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7

I am supposed to determine whether this function is syntactically correct:

int f3(int i, int j) { const int& k=i; ++i; return k; }

I have tested it out and it compiles with my main function.

I do not understand why this is so.

Surely by calling the function f3 I create copies of the variables iand j in a new memory space and setting const int& k=i I am setting the memory space of the newly created k to the exact the same space of the memory space of the copied i, therefore any change, i.e. the increment ++iwill result in ++k which is not possible given that it was set const

Any help is greatly appreciated

c++ syntax reference const

share|improve this question

asked 7 hours ago

user9078057user9078057

1168

add a comment | 

7

I am supposed to determine whether this function is syntactically correct:

int f3(int i, int j) { const int& k=i; ++i; return k; }

I have tested it out and it compiles with my main function.

I do not understand why this is so.

Surely by calling the function f3 I create copies of the variables iand j in a new memory space and setting const int& k=i I am setting the memory space of the newly created k to the exact the same space of the memory space of the copied i, therefore any change, i.e. the increment ++iwill result in ++k which is not possible given that it was set const

Any help is greatly appreciated

c++ syntax reference const

share|improve this question

asked 7 hours ago

user9078057user9078057

1168

add a comment | 

I am supposed to determine whether this function is syntactically correct:

int f3(int i, int j) { const int& k=i; ++i; return k; }

I have tested it out and it compiles with my main function.

I do not understand why this is so.

Surely by calling the function f3 I create copies of the variables iand j in a new memory space and setting const int& k=i I am setting the memory space of the newly created k to the exact the same space of the memory space of the copied i, therefore any change, i.e. the increment ++iwill result in ++k which is not possible given that it was set const

Any help is greatly appreciated

c++ syntax reference const

share|improve this question

asked 7 hours ago

user9078057user9078057

1168

share|improve this question

asked 7 hours ago

user9078057user9078057

1168

share|improve this question

asked 7 hours ago

user9078057user9078057

1168

asked 7 hours ago

user9078057user9078057

1168

asked 7 hours ago

user9078057user9078057

1168

1 Answer
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15

the increment ++iwill result in ++k which is not possible given that it was set const

That's a misunderstanding.

You may not change the value of the object through k but it can still be changed through other means. In other words, ++k is not allowed but ++i is still allowed, which will indirectly modify the value of k.

Here's a far-fetched analogy.

You may look through the window of a store and see what's inside but you won't be able to change what's inside the store. However, an employee, who is inside the store, can change the contents of the store. You will see that change from
outside. You have const access or view access to the store while the employee has non-const access or change access to the store.

share|improve this answer

edited 7 hours ago

answered 7 hours ago

R SahuR Sahu

169k1294193

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  Not so far-fetched. That's a good analogy.
  
  – user4581301
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.
  
  – David Schwartz
  7 hours ago
  

  
  
  
  

add a comment | 

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15

the increment ++iwill result in ++k which is not possible given that it was set const

That's a misunderstanding.

You may not change the value of the object through k but it can still be changed through other means. In other words, ++k is not allowed but ++i is still allowed, which will indirectly modify the value of k.

Here's a far-fetched analogy.

You may look through the window of a store and see what's inside but you won't be able to change what's inside the store. However, an employee, who is inside the store, can change the contents of the store. You will see that change from
outside. You have const access or view access to the store while the employee has non-const access or change access to the store.

share|improve this answer

edited 7 hours ago

answered 7 hours ago

R SahuR Sahu

169k1294193

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  Not so far-fetched. That's a good analogy.
  
  – user4581301
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.
  
  – David Schwartz
  7 hours ago
  

  
  
  
  

add a comment | 

15

the increment ++iwill result in ++k which is not possible given that it was set const

That's a misunderstanding.

You may not change the value of the object through k but it can still be changed through other means. In other words, ++k is not allowed but ++i is still allowed, which will indirectly modify the value of k.

Here's a far-fetched analogy.

You may look through the window of a store and see what's inside but you won't be able to change what's inside the store. However, an employee, who is inside the store, can change the contents of the store. You will see that change from
outside. You have const access or view access to the store while the employee has non-const access or change access to the store.

share|improve this answer

edited 7 hours ago

answered 7 hours ago

R SahuR Sahu

169k1294193

• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  
  Not so far-fetched. That's a good analogy.
  
  – user4581301
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Note that this is so because i itself is not const. So we have a const reference a non-const object. Const objects are different.
  
  – David Schwartz
  7 hours ago
  

  
  
  
  

add a comment | 

the increment ++iwill result in ++k which is not possible given that it was set const

That's a misunderstanding.

You may not change the value of the object through k but it can still be changed through other means. In other words, ++k is not allowed but ++i is still allowed, which will indirectly modify the value of k.

Here's a far-fetched analogy.

You may look through the window of a store and see what's inside but you won't be able to change what's inside the store. However, an employee, who is inside the store, can change the contents of the store. You will see that change from
outside. You have const access or view access to the store while the employee has non-const access or change access to the store.

share|improve this answer

edited 7 hours ago

answered 7 hours ago

R SahuR Sahu

169k1294193

share|improve this answer

edited 7 hours ago

answered 7 hours ago

R SahuR Sahu

169k1294193

answered 7 hours ago

R SahuR Sahu

169k1294193

answered 7 hours ago

R SahuR Sahu

169k1294193