CODEWITHSUNDEEP
c++cprintf
vhbsouza
vhbsouza

Reputation: 407

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

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?

Upvotes: 13

Views: 41614

Answers (3)

Ruslan Yushchenko
Ruslan Yushchenko

Reputation: 636

Here is my approach. It seems ugly but does work in most cases e.g. 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);

Upvotes: 2

laifjei
laifjei

Reputation: 646

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

Upvotes: 0

Kerrek SB
Kerrek SB

Reputation: 476950

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

Upvotes: 27

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