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Aleksandr [31]
3 years ago
6

Given the following structure and variable definitions, struct customer { char lastName[ 15 ]; char firstName[ 15 ]; unsigned in

t customerNumber; struct { char phoneNumber[ 11 ]; char address[ 50 ]; char city[ 15 ]; char state[ 3 ]; char zipCode[ 6 ]; } personal; } customerRecord, *customerPtr; customerPtr = &customerRecord; write an expression that can be used to access the structure members in each of the following parts: a) Member lastName of structure customerRecord. b) Member lastName of the structure pointed to by customerPtr. c) Member firstName of structure customerRecord. d) Member firstName of the structure pointed to by customerPtr. e) Member customerNumber of structure customerRecord. f) Member customerNumber of the structure pointed to by customerPtr. g) Member phoneNumber of member personal of structure customerRecord. h) Member phoneNumber of member personal of the structure pointed to by customerPtr. i) Member address of member personal of structure customerRecord. j) Member address of member personal of the structure pointed to by customerPtr. k) Member city of member personal of structure customerRecord. l) Member city of member personal of the structure pointed to by customerPtr.
Computers and Technology
1 answer:
garri49 [273]3 years ago
6 0

Answer:

see explaination

Explanation:

a)

customerRecord.lastName

b)

customerPtr->lastName or (*customerPtr).lastName

c)

customerRecord.firstName

d)

customerPtr->firstName or (*customerPtr).firstName

e)

customerRecord.customerNumber

f)

customerPtr->customerNumber or (*customerPtr).customerNumber

g)

customerRecord.personal.phoneNumber

h)

customerPtr->personal.phoneNumber or (*customerPtr).personal.phoneNumber

i)

customerRecord.personal.address

j)

customerPtr->personal.address or (*customerPtr).personal.address

k)

customerRecord.personal.city

l)

customerPtr->personal.city or (*customerPtr).personal.city

m)

customerRecord.personal.state

n)

customerPtr->personal.state or (*customerPtr).personal.state

o)

customerRecord.personal.zipCode

p)

customerPtr->personal.zipCode or (*customerPtr).personal.zipCode

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