Zrjydtj

Your magic is very sketchy

Is it okay / does it make sense for another player to join a running game of Munchkin?

Is there any reason not to eat food that's been dropped on the surface of the moon?

How do I define a right arrow with bar in LaTeX?

Are there any comparative studies done between Ashtavakra Gita and Buddhim?

Can I use my Chinese passport to enter China after I acquired another citizenship?

Can criminal fraud exist without damages?

How to be diplomatic in refusing to write code that breaches the privacy of our users

Can somebody explain Brexit in a few child-proof sentences?

Understanding "audieritis" in Psalm 94

Student evaluations of teaching assistants

Is the destination of a commercial flight important for the pilot?

How can I use the arrow sign in my bash prompt?

Why "be dealt cards" rather than "be dealing cards"?

Lay out the Carpet

If you attempt to grapple an opponent that you are hidden from, do they roll at disadvantage?

What would happen if the UK refused to take part in EU Parliamentary elections?

Why is `const int& k = i; ++i; ` possible?

Time travel short story where a man arrives in the late 19th century in a time machine and then sends the machine back into the past

How do I keep an essay about "feeling flat" from feeling flat?

Failed to fetch jessie backports repository

Is there a good way to store credentials outside of a password manager?

Will it be accepted, if there is no ''Main Character" stereotype?

Ways to speed up user implemented RK4

Why is `const int& k = i; ++i; ` possible?

What is the difference between const and readonly?How to convert a std::string to const char* or char*?Why can templates only be implemented in the header file?Meaning of 'const' last in a function declaration of a class?What is the difference between const int*, const int * const, and int const *?Why is “using namespace std” considered bad practice?define() vs constWhy does ++[[]][+[]]+[+[]] return the string “10”?Why are elementwise additions much faster in separate loops than in a combined loop?Why is it faster to process a sorted array than an unsorted array?

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
1

active

oldest

votes

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 | 

Your Answer

StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55367062%2fwhy-is-const-int-k-i-i-possible%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

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