1. EECE.2160 ECE Application Programming
Instructors:
Dr. Michael Geiger
Spring 2018
Lecture 25
Structure examples; nested structures

2. ECE Application Programming: Lecture 25
Lecture outline
Announcements/reminders
Program 5 regrades due Monday, 4/9
Program 6 due today
Program 7 due Friday, 4/13
Today’s lecture
Review: Structures
Structure examples
Nested structures
9/20/2018
ECE Application Programming: Lecture 25

3. ECE Application Programming: Lecture 25
Review: Structures
User-defined types; example:
typedef struct {
char first[50];
char middle;
char last[50];
unsigned int ID;
double GPA;
} StudentInfo;
Can define variables of that type
Scalar: StudentInfo student1;
Array: StudentInfo classList[10];
Pointer: StudentInfo *sPtr = &student1;
Access members using
Dot operator: student1.middle = ‘J’;
Arrow (if pointers): sPtr->GPA = 3.5;
Typically passed to functions by address
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ECE Application Programming: Lecture 25

4. Example: Using structures
What does the following print?
typedef struct {
double real;
double imag;
} Complex;
int main() {
Complex a = {1, 2};
Complex b = {3.4, 5.6};
Complex c, d, e;
printf("A = %.2lf+%.2lfi\n",
a.real, a.imag);
printf("B = %.2lf+%.2lfi\n", b.real, b.imag);
c = a;
d.real = a.real + b.real;
d.imag = a.imag + b.imag;
e.real = a.real - b.real;
e.imag = a.imag - b.imag;
printf("C = %.2lf+%.2lfi\n", c.real, c.imag);
printf("D = %.2lf+%.2lfi\n", d.real, d.imag);
printf("E = %.2lf+%.2lfi\n", e.real, e.imag);
return 0; }
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ECE Application Programming: Lecture 25

5. ECE Application Programming: Lecture 25
Example solution
A = i B = i C = i D = i E = i Note: code in handout has spaces before and after ‘+’ for readability; code on previous slide doesn’t because it wouldn’t fit!
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ECE Application Programming: Lecture 25

6. Structures and functions
Can pass structures to functions
int f(StudentInfo s);
Structures consume significant memory
Usually much more efficient to simply pass pointer
int g(StudentInfo *p);
Access structure through pointer: -> operator
Handles dereferencing and field access
Example: p->GPA = 3.0;
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ECE Application Programming: Lecture 25

7. Example: Structures and functions
Write the following functions that use the StudentInfo structure
Given a pointer to a single StudentInfo variable, print all of the student info to the screen using the following format:
Michael J. Geiger
ID #
GPA: 1.23
Given an array of StudentInfo variables and the size of the array, compute and return the average GPA of all students in the list
Prompt the user to enter 3 lines of input (using the format below), read the appropriate values into StudentInfo elements, and return a value of type StudentInfo
Format (user input underlined)
Enter name: Michael J. Geiger
Enter ID #:
Enter GPA: 1.23
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ECE Application Programming: Lecture 25

8. ECE Application Programming: Lecture 25
Example solution
void printStudent(StudentInfo *s) { printf(“%s %c. %s\n”, s->first, s->middle, s->last); printf(“ID #%u\n”, s->ID); printf(“GPA %.2lf\n”, s->GPA); }
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ECE Application Programming: Lecture 25

9. Example solution (cont.)
double avgGPA(StudentInfo list[], int n) { int i; double sum = 0; for (i = 0; i < n; i++) sum += list[i].GPA; return sum / n; }
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ECE Application Programming: Lecture 25

10. Example solution (cont.)
StudentInfo readStudent() { StudentInfo s; printf(“Enter name: ”); scanf(“%s %c. %s”, s.first, &s.middle, s.last); printf(“Enter ID #: ”); scanf(“%u”, &s.ID); printf(“Enter GPA: ”); scanf(“%lf”, &s.GPA); return s; }
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ECE Application Programming: Lecture 25

11. ECE Application Programming: Lecture 25
Nested structures
Structures can contain other structures:
typedef struct {
char first[50]; // First name
char middle; // Middle initial
char last[50]; // Last name
} Name;
Name sname; // Student name
unsigned int ID; // ID #
double GPA; // Grade point
} SINew;
Will need multiple dot operators to access field within nested structure
Given SINew s1;
s1.sname  Name structure within s1
s1.sname.middle  middle initial of name within s1
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ECE Application Programming: Lecture 25

12. ECE Application Programming: Lecture 25
PE4 (Structures)
Given header files, main program
Complete specified functions
For the Name structure
void printName(Name *n): Print the name pointed to by n, using format <first> <middle>. <last>
void readName(Name *n): Prompt for and read a first, middle, and last name, and store them in the structure pointed to by n
SINew functions on next slide
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ECE Application Programming: Lecture 25

13. Today’s exercise (continued)
Given header files, main program
Complete specified functions
Name functions on previous slide
For the SINew structure
void printStudent(SINew *s): Print information about the student pointed to by s
void readStudent(SINew *s): Prompt for and read information into the student pointed to by s
void printList(SINew list[], int n): Print the contents of an array list that contains n StudentInfo structures
int findByLName(SINew list[], int n, char lname[]): Search for the student with last name lname in the array list. Return the index of the structure containing that last name, or -1 if not found
int findByID(SINew list[], int n, unsigned int sID): Search for the student with ID # sID in the array list. Return the index of the structure containing that last name, or -1 if not found
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ECE Application Programming: Lecture 25

14. ECE Application Programming: Lecture 25
Final notes
Next time
Exam 2 Review
Reminders:
Program 5 regrades due Monday, 4/9
Program 6 due today
Program 7 due Friday, 4/13
9/20/2018
ECE Application Programming: Lecture 25