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Overloading Unary Operators Using Friend Function
class sign
{
int a,b,c;
public:
sign(){}; sign(int,int,int); void putdata(void);
friend void operator-(sign &);
};
void operator-(sign &s)
{s.a=-s.a;s.b=-s.b;s.c=-s.c; }
void sign::putdata(void)
{cout<<"a is: "<<a<<"\n"; cout<<"b is "<"\n"; cout<<"c is "<<c<<"\n";
}
sign::sign(int x,int y, int z)
{a=x;b=y;c=z; }
int main()
{sign s; //Implicit constructor calling s=sign(1000,2,10); //Explicit constructor calling s.putdata();
-s;
s.putdata();
cout<<endl;
s=sign(-2000,2,-5);
return 0;
#A palindrome is a string that reads the same from both the ends. Given a string S convert it to a palindrome by doing character replacement. Your task is to convert S to palindrom
hwat is the area fsdjlakl;aeklfjtealrtl;gka
P o i n t e r d e c l a r a t i o n f o r d a t a m e m b e r : M e t h o d 1 : i n t M : : * p x ; / / T h i s
C program to count the words: void CountWords(); void main() { printf("\n\tcount the words and enter string\n\n\n"); Count
decodethecode
What if I wish a local to "die" before the close} of the scope wherein it was created? Can I call a destructor on a local if I want to?
demonstrates shearing about origin
count the number of string in n-th padovan string
A: By keeping along with the C++ tradition of "there's more than one method to do that" (translation: "give programmers options & tradeoffs so they can choose what's best for them
write a c program to find input string using strlen(), strcpy(), strcat(),strncat(), strcmp().
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