exatly the same as the previous line
But, in the previous line they r assigning the adress of the 5th element in the array to j right?
well yeah, its doing the same but assigning to k
The name of the array is also a pointer to its first position, so when you add + 4 to the first position it makes it go 4 steps inside the array
Arr+4=&arr[4]
So arr+0 is 10?
no, arr+0 = arr
&arr[0]
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