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8

1

I'm trying to decompose an array of 2 integers given to a function into x, y

It doesn't work when using int init[2] as a parameter. But it does when I change it to int (&init)[2].

vector<vector<State>> Search(vector<vector<State>> board,
 int init[2], int goal[2]) 
 auto [x, y] = init;

What does (&init) mean here? And why it doesn't work when using int init[2]?

c++ c++17

share|improve this question

edited 3 hours ago

StoryTeller

108k16227290

asked 4 hours ago

BrunoBruno

14414

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Don't do using namespace std;.
  
  – Williham Totland
  1 hour ago
  
  

  
  
  
  

add a comment | 

8

1

I'm trying to decompose an array of 2 integers given to a function into x, y

It doesn't work when using int init[2] as a parameter. But it does when I change it to int (&init)[2].

vector<vector<State>> Search(vector<vector<State>> board,
 int init[2], int goal[2]) 
 auto [x, y] = init;

What does (&init) mean here? And why it doesn't work when using int init[2]?

c++ c++17

share|improve this question

edited 3 hours ago

StoryTeller

108k16227290

asked 4 hours ago

BrunoBruno

14414

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Don't do using namespace std;.
  
  – Williham Totland
  1 hour ago
  
  

  
  
  
  

add a comment | 

I'm trying to decompose an array of 2 integers given to a function into x, y

It doesn't work when using int init[2] as a parameter. But it does when I change it to int (&init)[2].

vector<vector<State>> Search(vector<vector<State>> board,
 int init[2], int goal[2]) 
 auto [x, y] = init;

What does (&init) mean here? And why it doesn't work when using int init[2]?

c++ c++17

share|improve this question

edited 3 hours ago

StoryTeller

108k16227290

asked 4 hours ago

BrunoBruno

14414

vector<vector<State>> Search(vector<vector<State>> board,
 int init[2], int goal[2]) 
 auto [x, y] = init;

share|improve this question

edited 3 hours ago

StoryTeller

108k16227290

asked 4 hours ago

BrunoBruno

14414

share|improve this question

edited 3 hours ago

StoryTeller

108k16227290

asked 4 hours ago

BrunoBruno

14414

edited 3 hours ago

StoryTeller

108k16227290

edited 3 hours ago

StoryTeller

108k16227290

asked 4 hours ago

BrunoBruno

14414

asked 4 hours ago

BrunoBruno

14414

1 Answer
1

active

oldest

votes

8

int (&init)[2] is a reference to an array of two integers. int init[2] as a function parameter is a leftover from C++'s C heritage. It doesn't declare the function as taking an array. The type of the parameter is adjusted to int* and all size information for an array being passed into the function is lost.

A function taking int init[2] can be called with an array of any size, on account of actually taking a pointer. It may even be passed nullptr. While a function taking int(&)[2] may only be given a valid array of two as an argument.

Since in the working version init refers to a int[2] object, structured bindings can work with that array object. But a decayed pointer cannot be the subject of structured bindings, because the static type information available only gives access to a single element being pointed at.

share|improve this answer

answered 4 hours ago

StoryTellerStoryTeller

108k16227290

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Thanks!! I got it :)
  
  – Bruno
  3 hours ago
  

  
  
  
  

add a comment | 

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8

int (&init)[2] is a reference to an array of two integers. int init[2] as a function parameter is a leftover from C++'s C heritage. It doesn't declare the function as taking an array. The type of the parameter is adjusted to int* and all size information for an array being passed into the function is lost.

A function taking int init[2] can be called with an array of any size, on account of actually taking a pointer. It may even be passed nullptr. While a function taking int(&)[2] may only be given a valid array of two as an argument.

Since in the working version init refers to a int[2] object, structured bindings can work with that array object. But a decayed pointer cannot be the subject of structured bindings, because the static type information available only gives access to a single element being pointed at.

share|improve this answer

answered 4 hours ago

StoryTellerStoryTeller

108k16227290

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Thanks!! I got it :)
  
  – Bruno
  3 hours ago
  

  
  
  
  

add a comment | 

8

int (&init)[2] is a reference to an array of two integers. int init[2] as a function parameter is a leftover from C++'s C heritage. It doesn't declare the function as taking an array. The type of the parameter is adjusted to int* and all size information for an array being passed into the function is lost.

A function taking int init[2] can be called with an array of any size, on account of actually taking a pointer. It may even be passed nullptr. While a function taking int(&)[2] may only be given a valid array of two as an argument.

Since in the working version init refers to a int[2] object, structured bindings can work with that array object. But a decayed pointer cannot be the subject of structured bindings, because the static type information available only gives access to a single element being pointed at.

share|improve this answer

answered 4 hours ago

StoryTellerStoryTeller

108k16227290

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Thanks!! I got it :)
  
  – Bruno
  3 hours ago
  

  
  
  
  

add a comment | 

int (&init)[2] is a reference to an array of two integers. int init[2] as a function parameter is a leftover from C++'s C heritage. It doesn't declare the function as taking an array. The type of the parameter is adjusted to int* and all size information for an array being passed into the function is lost.

A function taking int init[2] can be called with an array of any size, on account of actually taking a pointer. It may even be passed nullptr. While a function taking int(&)[2] may only be given a valid array of two as an argument.

Since in the working version init refers to a int[2] object, structured bindings can work with that array object. But a decayed pointer cannot be the subject of structured bindings, because the static type information available only gives access to a single element being pointed at.

share|improve this answer

answered 4 hours ago

StoryTellerStoryTeller

108k16227290

share|improve this answer

answered 4 hours ago

StoryTellerStoryTeller

108k16227290

answered 4 hours ago

StoryTellerStoryTeller

108k16227290

answered 4 hours ago

StoryTellerStoryTeller

108k16227290