How to truncate a floating point number after a certain number of decimal places (no rounding)?

Question

I'm trying to print the number 684.545007 with 2 points precision in the sense that the number be truncated (not rounded) after 684.54 .

When I use

var = 684.545007;
printf("%.2f\n",var);

it outputs 684.55 , but what I'd like to get is 684.54 .

Does anyone knows how can I correct this?

Answer 1

What you're looking for is truncation . This should work (at least for numbers that aren't terribly large):

printf(".2f", ((int)(100 * var)) / 100.0);

The conversion to integer truncates the fractional part.

In C++11 or C99, you can use the dedicated function trunc for this purpose (from the header <cmath> or <math.h> . This will avoid the restriction to values that fit into an integral type.

std::trunc(100 * var) / 100     // no need for casts

Answer 2

Here is my approach. It seems ugly but does work in most cases eg var can be larger then int, can be zero or bizarre '-0'. It does not handle infinities and NaNs though.

double var = 684.545007; // or whatever
double var_trunc = var>=0. ? floor(var*100.)/100. : ceil(var*100.)/100.;
printf ("%g\n", var_trunc);

Answer 3

printf("%.2f\n", var - 0.005);