Why does the compiler prefer f(const void*) to f(const std::string &)?

up vote
30
down vote

favorite

Consider the following piece of code:

#include <iostream>

#include <string>



// void f(const char *) { std::cout << "const char *"; } // <-- comment on purpose

void f(const std::string &) { std::cout << "const std::string &"; }

void f(const void *) { std::cout << "const void *"; }



int main()

{

    f("hello");

    std::cout << std::endl;

}

I compiled this program using g++ (Ubuntu 6.5.0-1ubuntu1~16.04) 6.5.0 20181026:

$ g++ -std=c++11 strings_1.cpp -Wall

$ ./a.out



const void *

Note that the comment is there on purpose to test, otherwise the compiler uses f(const char *).

So, why does the compiler pick f(const void*) over f(const std::string &)?

edited 2 days ago

curiousguy

4,47722940

asked 2 days ago

omar

20918

New contributor

3

Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
– geza
2 days ago

@geza awesome. I was looking for it, thanks.
– omar
2 days ago

The overloading resolution rule here is simple and unchanged in the many C++ versions.
– curiousguy
2 days ago

3

Well, a string literal is not an std::string, it's a static array of chars, which decays to a pointer to its first character. This behavior is inherited from C which never had something like std::string, but ample amounts of code handling strings nonetheless.
– cmaster
2 days ago

1

If you specifically want a std::string literal you can achieve that by adding a s behind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s
– henje
yesterday

|
show 2 more comments

up vote
30
down vote

favorite

Consider the following piece of code:

#include <iostream>

#include <string>



// void f(const char *) { std::cout << "const char *"; } // <-- comment on purpose

void f(const std::string &) { std::cout << "const std::string &"; }

void f(const void *) { std::cout << "const void *"; }



int main()

{

    f("hello");

    std::cout << std::endl;

}

I compiled this program using g++ (Ubuntu 6.5.0-1ubuntu1~16.04) 6.5.0 20181026:

$ g++ -std=c++11 strings_1.cpp -Wall

$ ./a.out



const void *

Note that the comment is there on purpose to test, otherwise the compiler uses f(const char *).

So, why does the compiler pick f(const void*) over f(const std::string &)?

edited 2 days ago

curiousguy

4,47722940

asked 2 days ago

omar

20918

New contributor

3

Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
– geza
2 days ago

@geza awesome. I was looking for it, thanks.
– omar
2 days ago

The overloading resolution rule here is simple and unchanged in the many C++ versions.
– curiousguy
2 days ago

3

Well, a string literal is not an std::string, it's a static array of chars, which decays to a pointer to its first character. This behavior is inherited from C which never had something like std::string, but ample amounts of code handling strings nonetheless.
– cmaster
2 days ago

1

If you specifically want a std::string literal you can achieve that by adding a s behind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s
– henje
yesterday

|
show 2 more comments

up vote
30
down vote

favorite

Consider the following piece of code:

#include <iostream>

#include <string>



// void f(const char *) { std::cout << "const char *"; } // <-- comment on purpose

void f(const std::string &) { std::cout << "const std::string &"; }

void f(const void *) { std::cout << "const void *"; }



int main()

{

    f("hello");

    std::cout << std::endl;

}

I compiled this program using g++ (Ubuntu 6.5.0-1ubuntu1~16.04) 6.5.0 20181026:

$ g++ -std=c++11 strings_1.cpp -Wall

$ ./a.out



const void *

Note that the comment is there on purpose to test, otherwise the compiler uses f(const char *).

So, why does the compiler pick f(const void*) over f(const std::string &)?

edited 2 days ago

curiousguy

4,47722940

asked 2 days ago

omar

20918

New contributor

Consider the following piece of code:

#include <iostream>

#include <string>



// void f(const char *) { std::cout << "const char *"; } // <-- comment on purpose

void f(const std::string &) { std::cout << "const std::string &"; }

void f(const void *) { std::cout << "const void *"; }



int main()

{

    f("hello");

    std::cout << std::endl;

}

I compiled this program using g++ (Ubuntu 6.5.0-1ubuntu1~16.04) 6.5.0 20181026:

$ g++ -std=c++11 strings_1.cpp -Wall

$ ./a.out



const void *

Note that the comment is there on purpose to test, otherwise the compiler uses f(const char *).

So, why does the compiler pick f(const void*) over f(const std::string &)?

c++ stdstring string-literals function-overloading overload-resolution

edited 2 days ago

curiousguy

4,47722940

asked 2 days ago

omar

20918

New contributor

edited 2 days ago

curiousguy

4,47722940

asked 2 days ago

omar

20918

New contributor

edited 2 days ago

curiousguy

4,47722940

edited 2 days ago

curiousguy

4,47722940

edited 2 days ago

curiousguy

4,47722940

asked 2 days ago

omar

20918

New contributor

asked 2 days ago

omar

20918

asked 2 days ago

omar

20918

New contributor

omar is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

3

Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
– geza
2 days ago

@geza awesome. I was looking for it, thanks.
– omar
2 days ago

The overloading resolution rule here is simple and unchanged in the many C++ versions.
– curiousguy
2 days ago

3

Well, a string literal is not an std::string, it's a static array of chars, which decays to a pointer to its first character. This behavior is inherited from C which never had something like std::string, but ample amounts of code handling strings nonetheless.
– cmaster
2 days ago

1

If you specifically want a std::string literal you can achieve that by adding a s behind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s
– henje
yesterday

|
show 2 more comments

3

Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
– geza
2 days ago

@geza awesome. I was looking for it, thanks.
– omar
2 days ago

The overloading resolution rule here is simple and unchanged in the many C++ versions.
– curiousguy
2 days ago

3

Well, a string literal is not an std::string, it's a static array of chars, which decays to a pointer to its first character. This behavior is inherited from C which never had something like std::string, but ample amounts of code handling strings nonetheless.
– cmaster
2 days ago

1

If you specifically want a std::string literal you can achieve that by adding a s behind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s
– henje
yesterday

Here's the relevant part of the standard: eel.is/c++draft/over.ics.rank#2.1
– geza
2 days ago

@geza awesome. I was looking for it, thanks.
– omar
2 days ago

The overloading resolution rule here is simple and unchanged in the many C++ versions.
– curiousguy
2 days ago

Well, a string literal is not an std::string, it's a static array of chars, which decays to a pointer to its first character. This behavior is inherited from C which never had something like std::string, but ample amounts of code handling strings nonetheless.
– cmaster
2 days ago

If you specifically want a std::string literal you can achieve that by adding a s behind the literal. This is a user-defined literal which is available since C++14. en.cppreference.com/w/cpp/string/basic_string/operator%22%22s
– henje
yesterday

|
show 2 more comments

1 Answer
1

active

oldest

votes

up vote
36
down vote

accepted

Converting to a std::string requires a "user defined conversion".

Converting to void const* does not.

User defined conversions are ordered behind built in ones.

answered 2 days ago

Yakk - Adam Nevraumont

178k19184364

add a comment |

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
36
down vote

accepted

Converting to a std::string requires a "user defined conversion".

Converting to void const* does not.

User defined conversions are ordered behind built in ones.

answered 2 days ago

Yakk - Adam Nevraumont

178k19184364

add a comment |

up vote
36
down vote

accepted

Converting to a std::string requires a "user defined conversion".

Converting to void const* does not.

User defined conversions are ordered behind built in ones.

answered 2 days ago

Yakk - Adam Nevraumont

178k19184364

add a comment |

up vote
36
down vote

accepted

Converting to a std::string requires a "user defined conversion".

Converting to void const* does not.

User defined conversions are ordered behind built in ones.

answered 2 days ago

Yakk - Adam Nevraumont

178k19184364

Converting to a std::string requires a "user defined conversion".

Converting to void const* does not.

User defined conversions are ordered behind built in ones.

answered 2 days ago

Yakk - Adam Nevraumont

178k19184364

answered 2 days ago

Yakk - Adam Nevraumont

178k19184364

answered 2 days ago

Yakk - Adam Nevraumont

178k19184364

answered 2 days ago

Yakk - Adam Nevraumont

178k19184364

add a comment |

omar is a new contributor. Be nice, and check out our Code of Conduct.

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