Why is `const T&` not sure to be const?

template<typename T>

void f(T a, const T& b)

{

    ++a; // ok

    ++b; // also ok!

}



template<typename T>

void g(T n)

{

    f<T>(n, n);

}



int main()

{

    int n{};

    g<int&>(n);

}

Please note: b is of const T& and ++b is ok!

Why is const T& not sure to be const?

edited Feb 2 at 3:30

curiousguy

4,63623046

asked Feb 1 at 13:32

xmllmx

13.9k988212

add a comment |

template<typename T>

void f(T a, const T& b)

{

    ++a; // ok

    ++b; // also ok!

}



template<typename T>

void g(T n)

{

    f<T>(n, n);

}



int main()

{

    int n{};

    g<int&>(n);

}

Please note: b is of const T& and ++b is ok!

Why is const T& not sure to be const?

edited Feb 2 at 3:30

curiousguy

4,63623046

asked Feb 1 at 13:32

xmllmx

13.9k988212

add a comment |

template<typename T>

void f(T a, const T& b)

{

    ++a; // ok

    ++b; // also ok!

}



template<typename T>

void g(T n)

{

    f<T>(n, n);

}



int main()

{

    int n{};

    g<int&>(n);

}

Please note: b is of const T& and ++b is ok!

Why is const T& not sure to be const?

edited Feb 2 at 3:30

curiousguy

4,63623046

asked Feb 1 at 13:32

xmllmx

13.9k988212

template<typename T>

void f(T a, const T& b)

{

    ++a; // ok

    ++b; // also ok!

}



template<typename T>

void g(T n)

{

    f<T>(n, n);

}



int main()

{

    int n{};

    g<int&>(n);

}

Please note: b is of const T& and ++b is ok!

Why is const T& not sure to be const?

c++ c++11 const function-templates const-reference

edited Feb 2 at 3:30

curiousguy

4,63623046

asked Feb 1 at 13:32

xmllmx

13.9k988212

edited Feb 2 at 3:30

curiousguy

4,63623046

asked Feb 1 at 13:32

xmllmx

13.9k988212

edited Feb 2 at 3:30

curiousguy

4,63623046

edited Feb 2 at 3:30

curiousguy

4,63623046

edited Feb 2 at 3:30

curiousguy

4,63623046

asked Feb 1 at 13:32

xmllmx

13.9k988212

asked Feb 1 at 13:32

xmllmx

13.9k988212

asked Feb 1 at 13:32

xmllmx

13.9k988212

add a comment |

2 Answers
2

active

oldest

votes

Welcome to const and reference collapsing. When you have const T&, the reference gets applied to T, and so does the const. You call g like

g<int&>(n);

so you have specified that T is a int&. When we apply a reference to an lvalue reference, the two references collapse to a single one, so int& & becomes just int&. Then we get to the rule from [dcl.ref]/1, which states that if you apply const to a reference it is discarded, so int& const just becomes int& (note that you can't actually declare int& const, it has to come from a typedef or template). That means for

g<int&>(n);

you are actually calling

void f(int& a, int& b)

and you are not actually modifying a constant.

Had you called g as

g<int>(n);

// or just

g(n);

then T would be int, and f would have been stamped out as

void f(int a, const int& b)

Since T isn't a reference anymore, the const and the & get applied to it, and you would have received a compiler error for trying to modify a constant variable.

edited Feb 2 at 3:19

Fabio Turati

2,63452341

answered Feb 1 at 13:36

NathanOliver

95.3k16134205

7

This is why the type traits like std::add_lvalue_reference exist, to ensure that references are added in a predictable way to prevent just this sort of pain.

– Mgetz
Feb 1 at 14:00

9

One way to make it easier to understand is to write T const& instead of const T& (which is the same), and then replace T with int&.

– Ruslan
Feb 1 at 15:38

I would reorder some of the first part of this answer, since in the type const T&, first the const applies to T, and then the lvalue-reference applies to the result of that. (If the rule were the opposite, the "const applied to a reference type is ignored" rule would always kick in, and const T& would always mean the same as T&.)

– aschepler
Feb 1 at 22:22

@aschepler The rules stop T& const, not const T&/T const &

– NathanOliver
Feb 1 at 22:24

@NathanOliver I'm just suggesting discussing the effect of const before the effect of lvalue-reference, since using U = const T&; means the same as using tmp1 = const T; using U = tmp1&; but not the same as using tmp2 = T&; using U = const tmp2;

– aschepler
Feb 1 at 22:30

|
show 3 more comments

-3

I know that there is already an accepted answer which is correct but just to add to it a little bit, even outside the realm of templates and just in function declarations in general...

( const T& )

is not the same as

( const T )

In your example which matches the first, you have a const reference. If you truly want a const value that is not modifiable remove the reference as in the second example.

answered Feb 1 at 14:17

Francis Cugler

4,81411227

8

when T is int&, const T& and const T both give int&.

– Ben Voigt
Feb 1 at 14:52

I think there was a misconception on what I was trying to say; I'm using T here not as a template parameter. T was just meant to be used as any data type: int, float, double etc.. so T in my example above should never be int&. I did specifically state outside the realm of templates.

– Francis Cugler
Feb 1 at 22:08

Hmm, it sounds as if you are explaining that you can see this without templates, which is no problem. But then your final sentence claims to offer a solution to OP's problem, and that problem definitely involves templates. It's fine to offer a solution to a template question which is broader than just templates. But a solution to a template question that isn't accurate for templates, seems not to answer the question.

– Ben Voigt
Feb 1 at 22:16

This can also be an issue with no templates involved: using T = int&; void f(const T&); declares void f(int&);.

– aschepler
Feb 1 at 22:23

@aschepler true, but I wasn't referring to using clauses; just basic function declarations in general.

– Francis Cugler
Feb 2 at 0:20

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');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f54480599%2fwhy-is-const-t-not-sure-to-be-const%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Welcome to const and reference collapsing. When you have const T&, the reference gets applied to T, and so does the const. You call g like

g<int&>(n);

g<int&>(n);

you are actually calling

void f(int& a, int& b)

and you are not actually modifying a constant.

Had you called g as

g<int>(n);

// or just

g(n);

then T would be int, and f would have been stamped out as

void f(int a, const int& b)

Since T isn't a reference anymore, the const and the & get applied to it, and you would have received a compiler error for trying to modify a constant variable.

edited Feb 2 at 3:19

Fabio Turati

2,63452341

answered Feb 1 at 13:36

NathanOliver

95.3k16134205

7

This is why the type traits like std::add_lvalue_reference exist, to ensure that references are added in a predictable way to prevent just this sort of pain.

– Mgetz
Feb 1 at 14:00

9

One way to make it easier to understand is to write T const& instead of const T& (which is the same), and then replace T with int&.

– Ruslan
Feb 1 at 15:38

I would reorder some of the first part of this answer, since in the type const T&, first the const applies to T, and then the lvalue-reference applies to the result of that. (If the rule were the opposite, the "const applied to a reference type is ignored" rule would always kick in, and const T& would always mean the same as T&.)

– aschepler
Feb 1 at 22:22

@aschepler The rules stop T& const, not const T&/T const &

– NathanOliver
Feb 1 at 22:24

@NathanOliver I'm just suggesting discussing the effect of const before the effect of lvalue-reference, since using U = const T&; means the same as using tmp1 = const T; using U = tmp1&; but not the same as using tmp2 = T&; using U = const tmp2;

– aschepler
Feb 1 at 22:30

|
show 3 more comments

Welcome to const and reference collapsing. When you have const T&, the reference gets applied to T, and so does the const. You call g like

g<int&>(n);

g<int&>(n);

you are actually calling

void f(int& a, int& b)

and you are not actually modifying a constant.

Had you called g as

g<int>(n);

// or just

g(n);

then T would be int, and f would have been stamped out as

void f(int a, const int& b)

Since T isn't a reference anymore, the const and the & get applied to it, and you would have received a compiler error for trying to modify a constant variable.

edited Feb 2 at 3:19

Fabio Turati

2,63452341

answered Feb 1 at 13:36

NathanOliver

95.3k16134205

7

This is why the type traits like std::add_lvalue_reference exist, to ensure that references are added in a predictable way to prevent just this sort of pain.

– Mgetz
Feb 1 at 14:00

9

One way to make it easier to understand is to write T const& instead of const T& (which is the same), and then replace T with int&.

– Ruslan
Feb 1 at 15:38

I would reorder some of the first part of this answer, since in the type const T&, first the const applies to T, and then the lvalue-reference applies to the result of that. (If the rule were the opposite, the "const applied to a reference type is ignored" rule would always kick in, and const T& would always mean the same as T&.)

– aschepler
Feb 1 at 22:22

@aschepler The rules stop T& const, not const T&/T const &

– NathanOliver
Feb 1 at 22:24

@NathanOliver I'm just suggesting discussing the effect of const before the effect of lvalue-reference, since using U = const T&; means the same as using tmp1 = const T; using U = tmp1&; but not the same as using tmp2 = T&; using U = const tmp2;

– aschepler
Feb 1 at 22:30

|
show 3 more comments

Welcome to const and reference collapsing. When you have const T&, the reference gets applied to T, and so does the const. You call g like

g<int&>(n);

g<int&>(n);

you are actually calling

void f(int& a, int& b)

and you are not actually modifying a constant.

Had you called g as

g<int>(n);

// or just

g(n);

then T would be int, and f would have been stamped out as

void f(int a, const int& b)

Since T isn't a reference anymore, the const and the & get applied to it, and you would have received a compiler error for trying to modify a constant variable.

edited Feb 2 at 3:19

Fabio Turati

2,63452341

answered Feb 1 at 13:36

NathanOliver

95.3k16134205

Welcome to const and reference collapsing. When you have const T&, the reference gets applied to T, and so does the const. You call g like

g<int&>(n);

g<int&>(n);

you are actually calling

void f(int& a, int& b)

and you are not actually modifying a constant.

Had you called g as

g<int>(n);

// or just

g(n);

then T would be int, and f would have been stamped out as

void f(int a, const int& b)

Since T isn't a reference anymore, the const and the & get applied to it, and you would have received a compiler error for trying to modify a constant variable.

edited Feb 2 at 3:19

Fabio Turati

2,63452341

answered Feb 1 at 13:36

NathanOliver

95.3k16134205

edited Feb 2 at 3:19

Fabio Turati

2,63452341

edited Feb 2 at 3:19

Fabio Turati

2,63452341

edited Feb 2 at 3:19

Fabio Turati

2,63452341

answered Feb 1 at 13:36

NathanOliver

95.3k16134205

answered Feb 1 at 13:36

NathanOliver

95.3k16134205

answered Feb 1 at 13:36

NathanOliver

95.3k16134205

7

This is why the type traits like std::add_lvalue_reference exist, to ensure that references are added in a predictable way to prevent just this sort of pain.

– Mgetz
Feb 1 at 14:00

9

One way to make it easier to understand is to write T const& instead of const T& (which is the same), and then replace T with int&.

– Ruslan
Feb 1 at 15:38

I would reorder some of the first part of this answer, since in the type const T&, first the const applies to T, and then the lvalue-reference applies to the result of that. (If the rule were the opposite, the "const applied to a reference type is ignored" rule would always kick in, and const T& would always mean the same as T&.)

– aschepler
Feb 1 at 22:22

@aschepler The rules stop T& const, not const T&/T const &

– NathanOliver
Feb 1 at 22:24

@NathanOliver I'm just suggesting discussing the effect of const before the effect of lvalue-reference, since using U = const T&; means the same as using tmp1 = const T; using U = tmp1&; but not the same as using tmp2 = T&; using U = const tmp2;

– aschepler
Feb 1 at 22:30

|
show 3 more comments

7

This is why the type traits like std::add_lvalue_reference exist, to ensure that references are added in a predictable way to prevent just this sort of pain.

– Mgetz
Feb 1 at 14:00

9

One way to make it easier to understand is to write T const& instead of const T& (which is the same), and then replace T with int&.

– Ruslan
Feb 1 at 15:38

I would reorder some of the first part of this answer, since in the type const T&, first the const applies to T, and then the lvalue-reference applies to the result of that. (If the rule were the opposite, the "const applied to a reference type is ignored" rule would always kick in, and const T& would always mean the same as T&.)

– aschepler
Feb 1 at 22:22

@aschepler The rules stop T& const, not const T&/T const &

– NathanOliver
Feb 1 at 22:24

@NathanOliver I'm just suggesting discussing the effect of const before the effect of lvalue-reference, since using U = const T&; means the same as using tmp1 = const T; using U = tmp1&; but not the same as using tmp2 = T&; using U = const tmp2;

– aschepler
Feb 1 at 22:30

This is why the type traits like std::add_lvalue_reference exist, to ensure that references are added in a predictable way to prevent just this sort of pain.

– Mgetz
Feb 1 at 14:00

One way to make it easier to understand is to write T const& instead of const T& (which is the same), and then replace T with int&.

– Ruslan
Feb 1 at 15:38

I would reorder some of the first part of this answer, since in the type const T&, first the const applies to T, and then the lvalue-reference applies to the result of that. (If the rule were the opposite, the "const applied to a reference type is ignored" rule would always kick in, and const T& would always mean the same as T&.)

– aschepler
Feb 1 at 22:22

@aschepler The rules stop T& const, not const T&/T const &

– NathanOliver
Feb 1 at 22:24

@NathanOliver I'm just suggesting discussing the effect of const before the effect of lvalue-reference, since using U = const T&; means the same as using tmp1 = const T; using U = tmp1&; but not the same as using tmp2 = T&; using U = const tmp2;

– aschepler
Feb 1 at 22:30

|
show 3 more comments

-3

I know that there is already an accepted answer which is correct but just to add to it a little bit, even outside the realm of templates and just in function declarations in general...

( const T& )

is not the same as

( const T )

In your example which matches the first, you have a const reference. If you truly want a const value that is not modifiable remove the reference as in the second example.

answered Feb 1 at 14:17

Francis Cugler

4,81411227

8

when T is int&, const T& and const T both give int&.

– Ben Voigt
Feb 1 at 14:52

I think there was a misconception on what I was trying to say; I'm using T here not as a template parameter. T was just meant to be used as any data type: int, float, double etc.. so T in my example above should never be int&. I did specifically state outside the realm of templates.

– Francis Cugler
Feb 1 at 22:08

Hmm, it sounds as if you are explaining that you can see this without templates, which is no problem. But then your final sentence claims to offer a solution to OP's problem, and that problem definitely involves templates. It's fine to offer a solution to a template question which is broader than just templates. But a solution to a template question that isn't accurate for templates, seems not to answer the question.

– Ben Voigt
Feb 1 at 22:16

This can also be an issue with no templates involved: using T = int&; void f(const T&); declares void f(int&);.

– aschepler
Feb 1 at 22:23

@aschepler true, but I wasn't referring to using clauses; just basic function declarations in general.

– Francis Cugler
Feb 2 at 0:20

add a comment |

-3

I know that there is already an accepted answer which is correct but just to add to it a little bit, even outside the realm of templates and just in function declarations in general...

( const T& )

is not the same as

( const T )

In your example which matches the first, you have a const reference. If you truly want a const value that is not modifiable remove the reference as in the second example.

answered Feb 1 at 14:17

Francis Cugler

4,81411227

8

when T is int&, const T& and const T both give int&.

– Ben Voigt
Feb 1 at 14:52

I think there was a misconception on what I was trying to say; I'm using T here not as a template parameter. T was just meant to be used as any data type: int, float, double etc.. so T in my example above should never be int&. I did specifically state outside the realm of templates.

– Francis Cugler
Feb 1 at 22:08

Hmm, it sounds as if you are explaining that you can see this without templates, which is no problem. But then your final sentence claims to offer a solution to OP's problem, and that problem definitely involves templates. It's fine to offer a solution to a template question which is broader than just templates. But a solution to a template question that isn't accurate for templates, seems not to answer the question.

– Ben Voigt
Feb 1 at 22:16

This can also be an issue with no templates involved: using T = int&; void f(const T&); declares void f(int&);.

– aschepler
Feb 1 at 22:23

@aschepler true, but I wasn't referring to using clauses; just basic function declarations in general.

– Francis Cugler
Feb 2 at 0:20

add a comment |

-3

I know that there is already an accepted answer which is correct but just to add to it a little bit, even outside the realm of templates and just in function declarations in general...

( const T& )

is not the same as

( const T )

In your example which matches the first, you have a const reference. If you truly want a const value that is not modifiable remove the reference as in the second example.

answered Feb 1 at 14:17

Francis Cugler

4,81411227

I know that there is already an accepted answer which is correct but just to add to it a little bit, even outside the realm of templates and just in function declarations in general...

( const T& )

is not the same as

( const T )

In your example which matches the first, you have a const reference. If you truly want a const value that is not modifiable remove the reference as in the second example.

answered Feb 1 at 14:17

Francis Cugler

4,81411227

answered Feb 1 at 14:17

Francis Cugler

4,81411227

answered Feb 1 at 14:17

Francis Cugler

4,81411227

answered Feb 1 at 14:17

Francis Cugler

4,81411227

8

when T is int&, const T& and const T both give int&.

– Ben Voigt
Feb 1 at 14:52

I think there was a misconception on what I was trying to say; I'm using T here not as a template parameter. T was just meant to be used as any data type: int, float, double etc.. so T in my example above should never be int&. I did specifically state outside the realm of templates.

– Francis Cugler
Feb 1 at 22:08

Hmm, it sounds as if you are explaining that you can see this without templates, which is no problem. But then your final sentence claims to offer a solution to OP's problem, and that problem definitely involves templates. It's fine to offer a solution to a template question which is broader than just templates. But a solution to a template question that isn't accurate for templates, seems not to answer the question.

– Ben Voigt
Feb 1 at 22:16

This can also be an issue with no templates involved: using T = int&; void f(const T&); declares void f(int&);.

– aschepler
Feb 1 at 22:23

@aschepler true, but I wasn't referring to using clauses; just basic function declarations in general.

– Francis Cugler
Feb 2 at 0:20

add a comment |

8

when T is int&, const T& and const T both give int&.

– Ben Voigt
Feb 1 at 14:52

I think there was a misconception on what I was trying to say; I'm using T here not as a template parameter. T was just meant to be used as any data type: int, float, double etc.. so T in my example above should never be int&. I did specifically state outside the realm of templates.

– Francis Cugler
Feb 1 at 22:08

Hmm, it sounds as if you are explaining that you can see this without templates, which is no problem. But then your final sentence claims to offer a solution to OP's problem, and that problem definitely involves templates. It's fine to offer a solution to a template question which is broader than just templates. But a solution to a template question that isn't accurate for templates, seems not to answer the question.

– Ben Voigt
Feb 1 at 22:16

This can also be an issue with no templates involved: using T = int&; void f(const T&); declares void f(int&);.

– aschepler
Feb 1 at 22:23

@aschepler true, but I wasn't referring to using clauses; just basic function declarations in general.

– Francis Cugler
Feb 2 at 0:20

when T is int&, const T& and const T both give int&.

– Ben Voigt
Feb 1 at 14:52

I think there was a misconception on what I was trying to say; I'm using T here not as a template parameter. T was just meant to be used as any data type: int, float, double etc.. so T in my example above should never be int&. I did specifically state outside the realm of templates.

– Francis Cugler
Feb 1 at 22:08

Hmm, it sounds as if you are explaining that you can see this without templates, which is no problem. But then your final sentence claims to offer a solution to OP's problem, and that problem definitely involves templates. It's fine to offer a solution to a template question which is broader than just templates. But a solution to a template question that isn't accurate for templates, seems not to answer the question.

– Ben Voigt
Feb 1 at 22:16

This can also be an issue with no templates involved: using T = int&; void f(const T&); declares void f(int&);.

– aschepler
Feb 1 at 22:23

@aschepler true, but I wasn't referring to using clauses; just basic function declarations in general.

– Francis Cugler
Feb 2 at 0:20

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

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

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

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

Post as a guest

Name

Required, but never shown

Sign up or log in

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

Post as a guest

Name

Required, but never shown

Sign up or log in

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

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Dtyjlui