
– Stargateur
Nov 13 '18 at 0:51

1

The presented code has numerous non-idiomatic aspects, so I would not trust that tutorial strongly.
– Shepmaster
Nov 13 '18 at 2:50

add a comment |

While following a basic tutorial, I'm faced with this function:

use std::collections::LinkedList;

// ...

pub fn contains(&self, x: i32, y: i32) -> bool 
 let mut ch = 0;
 let list: &LinkedList<Block> = &self.body;

 for block in list 
 if block.x == x && block.y == y 
 return true;
 
 ch += 1;
 if ch == list.len() - 1 
 break;
 
 

 return false;

It felt obvious that I could get rid of the whole if ch == list.len() - 1 part and write it like so:

pub fn contains(&self, x: i32, y: i32) -> bool 
 for block in &self.body 
 if block.x == x && block.y == y 
 return true;
 
 
 return false;

It seems to work fine, but maybe there is something that I've missed? Is it just an unnecessary overhead that an author of the tutorial made by mistake?

edited Nov 13 '18 at 2:50

Shepmaster

149k12287422

asked Nov 12 '18 at 22:35

streletss

2,080421

1

Is this the LinkedList from the standard library? If not, perhaps it is a circularly linked list that iterates infinitely by linking the tail to the head.
– Benjamin Lindley
Nov 12 '18 at 23:12

Yes, it is. use std::collections::LinkedList;
– streletss
Nov 12 '18 at 23:14

4

even better self.body.iter().any(|block| block.x == x && block.y == y)
– Stargateur
Nov 13 '18 at 0:51

1

The presented code has numerous non-idiomatic aspects, so I would not trust that tutorial strongly.
– Shepmaster
Nov 13 '18 at 2:50

add a comment |

While following a basic tutorial, I'm faced with this function:

use std::collections::LinkedList;

// ...

pub fn contains(&self, x: i32, y: i32) -> bool 
 let mut ch = 0;
 let list: &LinkedList<Block> = &self.body;

 for block in list 
 if block.x == x && block.y == y 
 return true;
 
 ch += 1;
 if ch == list.len() - 1 
 break;
 
 

 return false;

It felt obvious that I could get rid of the whole if ch == list.len() - 1 part and write it like so:

pub fn contains(&self, x: i32, y: i32) -> bool 
 for block in &self.body 
 if block.x == x && block.y == y 
 return true;
 
 
 return false;

It seems to work fine, but maybe there is something that I've missed? Is it just an unnecessary overhead that an author of the tutorial made by mistake?

edited Nov 13 '18 at 2:50

Shepmaster

149k12287422

asked Nov 12 '18 at 22:35

streletss

2,080421

While following a basic tutorial, I'm faced with this function:

use std::collections::LinkedList;

// ...

pub fn contains(&self, x: i32, y: i32) -> bool 
 let mut ch = 0;
 let list: &LinkedList<Block> = &self.body;

 for block in list 
 if block.x == x && block.y == y 
 return true;
 
 ch += 1;
 if ch == list.len() - 1 
 break;
 
 

 return false;

It felt obvious that I could get rid of the whole if ch == list.len() - 1 part and write it like so:

pub fn contains(&self, x: i32, y: i32) -> bool 
 for block in &self.body 
 if block.x == x && block.y == y 
 return true;
 
 
 return false;

It seems to work fine, but maybe there is something that I've missed? Is it just an unnecessary overhead that an author of the tutorial made by mistake?

rust for-in-loop

edited Nov 13 '18 at 2:50

Shepmaster

149k12287422

asked Nov 12 '18 at 22:35

streletss

2,080421

edited Nov 13 '18 at 2:50

Shepmaster

149k12287422

asked Nov 12 '18 at 22:35

streletss

2,080421

edited Nov 13 '18 at 2:50

Shepmaster

149k12287422

edited Nov 13 '18 at 2:50

Shepmaster

149k12287422

edited Nov 13 '18 at 2:50

Shepmaster

149k12287422

asked Nov 12 '18 at 22:35

streletss

2,080421

asked Nov 12 '18 at 22:35

streletss

2,080421

asked Nov 12 '18 at 22:35

streletss

2,080421

1

Is this the LinkedList from the standard library? If not, perhaps it is a circularly linked list that iterates infinitely by linking the tail to the head.
– Benjamin Lindley
Nov 12 '18 at 23:12

Yes, it is. use std::collections::LinkedList;
– streletss
Nov 12 '18 at 23:14

4

even better self.body.iter().any(|block| block.x == x && block.y == y)
– Stargateur
Nov 13 '18 at 0:51

1

The presented code has numerous non-idiomatic aspects, so I would not trust that tutorial strongly.
– Shepmaster
Nov 13 '18 at 2:50

add a comment |

1

Is this the LinkedList from the standard library? If not, perhaps it is a circularly linked list that iterates infinitely by linking the tail to the head.
– Benjamin Lindley
Nov 12 '18 at 23:12

Yes, it is. use std::collections::LinkedList;
– streletss
Nov 12 '18 at 23:14

4

even better self.body.iter().any(|block| block.x == x && block.y == y)
– Stargateur
Nov 13 '18 at 0:51

1

The presented code has numerous non-idiomatic aspects, so I would not trust that tutorial strongly.
– Shepmaster
Nov 13 '18 at 2:50

Is this the LinkedList from the standard library? If not, perhaps it is a circularly linked list that iterates infinitely by linking the tail to the head.
– Benjamin Lindley
Nov 12 '18 at 23:12

Yes, it is. use std::collections::LinkedList;
– streletss
Nov 12 '18 at 23:14

even better self.body.iter().any(|block| block.x == x && block.y == y)
– Stargateur
Nov 13 '18 at 0:51

The presented code has numerous non-idiomatic aspects, so I would not trust that tutorial strongly.
– Shepmaster
Nov 13 '18 at 2:50

add a comment |

1 Answer
1

active

oldest

votes

As written in the original 'tutorial' version, it doesn't seem to look at the last element. Consider a list of length 2 where the second element is the one you're looking for.

After the first comparison, ch becomes 1. It's now equal to the list length minus 1, so you break out of the loop just before the loop would (if executed one more time) find the last element.

That doesn't make much sense, so I conclude yours is not only shorter but correct.

answered Nov 13 '18 at 0:55

dave

1032

add a comment |

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1 Answer
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active

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1 Answer
1

active

oldest

votes

As written in the original 'tutorial' version, it doesn't seem to look at the last element. Consider a list of length 2 where the second element is the one you're looking for.

After the first comparison, ch becomes 1. It's now equal to the list length minus 1, so you break out of the loop just before the loop would (if executed one more time) find the last element.

That doesn't make much sense, so I conclude yours is not only shorter but correct.

answered Nov 13 '18 at 0:55

dave

1032

add a comment |

As written in the original 'tutorial' version, it doesn't seem to look at the last element. Consider a list of length 2 where the second element is the one you're looking for.

After the first comparison, ch becomes 1. It's now equal to the list length minus 1, so you break out of the loop just before the loop would (if executed one more time) find the last element.

That doesn't make much sense, so I conclude yours is not only shorter but correct.

answered Nov 13 '18 at 0:55

dave

1032

add a comment |

As written in the original 'tutorial' version, it doesn't seem to look at the last element. Consider a list of length 2 where the second element is the one you're looking for.

After the first comparison, ch becomes 1. It's now equal to the list length minus 1, so you break out of the loop just before the loop would (if executed one more time) find the last element.

That doesn't make much sense, so I conclude yours is not only shorter but correct.

answered Nov 13 '18 at 0:55

dave

1032

As written in the original 'tutorial' version, it doesn't seem to look at the last element. Consider a list of length 2 where the second element is the one you're looking for.

After the first comparison, ch becomes 1. It's now equal to the list length minus 1, so you break out of the loop just before the loop would (if executed one more time) find the last element.

That doesn't make much sense, so I conclude yours is not only shorter but correct.

answered Nov 13 '18 at 0:55

dave

1032

answered Nov 13 '18 at 0:55

dave

1032

answered Nov 13 '18 at 0:55

dave

1032

answered Nov 13 '18 at 0:55

dave

1032

add a comment |

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