Why is `const int& k = i; ++i; ` possible?
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
asked 7 hours ago
user9078057user9078057
1118
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
asked 7 hours ago
user9078057user9078057
1118
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
asked 7 hours ago
user9078057user9078057
1118
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
c++ syntax reference const
asked 7 hours ago
user9078057user9078057
1118
asked 7 hours ago
user9078057user9078057
1118
asked 7 hours ago
user9078057user9078057
1118
asked 7 hours ago
user9078057user9078057
1118
asked 7 hours ago
user9078057user9078057
1118
1118
add a comment |
add a comment |
1 Answer
1
active
oldest
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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.
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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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
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.
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.
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.
answered 7 hours ago
R SahuR Sahu
169k1294193
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.
answered 7 hours ago
R SahuR Sahu
169k1294193
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
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 |
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
5
5
Not so far-fetched. That's a good analogy.
– user4581301
7 hours ago
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
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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