Hi,
to output the private data of a class I want to overload the operator<<. The
Output should be written in a string-variable.
To do this i have written the attached code:
But the program-compilation aborts with the following error message:
StreamKoord.cpp : In function `int main()':
StreamKoord.cpp :44: error: no match for 'operator<<' in 'std::operator< <
[with _Traits = std::char_trait s<char>]((&oss), "String: Scheitel=") <<
scheitel'
/usr/include/g++/bits/ostream.tcc:63: error: candidates are:
***std::basic_o stream<_CharT,* _Traits>&*std:: basic_ostream<_ CharT,
***_Traits>::op erator<<(std::b asic_ostream<_C harT,
***_Traits>&(*) (std::basic_ost ream<_CharT,*_T raits>&))*[with*_CharT*=*c har,
***_Traits*=*st d::char_traits< char>]
*...
A lot of further candidates are listed. But my additional method wasn't
listed.
What is wrong in my code and how can i solve this problem?
Many thanks in advance,
Andreas
PS: gcc: 3.3.4 / linux 2.6.8-24.11-smp 6 2446
Andreas skrev: Hi,
to output the private data of a class I want to overload the operator<<. The Output should be written in a string-variable.
<snip> ostringstream & operator<<(ostr ingstream & os, punkt & v)
std::ostream& operator<<(std: :ostream& os, punkt& v)
{ os << "("; os.width(4); os << v.p.x << ","; os.width(4); os << v.p.y << ","; os.width(4); os << v.p.z << ")"; return os; };
<snip>
--
TB @ SWEDEN
Hi TB,
many thanks for your help!
Whats the rule behind this? I thought that it is necessary to use the same
Output-type for operator<< as defined for oss (like my example).
Sincerely,
Andreas
TB wrote: Andreas skrev: Hi,
to output the private data of a class I want to overload the operator<<. The Output should be written in a string-variable. <snip> ostringstream & operator<<(ostr ingstream & os, punkt & v)
std::ostream& operator<<(std: :ostream& os, punkt& v)
{ os << "("; os.width(4); os << v.p.x << ","; os.width(4); os << v.p.y << ","; os.width(4); os << v.p.z << ")"; return os; };
<snip>
the best way is .. .
friend std::ostream& operator <<(std::ostream & , const <TYPE>);
don't miss the *friend* keyword
Andreas skrev: Hi TB,
many thanks for your help!
Whats the rule behind this? I thought that it is necessary to use the same Output-type for operator<< as defined for oss (like my example).
std::ostringstr eam::operator<< () returns std::ostream&.
(std::ostream is a common base class for all output stream classes)
Your original declaration was:
ostringstream & operator<<(ostr ingstream & os, punkt & v);
And it will work with this code:
std::ostringstr eam os;
punkt p;
os << p;
But not here:
os << p << p;
If you ask why, reread this reply from the top and look at the error
message your compiler produces.
TB wrote:
Andreas skrev: Hi,
to output the private data of a class I want to overload the operator<<. The Output should be written in a string-variable. <snip> ostringstream & operator<<(ostr ingstream & os, punkt & v) std::ostream& operator<<(std: :ostream& os, punkt& v)
{ os << "("; os.width(4); os << v.p.x << ","; os.width(4); os << v.p.y << ","; os.width(4); os << v.p.z << ")"; return os; }; <snip>
--
TB @ SWEDEN
TB skrev: Andreas skrev: Hi TB,
many thanks for your help!
Whats the rule behind this? I thought that it is necessary to use the same Output-type for operator<< as defined for oss (like my example).
std::ostringstr eam::operator<< () returns std::ostream&. (std::ostream is a common base class for all output stream classes)
Your original declaration was: ostringstream & operator<<(ostr ingstream & os, punkt & v);
And it will work with this code: std::ostringstr eam os; punkt p; os << p;
But not here: os << p << p;
I meant:
os << "text" << p;
--
TB @ SWEDEN
Hi,
thanks!
Another small step for the mankind, but a big step for me.
I believe now its clear why it is necassary to use the baseclass to overload
operators!
Sincererly,
Andreas
TB wrote: TB skrev: Andreas skrev: Hi TB,
many thanks for your help!
Whats the rule behind this? I thought that it is necessary to use the same Output-type for operator<< as defined for oss (like my example).
std::ostringstr eam::operator<< () returns std::ostream&. (std::ostream is a common base class for all output stream classes)
Your original declaration was: ostringstream & operator<<(ostr ingstream & os, punkt & v);
And it will work with this code: std::ostringstr eam os; punkt p; os << p;
But not here: os << p << p;
I meant:
os << "text" << p;
--
Viele Grüße,
Andreas This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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