Re: Comparison based on type of value held by variables

They do match because their value is 0.

please also note, that you should remove leading zeros. Please compare the outputs:

print 5+010;
print 5+010.4;
print 5+ 10;
print 5+ 10.4;
[download]

13 134 15 15.4

s$$([},&%#}/&/]+}%&{})*;#$&&s&&$^X.($'^"%]=\&(|?*{%
+.+=%;.#_}\&"^"-+%*).}%:##%}={~=~:.")&e&&s""`$''`"e

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Re^2: Comparison based on type of value held by variables by ranjan_jajodia (Monk) on Oct 18, 2005 at 13:52 UTC
Hi Skeeve, So does that mean that all strings evaluate to 0 value in numerical comparison? Regards, Ranjan	[reply]
Re^3: Comparison based on type of value held by variables by Skeeve (Parson) on Oct 18, 2005 at 14:10 UTC
Yes. The numerical value of any string that is not a valid representation of a number is 0. `s$$([},&%#}/&/]+}%&{});#$&&s&&$^X.($'^"%]=\&(\|?{%` `+`.+=%;.#_}\&"^"-+%*).}%:##%}={~=~:.")&e&&s""`$''`"e	[reply] [d/l] [select]
Re^4: Comparison based on type of value held by variables by ranjan_jajodia (Monk) on Oct 18, 2005 at 14:11 UTC
Thanks Skeeve. I will keep that in mind. Regards, Ranjan	[reply]