Truncate a decimal value in C++

Question

What\'s the easiest way to truncate a C++ float variable that has a value of 0.6000002 to a value of 0.6000 and store it back in the variable?

Answer 1

Similar to other answers, BUT you must not forget that round, floor and trunc are different by definition. See the definition and output example of the following:

http://www.cplusplus.com/reference/cmath/trunc/

In this case we need to trucate with a precision of 4 decimals and get rid of non-significant decimals:

trunc(valueToTrunc*10000)/10000

or

value = (double)((int)(valueToTrunc*10000))/(double)10000

Answer 2

A good reference for why this happens can be found in What Every Computer Scientist Should Know About Floating Point Arithmetic by David Goldberg.

Answer 3

Here is a function using the advice in other answers and an example of its use:

#include <iostream>
#include <cmath>

static void Truncate(double& d, unsigned int numberOfDecimalsToKeep);

int main(int, char*[])
{

  double a = 1.23456789;
  unsigned int numDigits = 3;

  std::cout << a << std::endl;

  Truncate(a,3);

  std::cout << a << std::endl;

  return 0;
}

void Truncate(double& d, unsigned int numberOfDecimalsToKeep)
{
    d = roundf(d * powf(10, numberOfDecimalsToKeep)) / powf(10, numberOfDecimalsToKeep);
}

Answer 4

you can use this:

int n = 1.12378;
cout << fixed << setprecision(4) << n;

the output will be: 1.1238

not: 1.1237

it just sea the last decimal number not all: 12.123447 ==> 12.1234

be careful about that...