Is it possible to use function return type as an argument in declaration of another function in C++?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
23
down vote

favorite












I want something like this:



std::tuple<int, bool, double> MyFunction_1 (void);

void MyFunction_2 (decltype (MyFunction_1) &params);


Obviously, in this example a code pointer to function would be passed.



I want to have the equivalent of this:



void MyFunction_2 (std::tuple<int, bool, double> &params);


Is it possible to do so?






c++ c++11 decltype







share|improve this question



















  • 1




    About wording, I think you are asking if it is possible to use MyFunction_1 return type as an argument for MyFunction_2
    – Loreto
    1 hour ago










  • Yup. Thanks for correction
    – Dmitrii Motorygin
    38 mins ago














up vote
23
down vote

favorite












I want something like this:



std::tuple<int, bool, double> MyFunction_1 (void);

void MyFunction_2 (decltype (MyFunction_1) &params);


Obviously, in this example a code pointer to function would be passed.



I want to have the equivalent of this:



void MyFunction_2 (std::tuple<int, bool, double> &params);


Is it possible to do so?






c++ c++11 decltype







share|improve this question



















  • 1




    About wording, I think you are asking if it is possible to use MyFunction_1 return type as an argument for MyFunction_2
    – Loreto
    1 hour ago










  • Yup. Thanks for correction
    – Dmitrii Motorygin
    38 mins ago












up vote
23
down vote

favorite









up vote
23
down vote

favorite











I want something like this:



std::tuple<int, bool, double> MyFunction_1 (void);

void MyFunction_2 (decltype (MyFunction_1) &params);


Obviously, in this example a code pointer to function would be passed.



I want to have the equivalent of this:



void MyFunction_2 (std::tuple<int, bool, double> &params);


Is it possible to do so?






c++ c++11 decltype







share|improve this question















I want something like this:



std::tuple<int, bool, double> MyFunction_1 (void);

void MyFunction_2 (decltype (MyFunction_1) &params);


Obviously, in this example a code pointer to function would be passed.



I want to have the equivalent of this:



void MyFunction_2 (std::tuple<int, bool, double> &params);


Is it possible to do so?






c++ c++11 decltype




c++ c++11 decltype






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 37 mins ago

























asked 12 hours ago









Dmitrii Motorygin

1357




1357







  • 1




    About wording, I think you are asking if it is possible to use MyFunction_1 return type as an argument for MyFunction_2
    – Loreto
    1 hour ago










  • Yup. Thanks for correction
    – Dmitrii Motorygin
    38 mins ago












  • 1




    About wording, I think you are asking if it is possible to use MyFunction_1 return type as an argument for MyFunction_2
    – Loreto
    1 hour ago










  • Yup. Thanks for correction
    – Dmitrii Motorygin
    38 mins ago







1




1




About wording, I think you are asking if it is possible to use MyFunction_1 return type as an argument for MyFunction_2
– Loreto
1 hour ago




About wording, I think you are asking if it is possible to use MyFunction_1 return type as an argument for MyFunction_2
– Loreto
1 hour ago












Yup. Thanks for correction
– Dmitrii Motorygin
38 mins ago




Yup. Thanks for correction
– Dmitrii Motorygin
38 mins ago












2 Answers
2






active

oldest

votes

















up vote
29
down vote













decltype (MyFunction_1) will give you the type of MyFunction_1 (i.e. the function type std::tuple<int, bool, double> ()), you need to emulate a function calling 1 (via adding ()) to get the return type (i.e. std::tuple<int, bool, double>), e.g.



void MyFunction_2 (decltype (MyFunction_1()) &params);
// ^^


1 The expression is evaluated at compile-time, the function won't be called at run-time actually.






share|improve this answer


















  • 2




    Because....? FGITWing is nice and all, but an answer should explain to the OP and future readers why adding two characters solves the problem here.
    – StoryTeller
    12 hours ago






  • 2




    Would "you need to emulate a function call in the decltype()" be a better wording? :P
    – Rerito
    12 hours ago






  • 9




    Another approach (maybe more straightforward): std::result_of
    – Ptaq666
    12 hours ago






  • 5




    @Ptaq666 That's an answer and should be posted as one.
    – pipe
    9 hours ago

















up vote
15
down vote













The type of MyFunction_1 is not std::tuple<int, bool, double> - informally you can think of it as a function pointer. In fact &MyFunction_1 decays to itself and is certainly a pointer.



So decltype(MyFunction_1) is not what you want.



The solution is to write decltype(MyFunction_1()). The type of MyFunction_1() is std::tuple<int, bool, double>. Note that this doesn't actually call the function; it's rather like sizeof in that respect.






share|improve this answer
















  • 1




    MyFunction_1 decays to a pointer, &MyFunction_1 does not -- it is one already :)
    – Quentin
    7 hours ago










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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f52811378%2fis-it-possible-to-use-function-return-type-as-an-argument-in-declaration-of-anot%23new-answer', 'question_page');

);

Post as a guest

















2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
29
down vote













decltype (MyFunction_1) will give you the type of MyFunction_1 (i.e. the function type std::tuple<int, bool, double> ()), you need to emulate a function calling 1 (via adding ()) to get the return type (i.e. std::tuple<int, bool, double>), e.g.



void MyFunction_2 (decltype (MyFunction_1()) &params);
// ^^


1 The expression is evaluated at compile-time, the function won't be called at run-time actually.






share|improve this answer


















  • 2




    Because....? FGITWing is nice and all, but an answer should explain to the OP and future readers why adding two characters solves the problem here.
    – StoryTeller
    12 hours ago






  • 2




    Would "you need to emulate a function call in the decltype()" be a better wording? :P
    – Rerito
    12 hours ago






  • 9




    Another approach (maybe more straightforward): std::result_of
    – Ptaq666
    12 hours ago






  • 5




    @Ptaq666 That's an answer and should be posted as one.
    – pipe
    9 hours ago














up vote
29
down vote













decltype (MyFunction_1) will give you the type of MyFunction_1 (i.e. the function type std::tuple<int, bool, double> ()), you need to emulate a function calling 1 (via adding ()) to get the return type (i.e. std::tuple<int, bool, double>), e.g.



void MyFunction_2 (decltype (MyFunction_1()) &params);
// ^^


1 The expression is evaluated at compile-time, the function won't be called at run-time actually.






share|improve this answer


















  • 2




    Because....? FGITWing is nice and all, but an answer should explain to the OP and future readers why adding two characters solves the problem here.
    – StoryTeller
    12 hours ago






  • 2




    Would "you need to emulate a function call in the decltype()" be a better wording? :P
    – Rerito
    12 hours ago






  • 9




    Another approach (maybe more straightforward): std::result_of
    – Ptaq666
    12 hours ago






  • 5




    @Ptaq666 That's an answer and should be posted as one.
    – pipe
    9 hours ago












up vote
29
down vote










up vote
29
down vote









decltype (MyFunction_1) will give you the type of MyFunction_1 (i.e. the function type std::tuple<int, bool, double> ()), you need to emulate a function calling 1 (via adding ()) to get the return type (i.e. std::tuple<int, bool, double>), e.g.



void MyFunction_2 (decltype (MyFunction_1()) &params);
// ^^


1 The expression is evaluated at compile-time, the function won't be called at run-time actually.






share|improve this answer














decltype (MyFunction_1) will give you the type of MyFunction_1 (i.e. the function type std::tuple<int, bool, double> ()), you need to emulate a function calling 1 (via adding ()) to get the return type (i.e. std::tuple<int, bool, double>), e.g.



void MyFunction_2 (decltype (MyFunction_1()) &params);
// ^^


1 The expression is evaluated at compile-time, the function won't be called at run-time actually.







share|improve this answer














share|improve this answer



share|improve this answer








edited 11 hours ago

























answered 12 hours ago









songyuanyao

86k9166226




86k9166226







  • 2




    Because....? FGITWing is nice and all, but an answer should explain to the OP and future readers why adding two characters solves the problem here.
    – StoryTeller
    12 hours ago






  • 2




    Would "you need to emulate a function call in the decltype()" be a better wording? :P
    – Rerito
    12 hours ago






  • 9




    Another approach (maybe more straightforward): std::result_of
    – Ptaq666
    12 hours ago






  • 5




    @Ptaq666 That's an answer and should be posted as one.
    – pipe
    9 hours ago












  • 2




    Because....? FGITWing is nice and all, but an answer should explain to the OP and future readers why adding two characters solves the problem here.
    – StoryTeller
    12 hours ago






  • 2




    Would "you need to emulate a function call in the decltype()" be a better wording? :P
    – Rerito
    12 hours ago






  • 9




    Another approach (maybe more straightforward): std::result_of
    – Ptaq666
    12 hours ago






  • 5




    @Ptaq666 That's an answer and should be posted as one.
    – pipe
    9 hours ago







2




2




Because....? FGITWing is nice and all, but an answer should explain to the OP and future readers why adding two characters solves the problem here.
– StoryTeller
12 hours ago




Because....? FGITWing is nice and all, but an answer should explain to the OP and future readers why adding two characters solves the problem here.
– StoryTeller
12 hours ago




2




2




Would "you need to emulate a function call in the decltype()" be a better wording? :P
– Rerito
12 hours ago




Would "you need to emulate a function call in the decltype()" be a better wording? :P
– Rerito
12 hours ago




9




9




Another approach (maybe more straightforward): std::result_of
– Ptaq666
12 hours ago




Another approach (maybe more straightforward): std::result_of
– Ptaq666
12 hours ago




5




5




@Ptaq666 That's an answer and should be posted as one.
– pipe
9 hours ago




@Ptaq666 That's an answer and should be posted as one.
– pipe
9 hours ago












up vote
15
down vote













The type of MyFunction_1 is not std::tuple<int, bool, double> - informally you can think of it as a function pointer. In fact &MyFunction_1 decays to itself and is certainly a pointer.



So decltype(MyFunction_1) is not what you want.



The solution is to write decltype(MyFunction_1()). The type of MyFunction_1() is std::tuple<int, bool, double>. Note that this doesn't actually call the function; it's rather like sizeof in that respect.






share|improve this answer
















  • 1




    MyFunction_1 decays to a pointer, &MyFunction_1 does not -- it is one already :)
    – Quentin
    7 hours ago














up vote
15
down vote













The type of MyFunction_1 is not std::tuple<int, bool, double> - informally you can think of it as a function pointer. In fact &MyFunction_1 decays to itself and is certainly a pointer.



So decltype(MyFunction_1) is not what you want.



The solution is to write decltype(MyFunction_1()). The type of MyFunction_1() is std::tuple<int, bool, double>. Note that this doesn't actually call the function; it's rather like sizeof in that respect.






share|improve this answer
















  • 1




    MyFunction_1 decays to a pointer, &MyFunction_1 does not -- it is one already :)
    – Quentin
    7 hours ago












up vote
15
down vote










up vote
15
down vote









The type of MyFunction_1 is not std::tuple<int, bool, double> - informally you can think of it as a function pointer. In fact &MyFunction_1 decays to itself and is certainly a pointer.



So decltype(MyFunction_1) is not what you want.



The solution is to write decltype(MyFunction_1()). The type of MyFunction_1() is std::tuple<int, bool, double>. Note that this doesn't actually call the function; it's rather like sizeof in that respect.






share|improve this answer












The type of MyFunction_1 is not std::tuple<int, bool, double> - informally you can think of it as a function pointer. In fact &MyFunction_1 decays to itself and is certainly a pointer.



So decltype(MyFunction_1) is not what you want.



The solution is to write decltype(MyFunction_1()). The type of MyFunction_1() is std::tuple<int, bool, double>. Note that this doesn't actually call the function; it's rather like sizeof in that respect.







share|improve this answer












share|improve this answer



share|improve this answer










answered 12 hours ago









Bathsheba

170k26239362




170k26239362







  • 1




    MyFunction_1 decays to a pointer, &MyFunction_1 does not -- it is one already :)
    – Quentin
    7 hours ago












  • 1




    MyFunction_1 decays to a pointer, &MyFunction_1 does not -- it is one already :)
    – Quentin
    7 hours ago







1




1




MyFunction_1 decays to a pointer, &MyFunction_1 does not -- it is one already :)
– Quentin
7 hours ago




MyFunction_1 decays to a pointer, &MyFunction_1 does not -- it is one already :)
– Quentin
7 hours ago

















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f52811378%2fis-it-possible-to-use-function-return-type-as-an-argument-in-declaration-of-anot%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What does second last employer means? [closed]

Image
Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
Read more

Installing NextGIS Connect into QGIS 3?

Image
Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugins nextgis share | improve this question edited 3 hours ago PolyGeo ♦ 51.9k 17 77 233 asked 4 hours ago IanGS 16 4 New contributor IanGS is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct. add a comment  |  up vote 1 down vote favorite Just acquired free version of NEXTGIS and trying to install the plugin for QGIS. It tells me it was not designed for version 3 of QGIS. How do I set up the QGIS-NEXTGIS interface then? qgis qgis-plugin...
Read more

One-line joke

A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
Read more