About round command?

  • Thread starter Thread starter hamster
  • Start date Start date
H

hamster

I don't know how to write the code for round command?
For example if more than .5 then round up,if less than .5
round down such as 4.3 round to 4.0 and 5.5 round to 6.0
 
There are lots of solutions to this.

If you would like to make your own round-function take a look at the modulo
function %.

Have you seen the
Math::Round(3.44, 1); //Returns 3.4.
Math::Round(3.45, 1); //Returns 3.4.
Math::Round(3.46, 1); //Returns 3.5.

in the .Net framework?
 
Lars-Inge Tønnessen said:
There are lots of solutions to this.

If you would like to make your own round-function take a look at the modulo
function %.
Click to expand...

% is an operator, not a function, and it is not defined for non-integral
types.

Gerhard Menzl
 
I don't know how to write the code for round command?
For example if more than .5 then round up,if less than .5
round down such as 4.3 round to 4.0 and 5.5 round to 6.0
Click to expand...

A simple trick is to add 0.5 to the value (if positive) or remove 0.5 (if
negative). By converting the value back into an integer it will
automatically round to the right value.

Ex:

4.0 + 0.5 = 4.5... truncated (cast) = 4.0
4.3 + 0.5 = 4.8... truncated (cast) = 4.0
4.5 + 0.5 = 5.0... truncated (cast) = 5.0
4.7 + 0.5 = 5.2... truncated (cast) = 5.0

Here's a macro that implements the "logic" :

#define ROUND_INT(x) ((int)((x)+(((x)>0)?0.5:-0.5)))

If you need it for a float/double, just cast back into one...

Alex.
 
This is where your programming creativity comes in when you develop your own
function/method/class. =:o)
 
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