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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.

See Question&Answers more detail:os

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1 Answer

/* 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.


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