Rounding a float and casting to an int

**pollypocket4eva** · 01-09-2009

Is this the right way to do this?

Code:

int input;

float temp = input * (2/3);

int output = roundf(temp);

Thank you

**laserlight** · 01-09-2009

Assuming that you actually give input a value, the problem would be that (2/3) results in integer division which truncates the result to 0. If you change it to (2.0f/3.0f) it should work. However, I wonder if this would be simpler yet give identical results:

Code:

int output = (input * 2) / 3;

**pollypocket4eva** · 01-09-2009

Ah, yes. thank you.
So, for
int output = (input * 2) / 3;
Is it truncating or rounding?

**laserlight** · 01-09-2009

Originally Posted by pollypocket4eva

So, for
int output = (input * 2) / 3;
Is it truncating or rounding?

Ah yes, that would truncate, not round.

**pollypocket4eva** · 01-09-2009

For my purpose, either is fine. But was curious for future reference.
Thank you

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