Crthuky

What to do with wrong results in talks?

Your magic is very sketchy

Valid Badminton Score?

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

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

At which point does a character regain all their Hit Dice?

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

What does this 7 mean above the f flat

Applicability of Single Responsibility Principle

The baby cries all morning

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

Is this Spell Mimic feat balanced?

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

Hide Select Output from T-SQL

Curses work by shouting - How to avoid collateral damage?

Efficiently merge handle parallel feature branches in SFDX

Go Pregnant or Go Home

Personal Teleportation as a Weapon

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

Confused about a passage in Harry Potter y la piedra filosofal

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

Is expanding the research of a group into machine learning as a PhD student risky?

apt-get update is failing in debian

How does residential electricity work?

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?

6

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

1118

add a comment | 

6

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

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

share|improve this question

asked 7 hours ago

user9078057user9078057

1118

share|improve this question

asked 7 hours ago

user9078057user9078057

1118

share|improve this question

asked 7 hours ago

user9078057user9078057

1118

asked 7 hours ago

user9078057user9078057

1118

asked 7 hours ago

user9078057user9078057

1118

1 Answer
1

active

oldest

votes

14

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
  6 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');

var $window = $(window),
onScroll = function(e)
var $elem = $('.new-login-left'),
docViewTop = $window.scrollTop(),
docViewBottom = docViewTop + $window.height(),
elemTop = $elem.offset().top,
elemBottom = elemTop + $elem.height();
if ((docViewTop elemBottom))
StackExchange.using('gps', function() StackExchange.gps.track('embedded_signup_form.view', location: 'question_page' ); );
$window.unbind('scroll', onScroll);

;
$window.on('scroll', onScroll);

);

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

14

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
  6 hours ago
  

  
  
  
  

add a comment | 

14

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
  6 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