How to truncate a floating point number after a certain number of decimal places (no rounding)?
c, c++, printf
Solution
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
Problem
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?