you're right Matt,
in ++t the comparison yields true.
In fact, unary operator (like ++) have a higher priority than binary ones.
This topic has been discussed many times here, but here it comes:
In t++, the value of t is stored and incremented right away. BUT the value used in the expression is the stored value not the result of the incrementation.
In ++t, the value is first incremented and then used so the result is available right away
HIH
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Valentin Crettaz
Sun Certified Programmer for
Java 2 Platform