Limit floating point precision?

Question

Is there a way to round floating points to 2 points? E.g.: 3576.7675745342556 becomes 3576.76.

Answer 1

round(x * 100) / 100.0

If you must keep things floats:

roundf(x * 100) / 100.0

Flexible version using standard library functions:

double GetFloatPrecision(double value, double precision)
{
    return (floor((value * pow(10, precision) + 0.5)) / pow(10, precision)); 
}

Answer 2

If you are printing it out, instead use whatever print formatting function available to you.

In c++

cout << setprecision(2) << f; 

For rounding to render to GUI, use std::ostringstream

Answer 3

Multiply by 100, round to integer (anyway you want), divide by 100. Note that since 1/100 cannot be represented precisely in floating point, consider keeping fixed-precision integers.

Answer 4

For those of you googling to format a float to money like I was:

#include <iomanip>
#include <sstream>
#include <string>

std::string money_format (float val)
{
    std::ostringstream oss;

    oss << std::fixed << std::setfill ('0') << std::setprecision (2) << val;

    return oss.str();
}
// 12.3456 --> "12.35"
// 1.2 --> "1.20"

You must return it as a string. Putting it back into a float will lose the precision.

Answer 5

Don't use floats. Use integers storing the number of cents and print a decimal point before the last 2 places if you want to print dollars. Floats are almost always wrong for money unless you're doing simplistic calculations (like naive economic mathematical models) where only the magnitude of the numbers really matters and you never subtract nearby numbers.

Answer 6

try use

std::cout<<std::setprecision(2)<<std::cout<<x;

should works and only 2 digit after the floating point appear.

Answer 7

To limit the precision:
If x is a float, no rounding:
(shift up by 2 decimal digits, strip the fraction, shift down by 2 decimal digits)

((int)(x*100.0)) / 100.0F

Float w/ rounding:

((int)(x*100.0 + 0.5F)) / 100.0F

Double w/o rounding:

((long int)(x*100.0)) / 100.0

Double w/ rounding:

((long int)(x*100.0 + 0.5)) / 100.0

Note: Because x is either a float or a double, the fractional part will always be there. It is the difference between how a # is represented (IEEE 754) and the #'s precision.
C99 supports round()

Answer 8

Try this, it works perfectly

float=3576.7675745342556;
printf("%.2f",float);

change some objects in it to see and learn the code.

Answer 9

yourFloatNumber= Float.Round(yourFloatNumber,2); // In C#