trunc an int

[TyzA]

Hi,

Can anybody tell me how I can truncate an int.
I use now this:
sprintf "%.0f", ($num_records / $num_rows);

If I have a value like 17.6 I will always get 18. If i have 17.4, I'll get 17. So it rounds.

Is there a way that I will just get 17, no mather what the decimal is.

Thanks for any response.

Tijs Programming is like sex: one mistake and you have to support it for the rest of your life.

 

[TyzA]

Hi again,

sorry, but I made a mistake. Instead of getting 17 I want to get 18. So it must always round up, even when you have somthing like 17.3

Thanks again.

Tijs Programming is like sex: one mistake and you have to support it for the rest of your life.

 

[justice41]

Try

$trun = 1 + int($num_records / $num_rows);

jaa

 

[goBoating]

I think that will always round up, even if records/rows happened to be an int. This might be a little closer.

a generic treatment...

#!perl
$in = $ARGV[0] || die "No input from command line.\n";

$num = ($in =~ /\.\d/) ? int($in + 1) : $in ;
print "NUM: $num\n";

'hope this helps

If you are new to Tek-Tips, please use descriptive titles, check the FAQs, and beware the evil typo.

 

[TyzA]

Thanks guys,

I've tried them both, but only the first one seems to work.

Thanks again,

Tijs Programming is like sex: one mistake and you have to support it for the rest of your life.

 

[justice41]

goBoating is correct, if $num_rows evenly divides $num_records then you will still get the next highest integer (for example if $num_records = 85 and $num_rows = 5 then you will get 18 as the rounded value even though $num_records/$num_rows == 17. Perhaps you didn't implement it correctl. This worked for me

my $in = $num_records / $num_rows;
my $num = ($in =~ /\.\d/) ? int($in + 1) : $in ;

jaa

 

[TyzA]

Thanks justice41

It works also like that now and goboating was right about the rounding.

Tijs Programming is like sex: one mistake and you have to support it for the rest of your life.