调用方法将输出返回到同一类C ++的另一个方法中

Question

The code below tries to adds two numbers. I'd like to build a class that does that , for the fun of it.

So, I have method add that adds the 2 numbers and returns result I'd like the method getAdd to print the return output of the add method.

How do I do this ??

Below is the code

#include <iostream>
#include <string>

class Addition 
{
    public:
        Addition(double a=0, double b=0);
        double add(double a, double b);
        void getAdd() const;

    private:
        double mA;
        double mB;
};

Addition::Addition(double a, double b) : mA(a), mB(b) {}

double Addition::add(double a, double b)
{
    double result = a + b;

    return result;

}    

void Addition::getAdd() const
{   

    std::cout << "Result is " << //print here the result output of add method << std::endl ;
}          

int main()
{
    Addition sum;
    double num_a;
    double num_b;
    std::cout << "Enter first number to be added\n";
    std::cin >> num_a;
    std::cout << "Enter second number to be added\n";
    std::cin >> num_b;
    sum.add(num_a,num_b);
    sum.getAdd();

    return 0;

} 

Answer 1

Pass the parameters to getAdd?

void Addition::getAdd(double a, double b) const
{
   std::cout << "Result is " << add(a,b);
}

Store the result?

class Addition 
{
    public:
        Addition(double a=0, double b=0);
        double add(double a, double b);
        void getAdd() const;

    private:
        double mResult;
};

double Addition::add(double a, double b)
{
    mResult = a + b;

    return mResult ;
}    
void Addition::getAdd() const
{
   std::cout << "Result is " << mResult ;
}

Answer 2

If I get you right, first you need to declare another variable (it can be double result; ). Instead of using a local variable called result , use the one you've just created. Your class would be:

class Addition 
{
    public:
        Addition(double a=0, double b=0);
        double add(double a, double b);
        void getAdd() const;

    private:
        double mA;
        double mB;
        double result;
};

Your add() method would be:

double Addition::add(double a, double b)
{
    result = a + b;

    return result;

}    

And your getAdd() would be:

void Addition::getAdd() const {   

    std::cout << "Result is " << result << std::endl ;
}  

You need this because your getAdd() is a const function, and it cannot call add() directly. =)

Answer 3

Define another class member double last_added; . Initialise it to something sensible in the constructor. You need to do that in case someone calls getAdd() before the first call of add() : the behaviour on reading an uninitialised variable is undefined in C++.

Then, in your add function, replace the return with

return last_added = result; /*the single = is deliberate*/

You could refactor getAdd() , to return that value:

return last_added;

or write the last value to standard output using std::cout << "Result is " << last_added;

This is one approach. Another one is to use the mA and mB members that you've already defined. The function body of add becomes

return mA = a + mB = b;

That's a real touchstone for your operator precedence knowledge. This has the advantage of your not needing an extra field, since you could write std::cout << "Result is " << mA + mB; . But then you're effectively computing the addition twice which is cumbersome.

Do also consider implications of multithreading.

Answer 4

We have two ways of achieving the desired result,

1. Call the function add() from getAdd(), for that you'll have to change the definition of getAdd() to include two parameters.

   void Addition::getAdd(double a, double b) const {
   std::cout << "Result is " << add(a,b)<< std::endl ; }

(or)

2. store the result in a variable and print it. There are two approaches to it

a.

 class Addition 
{
    public:
        Addition(double a=0, double b=0);
        double add(double a, double b);
        void getAdd() const;

    private:
        double mA;
        double mB;
        double add;
};

Addition::Addition(double a, double b) : mA(a), mB(b) {}

double Addition::add(double a, double b)
{
    add = a + b;

    return add;

}    

void Addition::getAdd() const
{   

    std::cout << "Result is " << add() << std::endl ;
} 

In this case, make sure to call add() first and then getAdd()

sum.add(num_a,num_b);
sum.getAdd();

Else garbage value will be printed.

Another option is as follows

 class Addition 
{
    public:
        Addition(double a=0, double b=0);
        double add(double a, double b);
        void getAdd() const;

    private:
        double mA;
        double mB;
        double add;
};

Addition::Addition(double a, double b) : mA(a), mB(b) {}

double Addition::add(double a, double b)
{
    double result =  a + b;

    return result;

}    

void Addition::getAdd(double a, double b) const
{   
    add = add(a,b)
    std::cout << "Result is " << add << std::endl ;
} 

and finally call getAdd(a,b);

Answer 5

I don't understand the meaning of your methods, and your Addtion class. It's strange.

Maybe your Addtion class is just a class have some methods without any members. The double add(double a, double b) function is enough for you to get the sum of two double variables. The void getAdd(double a, double b) function may print some info about sum of two double variables.

class Addtion
{
public:
    static double add(double a, double b);
    static void getAdd(double a, double b);
};

Or, maybe you can just use namespace. Your really need is just some functions, not a class with members.

namespace myaddtion
{
    double add(double a, double b);
    void addinfo(double a, double b);
}