how *t will still points to the last byte of memory in the char pointer outside the loop? I mean t++; is happening inside the while loop so how the value remains after the loop ends fot *t
t is effectively just a local variable and its scope is bigger than the scope of the loop. So all modifications of t are visible inside the function because the scope of this variable is the same as the scope of the function body basically here
Thanks bro I just tagged you in another Group but you replied in the untagged one 😅😅
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