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

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);