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Why does calling cout.operator

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6

I was exploring the ostream class in C++. I am stuck on the strange output of cout on string and integer data types.

When passing an integer or floating-point value, the output is exactly what I pass. For example cout.operator<<(10); prints 10. But when passing a string as an argument it is printing some hexadecimal values:

#include <iostream>

#include <string>



using namespace std;



int main() {

        const char* str = "aia";

        cout.operator<<(str);

        return 0;

}

Output: 0x4007e0.

c++ c++11 c++14

share|improve this question

edited 6 hours ago

Boann

38.5k1313 gold badges9393 silver badges123123 bronze badges

asked 8 hours ago

Pranjal KalerPranjal Kaler

4766 bronze badges

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Why are you using operator<< directly like that? What are you trying to do?
  
  – Nicol Bolas
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @NicolBolas I suspect that they don't know what to do.
  
  – john
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I am trying to understand how the cascading of operators actually works
  
  – Pranjal Kaler
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Assuming you don't have some particular reason for using that form of operator<<, just use cout << str; and you'll see the output you expect.
  
  – john
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  i tried and it prints the string but my question is why it is printing some hexa values when it can print same result for other data types
  
  – Pranjal Kaler
  8 hours ago
  

  
  
  
  

 | 
show 1 more comment

6

I was exploring the ostream class in C++. I am stuck on the strange output of cout on string and integer data types.

When passing an integer or floating-point value, the output is exactly what I pass. For example cout.operator<<(10); prints 10. But when passing a string as an argument it is printing some hexadecimal values:

#include <iostream>

#include <string>



using namespace std;



int main() {

        const char* str = "aia";

        cout.operator<<(str);

        return 0;

}

Output: 0x4007e0.

c++ c++11 c++14

share|improve this question

edited 6 hours ago

Boann

38.5k1313 gold badges9393 silver badges123123 bronze badges

asked 8 hours ago

Pranjal KalerPranjal Kaler

4766 bronze badges

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Why are you using operator<< directly like that? What are you trying to do?
  
  – Nicol Bolas
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @NicolBolas I suspect that they don't know what to do.
  
  – john
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  I am trying to understand how the cascading of operators actually works
  
  – Pranjal Kaler
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Assuming you don't have some particular reason for using that form of operator<<, just use cout << str; and you'll see the output you expect.
  
  – john
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  i tried and it prints the string but my question is why it is printing some hexa values when it can print same result for other data types
  
  – Pranjal Kaler
  8 hours ago
  

  
  
  
  

 | 
show 1 more comment

I was exploring the ostream class in C++. I am stuck on the strange output of cout on string and integer data types.

When passing an integer or floating-point value, the output is exactly what I pass. For example cout.operator<<(10); prints 10. But when passing a string as an argument it is printing some hexadecimal values:

#include <iostream>

#include <string>



using namespace std;



int main() {

        const char* str = "aia";

        cout.operator<<(str);

        return 0;

}

Output: 0x4007e0.

c++ c++11 c++14

share|improve this question

edited 6 hours ago

Boann

38.5k1313 gold badges9393 silver badges123123 bronze badges

asked 8 hours ago

Pranjal KalerPranjal Kaler

4766 bronze badges

#include <iostream>

#include <string>



using namespace std;



int main() {

        const char* str = "aia";

        cout.operator<<(str);

        return 0;

}

share|improve this question

edited 6 hours ago

Boann

38.5k1313 gold badges9393 silver badges123123 bronze badges

asked 8 hours ago

Pranjal KalerPranjal Kaler

4766 bronze badges

share|improve this question

edited 6 hours ago

Boann

38.5k1313 gold badges9393 silver badges123123 bronze badges

asked 8 hours ago

Pranjal KalerPranjal Kaler

4766 bronze badges

edited 6 hours ago

Boann

38.5k1313 gold badges9393 silver badges123123 bronze badges

edited 6 hours ago

Boann

38.5k1313 gold badges9393 silver badges123123 bronze badges

asked 8 hours ago

Pranjal KalerPranjal Kaler

4766 bronze badges

asked 8 hours ago

Pranjal KalerPranjal Kaler

4766 bronze badges

3 Answers
3

active

oldest

votes

9

When you do cout.operator<<(str) you call cout's operator << member function. If we look at what member functions overloads cout has we have

basic_ostream& operator<<( short value );

basic_ostream& operator<<( unsigned short value );



basic_ostream& operator<<( int value );

basic_ostream& operator<<( unsigned int value );



basic_ostream& operator<<( long value );

basic_ostream& operator<<( unsigned long value );



basic_ostream& operator<<( long long value );

basic_ostream& operator<<( unsigned long long value );



basic_ostream& operator<<( float value );

basic_ostream& operator<<( double value );

basic_ostream& operator<<( long double value );



basic_ostream& operator<<( bool value );



basic_ostream& operator<<( const void* value );



basic_ostream& operator<<( std::nullptr_t );



basic_ostream& operator<<( std::basic_streambuf<CharT, Traits>* sb);



basic_ostream& operator<<(

    std::ios_base& (*func)(std::ios_base&) );



basic_ostream& operator<<(

    std::basic_ios<CharT,Traits>& (*func)(std::basic_ios<CharT,Traits>&) );



basic_ostream& operator<<(

    std::basic_ostream<CharT,Traits>& (*func)(std::basic_ostream<CharT,Traits>&) );

If you notice, there isn't one for a const char*, but there is one for a const void*. So, your const char* is converted to a const void* and that version of the function prints the address held by the pointer.

What you need to do is call the non member function overload of operator<< and to do that you can use

cout << str;

share|improve this answer

edited 8 hours ago

answered 8 hours ago

NathanOliverNathanOliver

110k1919 gold badges167167 silver badges246246 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Always use cout << X, never use cout.operator<<(X) or operator<<(cout, X) directly (unless you have a REALLY GOOD reason to do so). Let the compiler decide which one to call based on context and data type.
  
  – Remy Lebeau
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @RemyLebeau I've removed the operator<<(cout, X) call. As you said, they should just use the operator syntax.
  
  – NathanOliver
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

8

The problem is that for some types operator<< is overloaded as a member of ostream and for some types it is overloaded as a global function. In the case of const char* it's a global function, so if you want to call the operator function explicitly you must write

operator<<(cout, str);

but for integer types you must write

cout.operator<<(num);

What's happening in the code you posted is that the overload for const void* is being called, which is why you see hexadecimal numbers.

share|improve this answer

answered 8 hours ago

johnjohn

43.1k22 gold badges2929 silver badges5050 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  this makes sense. thank you
  
  – Pranjal Kaler
  8 hours ago
  

  
  
  
  

add a comment | 

-1

Use cout << str instead of cout.operator<<(str);
It will work.

share|improve this answer

answered 7 hours ago

XurahbeelXurahbeel

711 bronze badge

New contributor

Xurahbeel 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 | 

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9

When you do cout.operator<<(str) you call cout's operator << member function. If we look at what member functions overloads cout has we have

basic_ostream& operator<<( short value );

basic_ostream& operator<<( unsigned short value );



basic_ostream& operator<<( int value );

basic_ostream& operator<<( unsigned int value );



basic_ostream& operator<<( long value );

basic_ostream& operator<<( unsigned long value );



basic_ostream& operator<<( long long value );

basic_ostream& operator<<( unsigned long long value );



basic_ostream& operator<<( float value );

basic_ostream& operator<<( double value );

basic_ostream& operator<<( long double value );



basic_ostream& operator<<( bool value );



basic_ostream& operator<<( const void* value );



basic_ostream& operator<<( std::nullptr_t );



basic_ostream& operator<<( std::basic_streambuf<CharT, Traits>* sb);



basic_ostream& operator<<(

    std::ios_base& (*func)(std::ios_base&) );



basic_ostream& operator<<(

    std::basic_ios<CharT,Traits>& (*func)(std::basic_ios<CharT,Traits>&) );



basic_ostream& operator<<(

    std::basic_ostream<CharT,Traits>& (*func)(std::basic_ostream<CharT,Traits>&) );

If you notice, there isn't one for a const char*, but there is one for a const void*. So, your const char* is converted to a const void* and that version of the function prints the address held by the pointer.

What you need to do is call the non member function overload of operator<< and to do that you can use

cout << str;

share|improve this answer

edited 8 hours ago

answered 8 hours ago

NathanOliverNathanOliver

110k1919 gold badges167167 silver badges246246 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Always use cout << X, never use cout.operator<<(X) or operator<<(cout, X) directly (unless you have a REALLY GOOD reason to do so). Let the compiler decide which one to call based on context and data type.
  
  – Remy Lebeau
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @RemyLebeau I've removed the operator<<(cout, X) call. As you said, they should just use the operator syntax.
  
  – NathanOliver
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

9

When you do cout.operator<<(str) you call cout's operator << member function. If we look at what member functions overloads cout has we have

basic_ostream& operator<<( short value );

basic_ostream& operator<<( unsigned short value );



basic_ostream& operator<<( int value );

basic_ostream& operator<<( unsigned int value );



basic_ostream& operator<<( long value );

basic_ostream& operator<<( unsigned long value );



basic_ostream& operator<<( long long value );

basic_ostream& operator<<( unsigned long long value );



basic_ostream& operator<<( float value );

basic_ostream& operator<<( double value );

basic_ostream& operator<<( long double value );



basic_ostream& operator<<( bool value );



basic_ostream& operator<<( const void* value );



basic_ostream& operator<<( std::nullptr_t );



basic_ostream& operator<<( std::basic_streambuf<CharT, Traits>* sb);



basic_ostream& operator<<(

    std::ios_base& (*func)(std::ios_base&) );



basic_ostream& operator<<(

    std::basic_ios<CharT,Traits>& (*func)(std::basic_ios<CharT,Traits>&) );



basic_ostream& operator<<(

    std::basic_ostream<CharT,Traits>& (*func)(std::basic_ostream<CharT,Traits>&) );

If you notice, there isn't one for a const char*, but there is one for a const void*. So, your const char* is converted to a const void* and that version of the function prints the address held by the pointer.

What you need to do is call the non member function overload of operator<< and to do that you can use

cout << str;

share|improve this answer

edited 8 hours ago

answered 8 hours ago

NathanOliverNathanOliver

110k1919 gold badges167167 silver badges246246 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Always use cout << X, never use cout.operator<<(X) or operator<<(cout, X) directly (unless you have a REALLY GOOD reason to do so). Let the compiler decide which one to call based on context and data type.
  
  – Remy Lebeau
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @RemyLebeau I've removed the operator<<(cout, X) call. As you said, they should just use the operator syntax.
  
  – NathanOliver
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

When you do cout.operator<<(str) you call cout's operator << member function. If we look at what member functions overloads cout has we have

basic_ostream& operator<<( short value );

basic_ostream& operator<<( unsigned short value );



basic_ostream& operator<<( int value );

basic_ostream& operator<<( unsigned int value );



basic_ostream& operator<<( long value );

basic_ostream& operator<<( unsigned long value );



basic_ostream& operator<<( long long value );

basic_ostream& operator<<( unsigned long long value );



basic_ostream& operator<<( float value );

basic_ostream& operator<<( double value );

basic_ostream& operator<<( long double value );



basic_ostream& operator<<( bool value );



basic_ostream& operator<<( const void* value );



basic_ostream& operator<<( std::nullptr_t );



basic_ostream& operator<<( std::basic_streambuf<CharT, Traits>* sb);



basic_ostream& operator<<(

    std::ios_base& (*func)(std::ios_base&) );



basic_ostream& operator<<(

    std::basic_ios<CharT,Traits>& (*func)(std::basic_ios<CharT,Traits>&) );



basic_ostream& operator<<(

    std::basic_ostream<CharT,Traits>& (*func)(std::basic_ostream<CharT,Traits>&) );

If you notice, there isn't one for a const char*, but there is one for a const void*. So, your const char* is converted to a const void* and that version of the function prints the address held by the pointer.

What you need to do is call the non member function overload of operator<< and to do that you can use

cout << str;

share|improve this answer

edited 8 hours ago

answered 8 hours ago

NathanOliverNathanOliver

110k1919 gold badges167167 silver badges246246 bronze badges

basic_ostream& operator<<( short value );

basic_ostream& operator<<( unsigned short value );



basic_ostream& operator<<( int value );

basic_ostream& operator<<( unsigned int value );



basic_ostream& operator<<( long value );

basic_ostream& operator<<( unsigned long value );



basic_ostream& operator<<( long long value );

basic_ostream& operator<<( unsigned long long value );



basic_ostream& operator<<( float value );

basic_ostream& operator<<( double value );

basic_ostream& operator<<( long double value );



basic_ostream& operator<<( bool value );



basic_ostream& operator<<( const void* value );



basic_ostream& operator<<( std::nullptr_t );



basic_ostream& operator<<( std::basic_streambuf<CharT, Traits>* sb);



basic_ostream& operator<<(

    std::ios_base& (*func)(std::ios_base&) );



basic_ostream& operator<<(

    std::basic_ios<CharT,Traits>& (*func)(std::basic_ios<CharT,Traits>&) );



basic_ostream& operator<<(

    std::basic_ostream<CharT,Traits>& (*func)(std::basic_ostream<CharT,Traits>&) );

cout << str;

share|improve this answer

edited 8 hours ago

answered 8 hours ago

NathanOliverNathanOliver

110k1919 gold badges167167 silver badges246246 bronze badges

answered 8 hours ago

NathanOliverNathanOliver

110k1919 gold badges167167 silver badges246246 bronze badges

answered 8 hours ago

NathanOliverNathanOliver

110k1919 gold badges167167 silver badges246246 bronze badges

8

The problem is that for some types operator<< is overloaded as a member of ostream and for some types it is overloaded as a global function. In the case of const char* it's a global function, so if you want to call the operator function explicitly you must write

operator<<(cout, str);

but for integer types you must write

cout.operator<<(num);

What's happening in the code you posted is that the overload for const void* is being called, which is why you see hexadecimal numbers.

share|improve this answer

answered 8 hours ago

johnjohn

43.1k22 gold badges2929 silver badges5050 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  this makes sense. thank you
  
  – Pranjal Kaler
  8 hours ago
  

  
  
  
  

add a comment | 

8

The problem is that for some types operator<< is overloaded as a member of ostream and for some types it is overloaded as a global function. In the case of const char* it's a global function, so if you want to call the operator function explicitly you must write

operator<<(cout, str);

but for integer types you must write

cout.operator<<(num);

What's happening in the code you posted is that the overload for const void* is being called, which is why you see hexadecimal numbers.

share|improve this answer

answered 8 hours ago

johnjohn

43.1k22 gold badges2929 silver badges5050 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  this makes sense. thank you
  
  – Pranjal Kaler
  8 hours ago
  

  
  
  
  

add a comment | 

The problem is that for some types operator<< is overloaded as a member of ostream and for some types it is overloaded as a global function. In the case of const char* it's a global function, so if you want to call the operator function explicitly you must write

operator<<(cout, str);

but for integer types you must write

cout.operator<<(num);

What's happening in the code you posted is that the overload for const void* is being called, which is why you see hexadecimal numbers.

share|improve this answer

answered 8 hours ago

johnjohn

43.1k22 gold badges2929 silver badges5050 bronze badges

operator<<(cout, str);

cout.operator<<(num);

share|improve this answer

answered 8 hours ago

johnjohn

43.1k22 gold badges2929 silver badges5050 bronze badges

answered 8 hours ago

johnjohn

43.1k22 gold badges2929 silver badges5050 bronze badges

answered 8 hours ago

johnjohn

43.1k22 gold badges2929 silver badges5050 bronze badges

-1

Use cout << str instead of cout.operator<<(str);
It will work.

share|improve this answer

answered 7 hours ago

XurahbeelXurahbeel

711 bronze badge

New contributor

Xurahbeel 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 | 

-1

Use cout << str instead of cout.operator<<(str);
It will work.

share|improve this answer

answered 7 hours ago

XurahbeelXurahbeel

711 bronze badge

New contributor

Xurahbeel 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 | 

Use cout << str instead of cout.operator<<(str);
It will work.

share|improve this answer

answered 7 hours ago

XurahbeelXurahbeel

711 bronze badge

New contributor

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

share|improve this answer

answered 7 hours ago

XurahbeelXurahbeel

711 bronze badge

New contributor

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

answered 7 hours ago

XurahbeelXurahbeel

711 bronze badge

New contributor

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

answered 7 hours ago

XurahbeelXurahbeel

711 bronze badge