Thanks LanX.Understood but in second case which is a working condition $i's scope is in outer loop but how it got tucked in to inside loop.My understanding is if my is before a variable it's existence is strictly with in that boundary. In case 2 my is associated with $i so i think it only resides outside or where it is defined but i am confused how it got transported to loop.

In reply to Re^2: scoping inside the loop by Anonymous Monk
in thread scoping inside the loop by vyeddula

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