hi im learning about overloading operators in my class and i dont really understand it. anyways i got an assignment to overload the + operator so i can add two strings togther into a third string and i was hoping someone could help me with it.
IE:
string1=happy
string2=birthday
string3=string1+string2
string3 would print happy birthday
thanks in advance any help is appreicated
I assume you mean that you are overloading the operator+ for your own string class, since the C++ string class already has operator+ overloaded.
Do you know how to write the prototype for the function? That would be a good place to start. See if you can get it to work just returning the first string instead of adding the second to it. Then move on to combining the strings. Post code and ask questions if you get stuck.
its like
friend operator+ ();?
You have to add a return value. You also have to add the parameters to the function. All three of those depend on the name of your string class.
Also, you don't need the friend keyword to overload operators. You've probably seen someone overload the << operator(refering to an already overloaded version of it), in which case, you're making it friend so you can access private ostream members. Also, understand that overloading operators are very restricted. You can't change the number of arguements, and atleast one of the operands must be user defined, etc etc...Code:myClass operator+(const myClass &oprnd) const;
Last edited by SlyMaelstrom; 04-17-2006 at 06:56 PM.
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That prototype isn't correct, you shouldn't be returning a reference to a temporary object, and operator+ shouldn't be modifying either of its operands.
As I was taught, you return a reference as some compilers have trouble chaining the operator without returning it as a reference. While I'm sure that isn't the case in modern compilers, it's simply how I write it because it's how I needed to write it to get a proper grade. It may yield a warning, though, on a modern compiler. Off of a guess, he just said that because of how the insertion and extraction operators work. He wasn't very good at what he did.
And yes, you're right, for const correctness, I should have made the operator function constant.
Last edited by SlyMaelstrom; 04-17-2006 at 06:41 PM.
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I don't see how making it a reference would make chaining easier. Regardless, it is wrong because you would have to be returning a temporary (the current object should not be changed by operator+).
Speaking of which, it should really be a non-member function anyway and can often easily be implemented in terms of operator+=.
I'm not disagreeing with you, but you do things enough, you kinda have to get out of the habit of it. I had to do that for a while.Originally Posted by Daved
Anyway, from what I understand, operators that return the the type of the class is better overloaded as a member.
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Do you remember why? I can't think of a reason, especially in this case where you are returning a temporary value anyway, not the actual object.
I think that in the case of operator+, it probably is better to overload operator+= as a member function, then implement operator+ in terms of operator+= as a non-member non-friend function.
It might be something like:
Code:MyClass& MyClass::operator+=(const MyClass& rhs) { // ... return *this; } MyClass operator+(const MyClass& lhs, const MyClass& rhs) { return MyClass(lhs) += rhs; }
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It all boils down to habits and programming practice. For me, I'd implement functions which return myclass objects by reference as member functions (so I can use this*), and all other functions as nonmember functions, possibly adding a friend declaration.
Code:#include <stdio.h> void J(char*a){int f,i=0,c='1';for(;a[i]!='0';++i)if(i==81){ puts(a);return;}for(;c<='9';++c){for(f=0;f<9;++f)if(a[i-i%27+i%9 /3*3+f/3*9+f%3]==c||a[i%9+f*9]==c||a[i-i%9+f]==c)goto e;a[i]=c;J(a);a[i] ='0';e:;}}int main(int c,char**v){int t=0;if(c>1){for(;v[1][ t];++t);if(t==81){J(v[1]);return 0;}}puts("sudoku [0-9]{81}");return 1;}
