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Understanding Rust function parameter type declaration

How do I print the type of a variable in Rust?Is it possible to declare the type of the variable in Rust for loops?Why doesn't println! work in Rust unit tests?Convert a String to int in Rust?Why are explicit lifetimes needed in Rust?In Rust how do you pass a function as a parameter?How to declare typed bitflags in Rust?How to understand this Rust function that returns another function?How does the dropping of temporary values work in rust?Cast Rust function declaration to the extern “C” declaration in Rust

1

0

I was reading the chapter on higher order functions of Rust by Example. Where they present the following canoncial example:

fn is_odd(n: u32) -> bool 
 n % 2 == 1


fn main() acc, n_squared

Simple enough. But I realized that I don't understand the type of parameter n_squared. Both take_while and filter accept a function that takes a parameter by reference. That makes sense to me, you want to borrow instead of consuming the values in the map.

However, if n_squared is a reference, why don't I have to dereference it before comparing its value to limit or equaly surprising; why can I pass it directly to is_odd() without dereferencing?

I.e. why isn't it?

 |&n_squared| *n_squared < upper

When I try that the compiler gives the following error:

error[E0614]: type `integer` cannot be dereferenced
 --> srchigherorder.rs:13:34
 |
13 | .take_while(|&n_squared| *n_squared <= upper)
 | 

Indicating that n_squared is an i32 and not &i32. Looks like some sort pattern matching/destructuring is happening here, but I was unable to find the relevant documentation.

rust

share|improve this question

edited Mar 8 at 9:38

djf

asked Mar 8 at 9:33

djfdjf

5,33463348

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If you write |n_squared|, you'll get the expected error and using *n_squared in the body will fix it. See reddit.com/r/rust/comments/16b9wh/…
  
  – Alexey Romanov
  Mar 8 at 9:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Since I edited the comment and you might not have seen it: I've added a link to an old Reddit discussion of exactly this issue reddit.com/r/rust/comments/16b9wh/…. While Rust has changed a lot, this seems to still be what's going happening.
  
  – Alexey Romanov
  Mar 8 at 9:53
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  The documentation is in doc.rust-lang.org/1.30.0/book/2018-edition/….`
  
  – Alexey Romanov
  Mar 8 at 11:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @AlexeyRomanov Perfect! That's what I was looking for.
  
  – djf
  Mar 8 at 11:48
  

  
  
  
  

add a comment | 

1

0

I was reading the chapter on higher order functions of Rust by Example. Where they present the following canoncial example:

fn is_odd(n: u32) -> bool 
 n % 2 == 1


fn main() acc, n_squared

Simple enough. But I realized that I don't understand the type of parameter n_squared. Both take_while and filter accept a function that takes a parameter by reference. That makes sense to me, you want to borrow instead of consuming the values in the map.

However, if n_squared is a reference, why don't I have to dereference it before comparing its value to limit or equaly surprising; why can I pass it directly to is_odd() without dereferencing?

I.e. why isn't it?

 |&n_squared| *n_squared < upper

When I try that the compiler gives the following error:

error[E0614]: type `integer` cannot be dereferenced
 --> srchigherorder.rs:13:34
 |
13 | .take_while(|&n_squared| *n_squared <= upper)
 | 

Indicating that n_squared is an i32 and not &i32. Looks like some sort pattern matching/destructuring is happening here, but I was unable to find the relevant documentation.

rust

share|improve this question

edited Mar 8 at 9:38

djf

asked Mar 8 at 9:33

djfdjf

5,33463348

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  If you write |n_squared|, you'll get the expected error and using *n_squared in the body will fix it. See reddit.com/r/rust/comments/16b9wh/…
  
  – Alexey Romanov
  Mar 8 at 9:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Since I edited the comment and you might not have seen it: I've added a link to an old Reddit discussion of exactly this issue reddit.com/r/rust/comments/16b9wh/…. While Rust has changed a lot, this seems to still be what's going happening.
  
  – Alexey Romanov
  Mar 8 at 9:53
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  The documentation is in doc.rust-lang.org/1.30.0/book/2018-edition/….`
  
  – Alexey Romanov
  Mar 8 at 11:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @AlexeyRomanov Perfect! That's what I was looking for.
  
  – djf
  Mar 8 at 11:48
  

  
  
  
  

add a comment | 

I was reading the chapter on higher order functions of Rust by Example. Where they present the following canoncial example:

fn is_odd(n: u32) -> bool 
 n % 2 == 1


fn main() acc, n_squared

Simple enough. But I realized that I don't understand the type of parameter n_squared. Both take_while and filter accept a function that takes a parameter by reference. That makes sense to me, you want to borrow instead of consuming the values in the map.

However, if n_squared is a reference, why don't I have to dereference it before comparing its value to limit or equaly surprising; why can I pass it directly to is_odd() without dereferencing?

I.e. why isn't it?

 |&n_squared| *n_squared < upper

When I try that the compiler gives the following error:

error[E0614]: type `integer` cannot be dereferenced
 --> srchigherorder.rs:13:34
 |
13 | .take_while(|&n_squared| *n_squared <= upper)
 | 

Indicating that n_squared is an i32 and not &i32. Looks like some sort pattern matching/destructuring is happening here, but I was unable to find the relevant documentation.

rust

share|improve this question

edited Mar 8 at 9:38

djf

asked Mar 8 at 9:33

djfdjf

5,33463348

fn is_odd(n: u32) -> bool 
 n % 2 == 1


fn main() acc, n_squared

 |&n_squared| *n_squared < upper

error[E0614]: type `integer` cannot be dereferenced
 --> srchigherorder.rs:13:34
 |
13 | .take_while(|&n_squared| *n_squared <= upper)
 | 

share|improve this question

edited Mar 8 at 9:38

djf

asked Mar 8 at 9:33

djfdjf

5,33463348

share|improve this question

edited Mar 8 at 9:38

djf

asked Mar 8 at 9:33

djfdjf

5,33463348

asked Mar 8 at 9:33

djfdjf

5,33463348

asked Mar 8 at 9:33

djfdjf

5,33463348

1 Answer
1

active

oldest

votes

5

You are using function parameter destructuring:

|&n_squared| n_squared < upper

is functionally equivalent to:

|n_squared| *n_squared < upper

To understand this better, imagine you're passing a tuple of type &(i32, i32) to a lambda:

|&(x, y) : &(i32, i32)| x + y

share|improve this answer

edited Mar 8 at 11:50

Svetlin Zarev

5,93332760

answered Mar 8 at 10:12

SirDariusSirDarius

30.6k66284

add a comment | 

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5

You are using function parameter destructuring:

|&n_squared| n_squared < upper

is functionally equivalent to:

|n_squared| *n_squared < upper

To understand this better, imagine you're passing a tuple of type &(i32, i32) to a lambda:

|&(x, y) : &(i32, i32)| x + y

share|improve this answer

edited Mar 8 at 11:50

Svetlin Zarev

5,93332760

answered Mar 8 at 10:12

SirDariusSirDarius

30.6k66284

add a comment | 

5

You are using function parameter destructuring:

|&n_squared| n_squared < upper

is functionally equivalent to:

|n_squared| *n_squared < upper

To understand this better, imagine you're passing a tuple of type &(i32, i32) to a lambda:

|&(x, y) : &(i32, i32)| x + y

share|improve this answer

edited Mar 8 at 11:50

Svetlin Zarev

5,93332760

answered Mar 8 at 10:12

SirDariusSirDarius

30.6k66284

add a comment | 

You are using function parameter destructuring:

|&n_squared| n_squared < upper

is functionally equivalent to:

|n_squared| *n_squared < upper

To understand this better, imagine you're passing a tuple of type &(i32, i32) to a lambda:

|&(x, y) : &(i32, i32)| x + y

share|improve this answer

edited Mar 8 at 11:50

Svetlin Zarev

5,93332760

answered Mar 8 at 10:12

SirDariusSirDarius

30.6k66284

|&n_squared| n_squared < upper

|n_squared| *n_squared < upper

|&(x, y) : &(i32, i32)| x + y

share|improve this answer

edited Mar 8 at 11:50

Svetlin Zarev

5,93332760

answered Mar 8 at 10:12

SirDariusSirDarius

30.6k66284

edited Mar 8 at 11:50

Svetlin Zarev

5,93332760

edited Mar 8 at 11:50

Svetlin Zarev

5,93332760

answered Mar 8 at 10:12

SirDariusSirDarius

30.6k66284

answered Mar 8 at 10:12

SirDariusSirDarius

30.6k66284