(C++) How would you go about rounding numbers?

Ok Basically when I am run the program it will add both numbers 1.67 + 1.32 and give the result 3.00. Clearly, the sum of the displayed numbers should be 2.99 and not 3.00. The problem is that although the values in a and b have been displayed with two decimal digits, they were added internal to the program as three-digit numbers. The solution is to round the values in a and b before they are added to the statement c = a + b;

Using the int cast, how would I be able to round the values in variables a and b to the nearest hundredth(penny value) before they are added?

Code:

---

#include <iostream.h>
#include <iomanip.h>
void main(void)
{
  float a, b, c;

a = 1.674;
b = 1.322;
cout << setprecision(3) << a << endl;
cout << setprecision(3) << b << endl;
cout << "----\n";
c = a + b;
cout << setiosflags(ios::showpoint)
    << setprecision(3) << c << endl;

}

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Here is one way to round using a class and overiding the = operator.

Code:

---

class Math
{
public:
        int x;

        operator int()
        {
                return x;
        }

        int operator=(const double &num)
        {
                double y;
                if ( num < 0 )
                        y=num-.5;
                else
                        y=num+.5;
                x=int(y);
                return x;
        }
};

---

You would need to change the return to a double for your case. Add or subtrace .005 instead of .5

Quote:

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Originally posted by jeffcoulter
The problem is that although the values in a and b have been displayed with two decimal digits, they were added internal to the program as three-digit numbers.

---

Forget that.

There's nothing strange with you program above. 3.00 is the correctly rounded value to 2.998

you mean 2.996

Yes

double a, b, c;
int temp_round;

a = 1.674;
b = 1.322;

cout.setf(ios::fixed);
cout << setprecision(2) <<"a="<<a<< endl;
cout <<"b="<<b<< endl;
cout << "----\n";

/*
The following statement multiplies the double identifier named "a" (value of 1.67by 100 resulting in 167.4 and stores the result in an integer variable named temp_round. Due to truncation (float to int) temp_round holds the value of 167 (the .4 is truncated)
*/

temp_round= int(a*100);

/*
The following statement converts the variable temp_round (value of 167) to a double (value of 167.00) and divides it by 100.0 (resulting in 1.67) and stores it back the double identifier named "a"
*/
a = double(temp_round)/100.0;

//used the same conversion
temp_round= int(b*100);
b = double(temp_round)/100.0;

c = a + b;

cout << setprecision(2)<< "c = " <<c<< endl;
system ("pause");