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

Use this:

floor(0.6000002*10000)/10000

Answer 2

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 3

Realistically that's not possible. It's not a C++ limitation, but just the way floating point works. For many values there are no precise representations, so you can't simply truncate to a number of digits.

You could truncate when printing using printf format strings.

If you really need to be able to store only a limited number of digits, I suggest you use a fixed-precision data type instead.

Answer 4

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 5

First it is important to know that floating point numbers are approximated. See the link provided by @Greg Hewgill to understand why this problem is not fully solvable.

But here are a couple of solutions to the problem that will probably meet your need:

Probably the better method but less efficient:

char sz[64];
double lf = 0.600000002;
sprintf(sz, "%.4lf\n", lf); //sz contains 0.6000

double lf2 = atof(sz);

//lf == 0.600000002;
//lf2 == 0.6000

printf("%.4lf", lf2); //print 0.6000

The more efficient way, but probably less precise:

double lf = 0.600000002;
int iSigned = lf > 0? 1: -1;
unsigned int uiTemp = (lf*pow(10, 4)) * iSigned; //Note I'm using unsigned int so that I can increase the precision of the truncate
lf = (((double)uiTemp)/pow(10,4) * iSigned);