Set the digits after decimal point

Question

I have a float number for example 12.12123 Is there a function which would display only number with 2 digits after decimal point 12.12 ?

Here is the code:

y1 = ( c1 - (a1 * x)) / b1;
 y2 = ( c2 - a2 * x) / b2;

if (y1 == y2)
  cout << "The same";

so if the y1 = 1.001 and the y2 = 1.002 they do not appear as the same.

I tried to add. cout.setf(ios::fixed, ios::floatfield); cout.precision(2);

but it does not seem to help.

Answer 1

/* The C way */
#include <stdio.h>
...
float f = 12.12123f;
printf("%.2f",f);

// The C++ way
#include <iostream>
...
float f = 12.12123f;
std::cout.setf(std::ios_base::fixed, std::ios_base::floatfield);
std::cout.precision(2);
std::cout << f;

// The alternative C++ way
#include <iostream>
#include <iomanip>
...
float f = 12.12123f;
std::cout << std::fixed << std::setprecision(2) << f;

In C, the 0 padding is added automatically to the right if there are not enough digits to print. In the C++ examples, instead, this is disabled; to enable this behavior, you should enable the fixed mode on the stream with std::fixed (or enabling the relevant stream flags with std::ios_base::setf()).

Edit: I remembered wrong; if fixed is not set, the precision setting says to the stream the total number of digits to display, including also the ones before the decimal point. So, in this case I think that the only way is to use the fixed mode (examples fixed), which will yield the same behavior of printf.

---

Links:

• std::ios_base::precision() reference;
• std::setprecision reference;
• printf reference.

Answer 2

You're looking for printf("%.2f", 12.12123); or:

#include <iostream>
#include <iomanip>

using namespace std;
cout << fixed << setprecision(2) << 12.12123;

EDIT: Question changed, so does the answer.

You never want to use direct equality with floating point, you always compare within epsilon tolerance. Your epsilon is just quite large.

Replace if (y1 == y2) with if (abs(y1 - y2) < 0.01).