Mixing
Function declaration with auto using new C++11 syntax but with auto& and without ->
consider the function definition below:
auto& Fnc1() { return someNonLocalVariable; }
Return type is not explicitly specified by -> in this case.
But there is the & after auto keyword.
Does this guarantee that a reference is returned instead of copy of the variable?
Is this a supported language feature (returning reference)?
With VS 2017, it works as I expect: Return a reference.
But I could not find any online source to verify.
c++ c++14 auto function-declaration return-type-deduction
edited Jan 1 at 20:51
Deduplicator
34.7k64990
asked Jan 1 at 20:27
mamimami
4916
add a comment |
consider the function definition below:
auto& Fnc1() { return someNonLocalVariable; }
Return type is not explicitly specified by -> in this case.
But there is the & after auto keyword.
Does this guarantee that a reference is returned instead of copy of the variable?
Is this a supported language feature (returning reference)?
With VS 2017, it works as I expect: Return a reference.
But I could not find any online source to verify.
c++ c++14 auto function-declaration return-type-deduction
edited Jan 1 at 20:51
Deduplicator
34.7k64990
asked Jan 1 at 20:27
mamimami
4916
This is how C++ works. If a function returns a reference, that's what it returns. There are no exceptions.
– Sam Varshavchik
Jan 1 at 20:29
2
That's a C++14+ code. Not C++11. RVO is guaranteed since C++17.
– Ron
Jan 1 at 20:29
usual syntax is said to be auto Fnc1() {...} Without &. I could not find any source that says I can use auto& beside auto in this context
– mami
Jan 1 at 20:33
2
The source
– Igor Tandetnik
Jan 1 at 20:36
autois for deducing the "base" type only, without CV-qualifiers or references.
– Some programmer dude
Jan 1 at 20:36
add a comment |
consider the function definition below:
auto& Fnc1() { return someNonLocalVariable; }
Return type is not explicitly specified by -> in this case.
But there is the & after auto keyword.
Does this guarantee that a reference is returned instead of copy of the variable?
Is this a supported language feature (returning reference)?
With VS 2017, it works as I expect: Return a reference.
But I could not find any online source to verify.
c++ c++14 auto function-declaration return-type-deduction
edited Jan 1 at 20:51
Deduplicator
34.7k64990
asked Jan 1 at 20:27
mamimami
4916
consider the function definition below:
auto& Fnc1() { return someNonLocalVariable; }
Return type is not explicitly specified by -> in this case.
But there is the & after auto keyword.
Does this guarantee that a reference is returned instead of copy of the variable?
Is this a supported language feature (returning reference)?
With VS 2017, it works as I expect: Return a reference.
But I could not find any online source to verify.
c++ c++14 auto function-declaration return-type-deduction
c++ c++14 auto function-declaration return-type-deduction
edited Jan 1 at 20:51
Deduplicator
34.7k64990
asked Jan 1 at 20:27
mamimami
4916
edited Jan 1 at 20:51
Deduplicator
34.7k64990
asked Jan 1 at 20:27
mamimami
4916
edited Jan 1 at 20:51
Deduplicator
34.7k64990
edited Jan 1 at 20:51
Deduplicator
34.7k64990
edited Jan 1 at 20:51
Deduplicator
34.7k64990
34.7k64990
asked Jan 1 at 20:27
mamimami
4916
asked Jan 1 at 20:27
mamimami
4916
asked Jan 1 at 20:27
mamimami
4916
4916
This is how C++ works. If a function returns a reference, that's what it returns. There are no exceptions.
– Sam Varshavchik
Jan 1 at 20:29
2
That's a C++14+ code. Not C++11. RVO is guaranteed since C++17.
– Ron
Jan 1 at 20:29
usual syntax is said to be auto Fnc1() {...} Without &. I could not find any source that says I can use auto& beside auto in this context
– mami
Jan 1 at 20:33
2
The source
– Igor Tandetnik
Jan 1 at 20:36
autois for deducing the "base" type only, without CV-qualifiers or references.
– Some programmer dude
Jan 1 at 20:36
add a comment |
This is how C++ works. If a function returns a reference, that's what it returns. There are no exceptions.
– Sam Varshavchik
Jan 1 at 20:29
2
That's a C++14+ code. Not C++11. RVO is guaranteed since C++17.
– Ron
Jan 1 at 20:29
usual syntax is said to be auto Fnc1() {...} Without &. I could not find any source that says I can use auto& beside auto in this context
– mami
Jan 1 at 20:33
2
The source
– Igor Tandetnik
Jan 1 at 20:36
autois for deducing the "base" type only, without CV-qualifiers or references.
– Some programmer dude
Jan 1 at 20:36
This is how C++ works. If a function returns a reference, that's what it returns. There are no exceptions.
– Sam Varshavchik
Jan 1 at 20:29
This is how C++ works. If a function returns a reference, that's what it returns. There are no exceptions.
– Sam Varshavchik
Jan 1 at 20:29
2
2
That's a C++14+ code. Not C++11. RVO is guaranteed since C++17.
– Ron
Jan 1 at 20:29
That's a C++14+ code. Not C++11. RVO is guaranteed since C++17.
– Ron
Jan 1 at 20:29
usual syntax is said to be auto Fnc1() {...} Without &. I could not find any source that says I can use auto& beside auto in this context
– mami
Jan 1 at 20:33
usual syntax is said to be auto Fnc1() {...} Without &. I could not find any source that says I can use auto& beside auto in this context
– mami
Jan 1 at 20:33
2
2
The source
– Igor Tandetnik
Jan 1 at 20:36
The source
– Igor Tandetnik
Jan 1 at 20:36
auto is for deducing the "base" type only, without CV-qualifiers or references.– Some programmer dude
Jan 1 at 20:36
auto is for deducing the "base" type only, without CV-qualifiers or references.– Some programmer dude
Jan 1 at 20:36
add a comment |
1 Answer
1
active
oldest
votes
Yes, this is the correct specification, and as Ron commented, became operational in C++14. C++11 did require trailing return type syntax (-> after the parameters).
See cppreference for more.
In a function declaration that does not use the trailing return type syntax, the keyword
autoindicates that the return type will be deduced from the operand of its return statement using the rules for template argument deduction.
See also auto type deduction where they include an example using auto&. You can think of this as being the same set of rules for declaring variables using the auto keyword.
answered Jan 1 at 20:36
rsjaffersjaffe
3,95671632
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
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active
oldest
votes
Yes, this is the correct specification, and as Ron commented, became operational in C++14. C++11 did require trailing return type syntax (-> after the parameters).
See cppreference for more.
In a function declaration that does not use the trailing return type syntax, the keyword
autoindicates that the return type will be deduced from the operand of its return statement using the rules for template argument deduction.
See also auto type deduction where they include an example using auto&. You can think of this as being the same set of rules for declaring variables using the auto keyword.
answered Jan 1 at 20:36
rsjaffersjaffe
3,95671632
add a comment |
Yes, this is the correct specification, and as Ron commented, became operational in C++14. C++11 did require trailing return type syntax (-> after the parameters).
See cppreference for more.
In a function declaration that does not use the trailing return type syntax, the keyword
autoindicates that the return type will be deduced from the operand of its return statement using the rules for template argument deduction.
See also auto type deduction where they include an example using auto&. You can think of this as being the same set of rules for declaring variables using the auto keyword.
answered Jan 1 at 20:36
rsjaffersjaffe
3,95671632
add a comment |
Yes, this is the correct specification, and as Ron commented, became operational in C++14. C++11 did require trailing return type syntax (-> after the parameters).
See cppreference for more.
In a function declaration that does not use the trailing return type syntax, the keyword
autoindicates that the return type will be deduced from the operand of its return statement using the rules for template argument deduction.
See also auto type deduction where they include an example using auto&. You can think of this as being the same set of rules for declaring variables using the auto keyword.
answered Jan 1 at 20:36
rsjaffersjaffe
3,95671632
Yes, this is the correct specification, and as Ron commented, became operational in C++14. C++11 did require trailing return type syntax (-> after the parameters).
See cppreference for more.
In a function declaration that does not use the trailing return type syntax, the keyword
autoindicates that the return type will be deduced from the operand of its return statement using the rules for template argument deduction.
See also auto type deduction where they include an example using auto&. You can think of this as being the same set of rules for declaring variables using the auto keyword.
answered Jan 1 at 20:36
rsjaffersjaffe
3,95671632
edited Jan 1 at 21:06
answered Jan 1 at 20:36
rsjaffersjaffe
3,95671632
answered Jan 1 at 20:36
rsjaffersjaffe
3,95671632
answered Jan 1 at 20:36
rsjaffersjaffe
3,95671632
3,95671632
add a comment |
add a comment |
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This is how C++ works. If a function returns a reference, that's what it returns. There are no exceptions.
– Sam Varshavchik
Jan 1 at 20:29
2
That's a C++14+ code. Not C++11. RVO is guaranteed since C++17.
– Ron
Jan 1 at 20:29
usual syntax is said to be auto Fnc1() {...} Without &. I could not find any source that says I can use auto& beside auto in this context
– mami
Jan 1 at 20:33
2
The source
– Igor Tandetnik
Jan 1 at 20:36
autois for deducing the "base" type only, without CV-qualifiers or references.– Some programmer dude
Jan 1 at 20:36