1.0100.

Given the record structure and that &student[2] is 7828:

Each record requires: 11 + 28 + 30 + 2 + 2 + 4 + 4 + 4 + 4 = 89 bytes of storage.

The base address of the array would be: 7828 – 2 * 89 = 7828 – 178 = 7650

The linked list would thus appear as:

Addr	ssn	name	address	age	IQ	gpa	balance	next	prev
7650	123456789	Gates, W.	99 Microsoft Ln	38	108	3.87	4500.00	7828	7739
7739	234567890	Ford, H.	17 Michigan Ave	125	115	3.10	250.00	7650	8006
7828	345678901	Trump, D,	1 Trump Tower	55	110	1.86	500.00	NULL	7650
7917	456789012	Astor, J.J.	20 Wall Street	76	100	2.76	1000.00	8006	NULL
8006	567890123	Carnegie, A.	100 Steel St.	81	79	2.25	0.00	7739	7917

And the contents of the variables first and last are:

first	last
7917	7828

Therefore:

Yields:

a. printf(“%lu\n”,first); 7917

b. printf(“%lu\n”,last); 7828

c. printf(“%s\n”, first->address); 20 Wall St.

d. printf(“%lu\n”,&student[1].balance); 7739 + 11 + 28 + 30 + 2 + 2 + 4 = 7816

e. printf(“%f5.2\n”,student[3].next->balance); 8006 -> 0.00 (Carnegie’s Balance)

f. printf(“%lu\n”, ++first); Prefix notation: 7917 + 89 = 8006

g. printf(“%d\n”,last->previous->age); 7828 -> 7650 -> 38 (Gate’s age)