Why friend complex sum(complex s1, complex s2); is written?

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1.00/5 (2 votes)

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why friend complex sum(complex s1, complex s2); is written?
why not void sum(complex s1, complex s2) ?
why object s3 is declared in sum function and why is it written complex s3?

What I have tried:

C++

#include<iostream>
using namespace std;

class complex
{
	int a,b;
	public:
		void take(int x, int y)
		{
			a=x;
			b=y;
		}
		
		friend complex sum(complex s1, complex s2);
		void display(void)
		{
			cout<<"A is:"<<a<<endl;
			cout<<"B is:"<<b<<endl;
			cout<<"complex number is:"<<a<<"+"<<b<<"i"<<endl;
			cout<<endl;
		}
};

complex sum(complex s1, complex s2)
{
	complex s3;

	s3.a=s1.a+ s2.a;
	s3.b=s1.b + s2.b;
	return s3;
}

int main()
{
	complex c1,c2,c3;
	
	c1.take(4,5);
	c1.display();
	
	c2.take(2,6);
	c2.display();
	
	c3=sum(c1,c2);
	c3.display();
	return 0;
}

Posted 6-Sep-21 3:19am

Member 14897676

Updated 6-Sep-21 4:53am

v2

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OriginalGriff · Answer 1 · 2021-09-06T03:38:00

Solution 1

Quote:
why
C++
friend complex sum(complex s1, complex s2);
is written?

Start here: friend declaration - cppreference.com[^]
You will probably need to understand classes fairly well before it makes any sense to you though.

Quote:
why not
C++
void sum(complex s1, complex s2)
?

Because if you don't provide a return type you would have to overwrite the values in one of the two parameters passed to the function in order to return the combined value - a function called sum should logically add the parameter values together, and return the combined value. Without a return value, it can't do that.

Quote:
why object s3 is declared in sum function and why is it written complex s3?

Haven't got a clue, can't see that code at all!

Posted 6-Sep-21 3:38am

OriginalGriff

Comments

Member 14897676 6-Sep-21 9:46am

I have updated the code ..pls check it

OriginalGriff 6-Sep-21 10:07am

It's declared to give you somewhere to store the result, and something to return!
Seriously, you need to take a couple of steps back and brush up on classes (and C++ basics as well) as this is pretty basic stuff!

Member 14897676 6-Sep-21 9:48am

but why the return value is complex ? bcs the two parameters are of complex data type?
what if I write return type as void?

OriginalGriff 6-Sep-21 10:05am

If the return type is void, you can't return anything ...

And the sum of two complex numbers is a complex number. Even if the imaginary parts have equal magnitude and opposite sign (in which case the result is technically a real number) since real numbers are a subset of complex numbers. Since most complex sums won't produce a real value, you need to be able to return a complex value, and you can't "swap types" at run time.
Hence the fucnction is declared as returning a complex value.

CPallini 6-Sep-21 10:53am

5.

CPallini · Answer 2 · 2021-09-06T04:53:00

In addition of what our OriginalGriff already said, the arguments are usually passed (for preformance reasones) as const references, namely:

C++

#include<iostream>
using namespace std;

class complex
{
  int a,b;
  public:
    void take(int x, int y)
    {
      a=x;
      b=y;
    }

    friend complex sum(const complex & s1, const complex & s2);

    void display(void)
    {
      cout<<"A is:"<<a<<endl;
      cout<<"B is:"<<b<<endl;
      cout<<"complex number is:"<<a<<"+"<<b<<"i"<<endl;
      cout<<endl;
    }
};

complex sum(const complex & s1, const complex & s2)
{
  complex s3;

  s3.a = s1.a + s2.a;
  s3.b = s1.b + s2.b;

  return s3;
}

int main()
{
  complex c1, c2, c3;

  c1.take(4,5);
  c1.display();

  c2.take(2,6);
  c2.display();

  c3=sum(c1,c2);
  c3.display();
}