Given the following section of code:

struct studinfo { char studentname[30]; ** This means that each record requires:

short age, credithours; 30 + 2 + 2 + 4 + 4 = 42 bytes

long weightindrams;

float gpa; };

struct studinfo student[6] = {{“Christie, Agatha”,, 96, 82, 37324, 3.05},

{“Shakespeare, Bill”, 256,12, 52253, 3.56}, {“Albee, Edward”, 76, 120, 72689, 2.78},

{“Voltaire”, 56, 82, 64766, 3.75}, {“Chaucer, Geoffry”, 420, 91, 41566, 3.25},

{“Grisham, John”, 46, 24, 66456, 1.57}} ;

The command: printf(“%lu\n”,&student[3].weightindrams); yields the output: 7686

Given an array of 6 records, each with 5 fields, our table would appear as:

.	Studentname	Age	credhrs	wgtindrams	gpa
Rec #\Bytes	30	2	2	4	4
0
1
2
3				7686
4
5

Where we are know that &student[3].weightindrams is 7686. Given that we know that address and we know how many bytes each field uses (and that each record requires 42 bytes of storage), we can determine the address of every field in the array of records:

.	Studentname	Age	credhrs	wgtindrams	gpa
Rec #\Bytes	30	2	2	4	4
0	7526	7556	7558	7560	7564
1	7568	7598	7600	7602	7606
2	7610	7640	7642	7644	7648
3	7652	7682	7684	7686	7690
4	7694	7724	7726	7728	7732
5	7736	7766	7768	7770	7774

If we fill in the table with the given data, the table would appear as:

.	Rec Addr:	Studentname	Age	credhrs	wgtindrams	gpa
Rec #\Bytes		30	2	2	4	4
0	7526	Christie, Agatha	96	82	37324	3.05
1	7568	Shakespeare, Bill	256	12	52253	3.56
2	7610	Albee, Edward	76	120	72689	2.78
3	7652	Voltaire	56	82	64766	3.75
4	7694	Chaucer, Geoffry	420	91	41566	3.25
5	7736	Grisham, John	46	24	66456	1.57

Now, to answer our questions:

a. What is the value of student[2].name? From the above table: Albee, Edward

b. What is the value of student[1].gpa? From the above table: 3.56

c. What is the value of student[6].age? Unknown OR Not a valid question

d. What is the value of &student[4]? From the above table: 7694

e. What the value of &student[0].credithours? From our address table: 7558 OR

From the above table: 7526 + 30 + 2 = 7558

f. What the value of &student[85].age Given the base address: 7526

&student[85].age = 7526 + 85 * 42 + 30

= 7526 + 3570 + 30 = 11126