void human::displayInfo()
{
cout<< "Name : " << getName() <<endl;
cout<< "Hair Colour : " << getHairColour() <<endl;
cout<< "Eye Colour : " << getEyeColour() <<endl;
cout<< "Skin Colour : " << getSkinColour() <<endl;
cout<< "Walk Speed : " << getWalkSpeed() <<endl;
cout<< "Run Speed : " << getRunSpeed() <<endl;
cout<< "Total Health : " << getHealth() <<endl;
}
function using cout and call to function question
Author
Hi Guys,
I am practicing my OOP in C++ and I have a question about my code in one of my functions.
Author
Quote:Original post by Black Knight
It works for basic data types because operator << is overloaded for them.
See here
Thanks for that info Black knight.
Kind Regards
David
In your case you can have something like this :
#include <iostream>#include <string>using std::cout;class Human{public: Human() { } // data members methods...};std::ostream& operator<< (std::ostream& out,Human& h ){ cout<< "Name : " << h.getName() <<endl; cout<< "Hair Colour : " << h.getHairColour() <<endl; cout<< "Eye Colour : " << h.getEyeColour() <<endl; cout<< "Skin Colour : " << h.getSkinColour() <<endl; cout<< "Walk Speed : " << h.getWalkSpeed() <<endl; cout<< "Run Speed : " << h.getRunSpeed() <<endl; cout<< "Total Health : " << h.getHealth() <<endl; return out;}int main(){ Human myHuman; std::cout << myHuman; return 0;}
Author
Yup, thats exactly how I have my class setup only with an overloaded constructor.
Kind Regards
David
Kind Regards
David
Quote:Original post by Black Knight
In your case you can have something like this :
*** Source Snippet Removed ***
You shall stream to out, not cout, in operator <<. Otherwise you'll always stream to the standard output, no matter what actual stream object you pass your Human object to.
Quote:Original post by bentaberryvoid human::displayInfo(){ cout<< "Name : " << getName() <<endl; cout<< "Hair Colour : " << getHairColour() <<endl; cout<< "Eye Colour : " << getEyeColour() <<endl; cout<< "Skin Colour : " << getSkinColour() <<endl; cout<< "Walk Speed : " << getWalkSpeed() <<endl; cout<< "Run Speed : " << getRunSpeed() <<endl; cout<< "Total Health : " << getHealth() <<endl;}
You shouldn't flush after each line (endl is a combined "\n" with a flush), this unecessarily costs you performance. Better:
cout<< "Name : " << getName() <<'\n'; cout<< "Hair Colour : " << getHairColour() <<'\n'; cout<< "Eye Colour : " << getEyeColour() <<'\n'; cout<< "Skin Colour : " << getSkinColour() <<'\n'; cout<< "Walk Speed : " << getWalkSpeed() <<'\n'; cout<< "Run Speed : " << getRunSpeed() <<'\n'; cout<< "Total Health : " << getHealth() <<endl;
Author
Quote:Original post by phresnel
You shouldn't flush after each line (endl is a combined "\n" with a flush), this unecessarily costs you performance. Better:cout<< "Name : " << getName() <<'\n'; cout<< "Hair Colour : " << getHairColour() <<'\n'; cout<< "Eye Colour : " << getEyeColour() <<'\n'; cout<< "Skin Colour : " << getSkinColour() <<'\n'; cout<< "Walk Speed : " << getWalkSpeed() <<'\n'; cout<< "Run Speed : " << getRunSpeed() <<'\n'; cout<< "Total Health : " << getHealth() <<endl;
What is a flush?
Kind Regards
David
Output is not directly written to the screen instead it is buffered and then written to the screen when a flush happens.So instead of flushing everything each line.You put everything in the buffer and then flush them with either endl our cout.flush().
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