is there a way to round down float type in c++?

Question

How could I round down float type variable? (using printf)

for example,  
float x = 3.35;  
printf("%.1f", x);

In this situation, it will print out 3.4 but i want it to print 3.3.
Is there any way to do it? (without using trunc or floor function)

Related and/or possible duplicate for changing the rounding mode generically: https://stackoverflow.com/q/6867693/10871073 — Adrian Mole, Mar 21 '22 at 12:11
Are you aware that `3.35` cannot be represented exactly in binary? It's approximated as 3+1/4+1/16+... — MSalters, Mar 21 '22 at 13:37
What's wrong with `floor()` and `trunc()` ? They exist for a reason. — Michaël Roy, Mar 21 '22 at 13:45
@MichaëlRoy: They'd round to `3.0f`, not `3.3f` for starters. And that's not simply solved by a `/10.0f` because `3.3f` cannot be represented in binary, unlike `3.0f`. — MSalters, Mar 21 '22 at 13:53
@MSalters `printf("%.1f", floorf(x*10.f)/10.f));` is the usual way to do it. — Michaël Roy, Mar 21 '22 at 15:11

score -1 · Answer 1 · answered Mar 21 '22 at 13:40

-1

No, you can't change the rounding mode of printf. It's implementation-defined.

answered Mar 21 '22 at 13:40

MSalters

Yes you can ([`fesetround`](https://en.cppreference.com/w/c/numeric/fenv/feround)), and it's standard since C99 and C++11. – Ben Voigt Mar 23 '22 at 22:17
@BenVoigt: I double-checked C99, and it just describes the `%f` format as "The value is rounded to the appropriate number of digits." in the _normative_ text. But "correctly rounded." is listed as a _recommended practice_, `fesetround` exists, but whether it accepts the requested "round down" mode depends on `FE_DOWNWARD` - which is implementation-defined. – MSalters Mar 23 '22 at 23:43

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