Re^2: dependable way to test for a float

I have not totally isolated when it does not work yet. I am testing for a floating point number to separate from both booleans and strings all of which are values in a hash.

I just had deja vu writing this....yet no recollection of doing this in a past life.

Comment on Re^2: dependable way to test for a float

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Re^3: dependable way to test for a float by Roy Johnson (Monsignor) on May 05, 2005 at 02:22 UTC
For your purposes, are the integers not floats? I can't think of any case where the `int($n) != $n` test wouldn't work. Or would it be a float if it were '0.0' (for example)? Though you haven't isolated when it doesn't work, can you give any example for which it doesn't work? Caution: Contents may have been coded under pressure.	[reply] [d/l]
Re^4: dependable way to test for a float by vindaloo (Acolyte) on May 05, 2005 at 02:51 UTC
I will look into it more, I am really curious as to why this is happening.	[reply]
Re^4: dependable way to test for a float by vindaloo (Acolyte) on May 06, 2005 at 17:54 UTC
The error was in the test and a variable being empty in the comparison. My code is working fine now. I did come across this though and wondered why: `my $var = 23123412341234.01222222; print "var=$var\n"; if(int($var) != $var){ print "is a float\n"; } else{ print "not a float\n"; }` [download] Prints this: `var=23123412341234 is a float` [download] The test is always working for me, so that is a good confirmation. However, I was wondering why it prints the integer before the int() function touches the variable?	[reply] [d/l] [select]
Re^5: dependable way to test for a float by Roy Johnson (Monsignor) on May 06, 2005 at 19:12 UTC
It's because the number is so long it can't be represented numerically. If you `use bignum;`, it will behave more like you probably expect. Caution: Contents may have been coded under pressure.	[reply] [d/l]