Styjun

"It will become the talk of Paris" - translation into French

Are there any vegetarian astronauts?

Going to get married soon, should I do it on Dec 31 or Jan 1?

What are the penalties for overstaying in USA?

How well known and how commonly used was Huffman coding in 1979?

How can I convince my reader that I will not use a certain trope?

Bash echo $-1 prints hb1. Why?

Does ultrasonic bath cleaning damage laboratory volumetric glassware calibration?

Is there any set of 2-6 notes that doesn't have a chord name?

Why does the A-4 Skyhawk sit nose-up when on ground?

Why is the Turkish president's surname spelt in Russian as Эрдоган, with г?

Short story with brother-sister conjoined twins as protagonists?

Ending: accusative or not?

Is it OK to bottle condition using previously contaminated bottles?

Find smallest index that is identical to the value in an array

Why does adding parentheses prevent an error?

Counting occurrence of words in table is slow

Does image quality of the lens affect "focus and recompose" technique?

How many satellites can stay in a Lagrange point?

Simple object validator with a new API

How risky is real estate?

Is my Rep in Stack-Exchange Form?

Using “sparkling” as a diminutive of “spark” in a poem

Analog is Obtuse!

Child class calling unexpected overloaded function

What are the rules for calling the superclass constructor?Use 'class' or 'typename' for template parameters?How to call a parent class function from derived class function?Function overloading in Javascript - Best practicesCall a parent class's method from child class in Python?Python class inherits objectWhat are the basic rules and idioms for operator overloading?Template function overloading with identical signatures, why does this work?Overloading base method in derived classUnexpected overload resolution in visual studio involving void*, string and const char[]

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

1

1

I'm struggling to understand why in the following code the supposedly wrong overload is being called.

// overload.h
 struct T1
 
 template<class... Args>
 void doFoo(Args&&... args)
 
 std::cout << "T1 doFoo generic"<< std::endl;
 

 void doFoo(int a)
 
 std::cout << "T1 doFoo INT "<< std::endl;
 

 void doFoo(double a)
 
 std::cout << "T1 doFoo Double "<< std::endl;
 

 template<class... Args>
 void foo(Args&&... args)
 
 doFoo(args...);
 
 ;


 struct T2 : public T1
 
 void doFoo(char c)
 
 std::cout << "T2 doFoo char " << std::endl;
 
 ;

 // main.cpp

 #include <overload.h>
 int main()
 
 T2 t2;
 t2.foo(3);
 t2.foo('A'); // This outputs: T1 doFoo generic 
 

I was expecting that t2.foo('A') had as output: "T2 doFoo char", but instead I got "T1 doFoo generic".

If I move T2::doFoo(char c) into T1, everything works as expected. What's the reason of this behavior? any workaround?

c++ templates inheritance overloading template-meta-programming

share|improve this question

edited Mar 25 at 13:54

Saturnu

asked Mar 25 at 11:12

SaturnuSaturnu

9777 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Do you call t2.foo('A'); or t2.doFoo('A');? I think if you would call t2.doFoo('A'); you wold get the correct result.
  
  – vahancho
  Mar 25 at 11:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  yes, that's true. My intention is though to make all the doFoo functions protected, so that the only interface is foo(). So I need to call foo(myargs)
  
  – Saturnu
  Mar 25 at 11:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I see, but it looks like you expect that T1::foo<...> call the doFoo of a derived class, like a virtual function, but it will not work in that way - this is a static binding.
  
  – vahancho
  Mar 25 at 11:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's not exaclty like a virtual function call. Statically, the compiler should be able to figure this overload out, as in this case it does not need runtime information. I am just looking for a way to make this happen without incurring in run-time overhead
  
  – Saturnu
  Mar 25 at 11:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why do you expect T1::foo to call a function in the T2 class? T1 doesn't know about T2. When it calls doFoo, the type of this is T1 *.
  
  – 1201ProgramAlarm
  Mar 25 at 15:02
  

  
  
  
  

 | 
show 2 more comments

1

1

I'm struggling to understand why in the following code the supposedly wrong overload is being called.

// overload.h
 struct T1
 
 template<class... Args>
 void doFoo(Args&&... args)
 
 std::cout << "T1 doFoo generic"<< std::endl;
 

 void doFoo(int a)
 
 std::cout << "T1 doFoo INT "<< std::endl;
 

 void doFoo(double a)
 
 std::cout << "T1 doFoo Double "<< std::endl;
 

 template<class... Args>
 void foo(Args&&... args)
 
 doFoo(args...);
 
 ;


 struct T2 : public T1
 
 void doFoo(char c)
 
 std::cout << "T2 doFoo char " << std::endl;
 
 ;

 // main.cpp

 #include <overload.h>
 int main()
 
 T2 t2;
 t2.foo(3);
 t2.foo('A'); // This outputs: T1 doFoo generic 
 

I was expecting that t2.foo('A') had as output: "T2 doFoo char", but instead I got "T1 doFoo generic".

If I move T2::doFoo(char c) into T1, everything works as expected. What's the reason of this behavior? any workaround?

c++ templates inheritance overloading template-meta-programming

share|improve this question

edited Mar 25 at 13:54

Saturnu

asked Mar 25 at 11:12

SaturnuSaturnu

9777 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Do you call t2.foo('A'); or t2.doFoo('A');? I think if you would call t2.doFoo('A'); you wold get the correct result.
  
  – vahancho
  Mar 25 at 11:25
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  yes, that's true. My intention is though to make all the doFoo functions protected, so that the only interface is foo(). So I need to call foo(myargs)
  
  – Saturnu
  Mar 25 at 11:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I see, but it looks like you expect that T1::foo<...> call the doFoo of a derived class, like a virtual function, but it will not work in that way - this is a static binding.
  
  – vahancho
  Mar 25 at 11:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's not exaclty like a virtual function call. Statically, the compiler should be able to figure this overload out, as in this case it does not need runtime information. I am just looking for a way to make this happen without incurring in run-time overhead
  
  – Saturnu
  Mar 25 at 11:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why do you expect T1::foo to call a function in the T2 class? T1 doesn't know about T2. When it calls doFoo, the type of this is T1 *.
  
  – 1201ProgramAlarm
  Mar 25 at 15:02
  

  
  
  
  

 | 
show 2 more comments

I'm struggling to understand why in the following code the supposedly wrong overload is being called.

// overload.h
 struct T1
 
 template<class... Args>
 void doFoo(Args&&... args)
 
 std::cout << "T1 doFoo generic"<< std::endl;
 

 void doFoo(int a)
 
 std::cout << "T1 doFoo INT "<< std::endl;
 

 void doFoo(double a)
 
 std::cout << "T1 doFoo Double "<< std::endl;
 

 template<class... Args>
 void foo(Args&&... args)
 
 doFoo(args...);
 
 ;


 struct T2 : public T1
 
 void doFoo(char c)
 
 std::cout << "T2 doFoo char " << std::endl;
 
 ;

 // main.cpp

 #include <overload.h>
 int main()
 
 T2 t2;
 t2.foo(3);
 t2.foo('A'); // This outputs: T1 doFoo generic 
 

I was expecting that t2.foo('A') had as output: "T2 doFoo char", but instead I got "T1 doFoo generic".

If I move T2::doFoo(char c) into T1, everything works as expected. What's the reason of this behavior? any workaround?

c++ templates inheritance overloading template-meta-programming

share|improve this question

edited Mar 25 at 13:54

Saturnu

asked Mar 25 at 11:12

SaturnuSaturnu

9777 bronze badges

// overload.h
 struct T1
 
 template<class... Args>
 void doFoo(Args&&... args)
 
 std::cout << "T1 doFoo generic"<< std::endl;
 

 void doFoo(int a)
 
 std::cout << "T1 doFoo INT "<< std::endl;
 

 void doFoo(double a)
 
 std::cout << "T1 doFoo Double "<< std::endl;
 

 template<class... Args>
 void foo(Args&&... args)
 
 doFoo(args...);
 
 ;


 struct T2 : public T1
 
 void doFoo(char c)
 
 std::cout << "T2 doFoo char " << std::endl;
 
 ;

 // main.cpp

 #include <overload.h>
 int main()
 
 T2 t2;
 t2.foo(3);
 t2.foo('A'); // This outputs: T1 doFoo generic 
 

share|improve this question

edited Mar 25 at 13:54

Saturnu

asked Mar 25 at 11:12

SaturnuSaturnu

9777 bronze badges

share|improve this question

edited Mar 25 at 13:54

Saturnu

asked Mar 25 at 11:12

SaturnuSaturnu

9777 bronze badges

asked Mar 25 at 11:12

SaturnuSaturnu

9777 bronze badges

asked Mar 25 at 11:12

SaturnuSaturnu

9777 bronze badges

1 Answer
1

active

oldest

votes

1

As already suggested in comment, since T1 doesn't know about the derived struct T2, T1::foo also can not find T2::doFoo(char c) and this static binding can not be achieved.

A simple workaround to spuriously overload T1::foo for char in T2 would be again declaring foo in T2 and overloading it as follows:

DEMO

struct T1 

 template<class... Args>
 void doFoo(Args&&... args)
 std::cout << "T1 doFoo generic"<< std::endl;
 

 void doFoo(int a)
 std::cout << "T1 doFoo INT "<< std::endl; 
 

 void doFoo(double a)
 std::cout << "T1 doFoo Double "<< std::endl;
 

 template<class... Args>
 void foo(Args&&... args)
 doFoo(std::forward<Args>(args)...);
 
;

struct T2 : public T1

 template<class... Args>
 void foo(Args&&... args)
 T1::foo(std::forward<Args>(args)...);
 

 void foo(char c)
 std::cout << "T2 foo char " << std::endl;
 
;

share|improve this answer

edited Mar 26 at 6:23

answered Mar 25 at 19:29

HirokiHiroki

2,47533 gold badges55 silver badges2222 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Really interesting solution, probably it's what I was looking for. I'll try to test it, thank you for sharing!
  
  – Saturnu
  Mar 26 at 10:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Do you think this solution could be extendable, I mean, something like creating a struct TFinal that inherits from T1, T3, T4 etc and that has a method foo() that in some ways knows all the overloads?
  
  – Saturnu
  Mar 26 at 10:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Saturnu Yes, this is definitely extendable like this. But this also looks verbose :). Hmm.. I think that there may be a more simple solution.
  
  – Hiroki
  Mar 26 at 13:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  interesting solution. Yep it seems extensible in that way, but it requires a fixed inheritance hierarchy while it could have been much better to have T2, T3, T4 etc and inherit only from the ones you need.. Difficult to achieve though
  
  – Saturnu
  Mar 26 at 14:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  no worries, your answer was more than good. The other would be the cherry on the cake eheh
  
  – Saturnu
  Mar 26 at 15:56
  

  
  
  
  

 | 
show 5 more comments

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%2f55336516%2fchild-class-calling-unexpected-overloaded-function%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

As already suggested in comment, since T1 doesn't know about the derived struct T2, T1::foo also can not find T2::doFoo(char c) and this static binding can not be achieved.

A simple workaround to spuriously overload T1::foo for char in T2 would be again declaring foo in T2 and overloading it as follows:

DEMO

struct T1 

 template<class... Args>
 void doFoo(Args&&... args)
 std::cout << "T1 doFoo generic"<< std::endl;
 

 void doFoo(int a)
 std::cout << "T1 doFoo INT "<< std::endl; 
 

 void doFoo(double a)
 std::cout << "T1 doFoo Double "<< std::endl;
 

 template<class... Args>
 void foo(Args&&... args)
 doFoo(std::forward<Args>(args)...);
 
;

struct T2 : public T1

 template<class... Args>
 void foo(Args&&... args)
 T1::foo(std::forward<Args>(args)...);
 

 void foo(char c)
 std::cout << "T2 foo char " << std::endl;
 
;

share|improve this answer

edited Mar 26 at 6:23

answered Mar 25 at 19:29

HirokiHiroki

2,47533 gold badges55 silver badges2222 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Really interesting solution, probably it's what I was looking for. I'll try to test it, thank you for sharing!
  
  – Saturnu
  Mar 26 at 10:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Do you think this solution could be extendable, I mean, something like creating a struct TFinal that inherits from T1, T3, T4 etc and that has a method foo() that in some ways knows all the overloads?
  
  – Saturnu
  Mar 26 at 10:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Saturnu Yes, this is definitely extendable like this. But this also looks verbose :). Hmm.. I think that there may be a more simple solution.
  
  – Hiroki
  Mar 26 at 13:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  interesting solution. Yep it seems extensible in that way, but it requires a fixed inheritance hierarchy while it could have been much better to have T2, T3, T4 etc and inherit only from the ones you need.. Difficult to achieve though
  
  – Saturnu
  Mar 26 at 14:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  no worries, your answer was more than good. The other would be the cherry on the cake eheh
  
  – Saturnu
  Mar 26 at 15:56
  

  
  
  
  

 | 
show 5 more comments

1

As already suggested in comment, since T1 doesn't know about the derived struct T2, T1::foo also can not find T2::doFoo(char c) and this static binding can not be achieved.

A simple workaround to spuriously overload T1::foo for char in T2 would be again declaring foo in T2 and overloading it as follows:

DEMO

struct T1 

 template<class... Args>
 void doFoo(Args&&... args)
 std::cout << "T1 doFoo generic"<< std::endl;
 

 void doFoo(int a)
 std::cout << "T1 doFoo INT "<< std::endl; 
 

 void doFoo(double a)
 std::cout << "T1 doFoo Double "<< std::endl;
 

 template<class... Args>
 void foo(Args&&... args)
 doFoo(std::forward<Args>(args)...);
 
;

struct T2 : public T1

 template<class... Args>
 void foo(Args&&... args)
 T1::foo(std::forward<Args>(args)...);
 

 void foo(char c)
 std::cout << "T2 foo char " << std::endl;
 
;

share|improve this answer

edited Mar 26 at 6:23

answered Mar 25 at 19:29

HirokiHiroki

2,47533 gold badges55 silver badges2222 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Really interesting solution, probably it's what I was looking for. I'll try to test it, thank you for sharing!
  
  – Saturnu
  Mar 26 at 10:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Do you think this solution could be extendable, I mean, something like creating a struct TFinal that inherits from T1, T3, T4 etc and that has a method foo() that in some ways knows all the overloads?
  
  – Saturnu
  Mar 26 at 10:51
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Saturnu Yes, this is definitely extendable like this. But this also looks verbose :). Hmm.. I think that there may be a more simple solution.
  
  – Hiroki
  Mar 26 at 13:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  interesting solution. Yep it seems extensible in that way, but it requires a fixed inheritance hierarchy while it could have been much better to have T2, T3, T4 etc and inherit only from the ones you need.. Difficult to achieve though
  
  – Saturnu
  Mar 26 at 14:12
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  no worries, your answer was more than good. The other would be the cherry on the cake eheh
  
  – Saturnu
  Mar 26 at 15:56
  

  
  
  
  

 | 
show 5 more comments

As already suggested in comment, since T1 doesn't know about the derived struct T2, T1::foo also can not find T2::doFoo(char c) and this static binding can not be achieved.

A simple workaround to spuriously overload T1::foo for char in T2 would be again declaring foo in T2 and overloading it as follows:

DEMO

struct T1 

 template<class... Args>
 void doFoo(Args&&... args)
 std::cout << "T1 doFoo generic"<< std::endl;
 

 void doFoo(int a)
 std::cout << "T1 doFoo INT "<< std::endl; 
 

 void doFoo(double a)
 std::cout << "T1 doFoo Double "<< std::endl;
 

 template<class... Args>
 void foo(Args&&... args)
 doFoo(std::forward<Args>(args)...);
 
;

struct T2 : public T1

 template<class... Args>
 void foo(Args&&... args)
 T1::foo(std::forward<Args>(args)...);
 

 void foo(char c)
 std::cout << "T2 foo char " << std::endl;
 
;

share|improve this answer

edited Mar 26 at 6:23

answered Mar 25 at 19:29

HirokiHiroki

2,47533 gold badges55 silver badges2222 bronze badges

struct T1 

 template<class... Args>
 void doFoo(Args&&... args)
 std::cout << "T1 doFoo generic"<< std::endl;
 

 void doFoo(int a)
 std::cout << "T1 doFoo INT "<< std::endl; 
 

 void doFoo(double a)
 std::cout << "T1 doFoo Double "<< std::endl;
 

 template<class... Args>
 void foo(Args&&... args)
 doFoo(std::forward<Args>(args)...);
 
;

struct T2 : public T1

 template<class... Args>
 void foo(Args&&... args)
 T1::foo(std::forward<Args>(args)...);
 

 void foo(char c)
 std::cout << "T2 foo char " << std::endl;
 
;

share|improve this answer

edited Mar 26 at 6:23

answered Mar 25 at 19:29

HirokiHiroki

2,47533 gold badges55 silver badges2222 bronze badges

answered Mar 25 at 19:29

HirokiHiroki

2,47533 gold badges55 silver badges2222 bronze badges

answered Mar 25 at 19:29

HirokiHiroki

2,47533 gold badges55 silver badges2222 bronze badges