MADE EASY TEST SERIES

Question

Consider the following C-fragment where size of int is 1 B

Which is the output of above program?

Comments

• *(S+2) here S decays to pointer to the first element. It becomes (int *)
• (&S+1) :: : TYPE of &S is (int *)[]..i.e. pointer to an array of 6 elem.
• Depending on the type of these pointers.. they get incremented by the amount of Bytes they point.

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Great explanation.

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For those who are having a hard time with this:
https://stackoverflow.com/questions/16312299/starting-address-of-array-a-and-a

Answer 1

Given that s[6]={10,20,30,40,50,60}

here *(s+2) = s[2]

and s[2]=30

and int * str =(int * )(&s +1)

here str pointer will point to the array s[]= {20,30,40,50,60}

hence *(str-2) = s[-2]

here due to cyclic nature of int

s[-2] will be 50 i.e *(str -2) = 50

hence output is 30,50

Comments

Nice one @ Sanjay