Transcript Chapter 16 - Structures

Chapter 16 Pointers and Structures
Pointers

What is the difference sptr and ssptr?
char* sptr[ ] = { "One", "Two", "Three" };
char** xsptr = sptr;
sptr
"One\0"
xsptr
"Two\0"
"Three\0"
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Concepts to Learn…
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enum
Structures
typedef’s
structs in structs
Array of structs
struct Pointers
Union
Bit Fields
Dynamic Memory Allocation
Linked List
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enum
enum
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Closely related to the #define preprocessor.
Define a list of aliases which represent integer
numbers.
The advantage of enum over #define is that it has
scope.
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Two types:
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Only visible within the block it was declared.
Named: enum greekType { ALPHA, BETA, GAMMA };
Unnamed: enum { HOMER, MARGE, BART, LISA };
Values start at zero, unless specified.
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enum
enum Examples
enum numbers { zero, one, two, three };
enum animals { cat=1, dog, cow=9, sheep, };
enum plants { grass, roses, cabbages, oaktree };
enum diseases { heart, skin, brain, circulatory };
enum quarks { charmed, strange, truth, beauty };
enum treasures { rubies, sapphires, gold, silver };
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Structures
Structures
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A structure is a collection of variables, possibly of
different types, grouped together under a single
name (tag).
Structures help organize complicated data.
A structure must be defined prior to a structure
variable being declared.
Structure definitions include a tag, member
elements, and a variable definition.
 The variable definition is optional.
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Structures
Structures

Structure definitions inform the compiler what the
structure will look like.
struct flightType
{
char flightNum[7];
int altitude;
int longitude;
int latitude;
int heading;
double airSpeed;
};

/* max 6 characters */
/* in meters */
/* in tenths of degrees */
/* in tenths of degrees */
/* in tenths of degrees */
/* in km/hr */
Structure definition does not allocate memory.
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Structures
Structures
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To allocate memory for a struct, we declare a variable
using the new structure definition type.
struct flightType plane;
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Structure members are laid
out in the order specified
by the definition.
Memory is now allocated,
and can be accessed as
individual members of this
variable with the “dot”
operator:
plane.altitude = 10000;
plane.heading = 800;
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plane.flightNum[0]
plane.flightNum[1]
plane.flightNum[2]
plane.flightNum[3]
plane.flightNum[4]
plane.flightNum[5]
plane.flightNum[6]
plane.altitude
plane.longitude
plane.latitude
plane.heading
plane.airspeed
Structures
0x0000(SP)
0x0002(SP)
0x0004(SP)
0x0006(SP)
0x0008(SP)
0x000a(SP)
0x000c(SP)
0x000e(SP)
0x0010(SP)
0x0012(SP)
0x0014(SP)
0x0016(SP)
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Structures
Structure Example
#include <stdio.h>
#include <string.h>
Does not allocate
memory
struct person
{
char name[80];
int ssn;
};
struct person barney, fred;
Allocates two memory
structs
int main()
{
strcpy(barney.name, "Rubble, Barney");
barney.ssn = 555234561;
strcpy(fred.name, "Flintstone, Fred");
fred.ssn = 123451234;
printf(“\n%s %d", fred.name, fred.ssn);
printf(“\n%s %d", barney.name, barney.ssn);
}
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typedef’s
Typedef’s
Using Naturally Named Data Types
typedef’s
Why typedef?
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You use variables with logical names.
Why not use data types with logical names?
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Is an “int” 8-bits, 16-bits, 32-bits?
What’s a “long”?
Better question: why memorize it?
Most integer data types are platform dependent!!!
typedef’s make your code more portable.
Syntax:
typedef <type> <name>;
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typedef’s
How To Use typedef’s
1) Create a logical data type scheme.


A signed 8-bit number could be "s8"
An unsigned 16-bit number could be "u16"
2) Create a “typedef.h” file for each microcontroller
platform you use.
3) #include "typedef.h" in each of your files.
4) Use your new data type names.
typedef unsigned int u16;
typedef unsigned char u8;
u16 number;
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typedef’s
typedef.h Example
typedef
typedef
typedef
typedef
typedef
typedef
unsigned char u8;
signed char s8;
unsigned short u16;
signed short s16;
unsigned long u32;
signed long s32;
Replace:
unsigned char variable;
with:
u8 variable;
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typedef’s
typedef’s
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Typedef declarations provide no additional
functionality.
Makes code more readable by giving applicationspecific names to variable types.
typedef int Color;
typedef struct flightType Flight;
Color pixels[500];
Flight plane1, plane2;
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structs in structs
Structures in Structures

One field of a struct can be another structure
struct addressStruct
{
char street[80];
char city[32];
int zipCode;
};
struct person
{
char initials[4];
int ssn;
int height;
int weight;
struct addressStruct address;
} tom;
person
initials
ssn
height
weight
address
street
city
int main()
{
tom.ssn = 555123456;
tom.weight = 150;
tom.address.zipCode = 84062;
…
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Array of structs
Arrays of Structures
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Can declare an array of structs:
typedef struct flightType
{
char flightNum[7];
int altitude;
int longitude;
int latitude;
int heading;
double airSpeed;
} Flight planes[100];
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/* max 6 characters */
/* in meters */
/* in tenths of degrees */
/* in tenths of degrees */
/* in tenths of degrees */
/* in km/hr */
Each array element is a structure.
To access a member of a particular element in
the array you can use the “.” dot operator:
planes[34].altitude = 10000;
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struct Pointers
Pointers and Structures

Pointers can point at structures
struct person
{
char name[80];
int ssn;
} barney, *rubble;
How Much Memory?
int main()
Not Common
{
rubble = &barney;
strcpy((*rubble).name, “Rubble, Barney”);
(*rubble).ssn = 555234561;
printf(“%s %d\n”, (*rubble).name, (*rubble).ssn);
}
More Common
strcpy(rubble->name, “Rubble, Barney”);
rubble->ssn = 555234561;
printf(“%s %d\n”, rubble->name, rubble->ssn);
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struct Pointers
Pointers and Structures

Since pointers can point to structures, then it’s
easy to make links between structures.
struct person
{
char initials[2];
int ssn;
int height;
struct person *father;
struct person *mother;
};
tom
father
bill
/* Declare variables and initialize them at the same time */
struct person tom = { "tj", 555235512, 74, NULL, NULL };
struct person bill = { "wj", 351003232, 75, NULL, NULL };
struct person susan = { "sd", 980332153, 70, NULL, NULL };
int main()
{
/* Set tom's parents pointers */
tom.father = &bill;
tom.mother = &susan;
printf(“\nTom's mother's height is: %d in", tom.mother->height);
}
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mother
susan
tom is a struct and mother is a
field in that struct, thus
tom.mother is correct.
mother is a pointer to a struct
and thus mother->height is
correct.
Combine them for
tom.mother->height
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Dynamic Memory Allocation
Memory Usage + Heaps
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Variable memory is allocated in
three areas:
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Global data section
Run-time stack
Dynamically allocated - heap
Global variables are allocated in
the global data section and are
accessible after declaration.
Local variables are allocated
during execution on the stack.
Dynamically allocated variables
are items created during runtime and are allocated on the
heap.
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malloc() – allocates memory
free() – frees memory
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0x0000
0x0100
Global Data Section
Heap
SP
0x0600
Run-time stack
0x8000
PC
Program
(Flash)
0xffff
Structures
SFR’s
Interrupt Vectors
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Dynamic Memory Allocation
Dynamic Memory Allocation

The sizeof() function determines how much
space is necessary for allocation.
#include <stdio.h>
#include <stdlib.h>
int main()
{
int *dynArray;
double *ddynArray;
/* Allocate space for 16 ints */
dynArray = (int *)malloc( 16 * sizeof(int) );
dynArray[6] = 65;
dynArray[12] = 2;
/* Allocate space for 20 doubles */
ddynArray = (double *)malloc( 20 * sizeof(double) );
}
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Dynamic Memory Allocation
Dynamic Memory Allocation
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Dynamic memory allocation using malloc() is
used for many kinds of programs.
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When data size is unknown or variable.
When building abstract structures like trees and linked
lists.
A NULL pointer returned from malloc() means it
failed and you are likely out of memory.
Dynamic allocation can be a source of bugs in C
code.

Memory leak - allocating memory and forgetting to free
it during program execution.
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Dynamic Memory Allocation
Dynamic Memory Allocation

Once allocated, the memory belongs to your
program until it terminates or is free()’d.
#include <stdio.h>
#include <stdlib.h>
main()
{
int *dynArray;
/* Allocate space for 16 ints */
dynArray = (int *)malloc( 16 * sizeof(int) );
dynArray[6] = 65;
dynArray[12] = 2;
doSomething( dynArray );
free( dynArray );
}
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Dynamic Memory Allocation
Pointers, Structures, & malloc()

Common to let malloc() create space for structures
struct person
{
char initials[2];
int ssn;
int height;
struct person *father;
struct person *mother;
} *tom, *bill, *susan;
int main()
{
tom = (struct person *)malloc( sizeof( struct person ) );
bill = (struct person *)malloc( sizeof( struct person ) );
susan = (struct person *)malloc( sizeof( struct person ) );
strncpy(tom->initials, "tj“, 2);
tom->ssn = 555235512;
tom->father = bill;
tom->mother = susan;
susan->height = 68;
/* Since tom is now a pointer, tom->mother->height is correct. */
printf(“\nTom's mother's height is: %d", tom->mother->height);
}
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union
union
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A union is a value that may have any of
several representations or formats; or a
data structure that consists of a variable
which may hold such a value.
Unions are defined like structures
(structs), except that each data member
begins at the same location in memory.
The union object occupies as much space
as the largest member.
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union
union Example
#include <stdio.h>
union
{
char c;
int i;
} x;
int main()
{
int i;
char* c = &x.c;
x.i = 0x12345678;
printf("\ni = %08x\n", x.i);
for (i=0; i<4; i++) printf(" %02x", c[i]);
}
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union
union Example
union
{
char c;
int i;
float f;
double d;
} x;
x.c
x.i
x.f
x.d
=
=
=
=
char* c = &x.c;
int* i = &x.i;
float* f = &x.f;
double* d = &x.d;
printf("\nc
printf("\ni
printf("\nf
printf("\nd
'z';
5180;
3.14;
3.141592653589793;
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=
=
=
=
%p",
%p",
%p",
%p",
c);
i);
f);
d);
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Bit Fields
Bit Fields
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Allow specification of some very small objects of a given
number of bits in length.
Allow specification of fields within some externally
produced data files.
Can only be declared inside a structure or a union.
No guarantee of the ordering of fields within machine
words, so programs will be non-portable and compilerdependent.
Be careful using them. It can require a surprising
amount of run-time code to manipulate these things and
you can end up using more space than they save.
Bit fields do not have addresses—you can't have
pointers to them or arrays of them.
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Bit Fields
Bit Fields Example
#pragma pack(push,1)
typedef struct
{
unsigned short sec: 5;
unsigned short min: 6;
unsigned short hour: 5;
} FATTime;
#pragma pack(pop)
// BYTE align in memory (no padding)
#pragma pack(push,1)
typedef struct
{
unsigned short day: 5;
unsigned short month: 4;
unsigned short year: 7;
} FATDate;
#pragma pack(pop)
// BYTE align in memory (no padding)
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//
//
//
//
(total 16 bits--a unsigned short)
low-order 5 bits are the seconds
next 6 bits are the minutes
high-order 5 bits are the hour
// End of strict alignment
//
//
//
//
(total 16 bits--a unsigned short)
low-order 5 bits are the day
next 4 bits are the month
high-order 7 bits are the year
// End of strict alignment
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Bit Fields
Bit Fields Example
typedef struct
{
FATTime time;
FATDate date;
} DirEntry;
void setTimeDate(DirEntry* dir)
{
time_t a;
struct tm *b;
time(&a);
b = localtime(&a);
dir->date.year = b->tm_year + 1900 - 1980;
dir->date.month = b->tm_mon;
dir->date.day = b->tm_wday;
dir->time.hour = b->tm_hour;
dir->time.min = b->tm_min;
dir->time.sec = b->tm_sec;
return;
} // end setDirTimeDate
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Linked List
Linked List Data Structure

A linked list is an collection of nodes, each of
which contains data and is connected using
pointers.



Each node points to the next node in the list.
The first node in the list is called the head.
The last node in the list is called the tail.
list
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75
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NULL
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Linked List
Simple Linked List Example
typedef struct element
{
int value;
struct element* next;
} Element;
Element *list;
list
Element* newElement(int v)
{
Element* tmp;
tmp = (Element*)malloc( sizeof(Element) );
tmp->value = v;
tmp->next = NULL;
return tmp;
}
50
75
99
NULL
int main()
{
/* Create linked list */
list = newElement(50);
list->next = newElement(75);
list->next->next = newElement(99);
}
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Linked List
Linked Lists vs Arrays
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Advantages of a linked list 
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
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Dynamic size.
Easy to add additional nodes.
Easy to add or remove nodes from the middle of the
list by adding or redirecting links.
Advantages of an array 
Can easily and quickly access arbitrary elements.

In a linked list in order to access the 5th element of the list you
must start at the head and follow the links through four other
nodes.
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Linked List
Linked List Creation

There are three steps required to create a linked
list 


Allocate space for predefined structure.
Fill in the structure fields.
Link the structure into the list using pointers.
struct element* newElement(int v)
{
struct element* tmp;
tmp = (struct element*)malloc( sizeof(struct element) );
tmp->value = v;
tmp->next = NULL;
return tmp;
}
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Linked List
Pre-pending Items to a Linked List
/* Prepend an item to start of oldList, returning ptr to new list */
struct element* prepend(struct element* item, struct element* oldList)
{
item->next = oldList;
return item;
}
int main()
{
struct element* tmp;
list = newElement(45);
/* Prepend item to list */
tmp = newElement(30);
list = prepend(tmp, list);
/* Can prepend in one statement */
list = prepend(newElement(10), list);
printList(list);
}
list
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Linked List
Appending Items to a Linked List
/* Append an item to the end of oldList, returning ptr to list */
struct element* append(struct element* item, struct element* oldList)
{
struct element* tmp;
if (oldList == NULL) return item; /* oldList is empty */
tmp = oldList;
while (tmp->next != NULL) tmp = tmp->next;
tmp->next = item;
return oldList;
/* Search for end of list */
}
int main()
{
/* Append some items to end of list */
list = append(newElement(200), list);
list = append(newElement(201), list);
list = append(newElement(202), list);
printList(list);
}
list
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201
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NULL
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Linked List
Inserting Items into a Linked List

Search to find the proper location


link the new record into it’s proper place.
multiple cases to consider

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
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list is empty
item belongs at front of existing list
item belongs somewhere in middle of existing list
item belongs at end of existing list
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Linked List
Inserting Items into a Linked List
struct element* insert(struct element* item, struct element* oldList)
{
struct element* tmp;
/* oldList is empty */
if (oldList == NULL) return item;
/* The new item goes first (pre-pend) */
if (item->value < oldList->value)
{ item->next = oldList;
return item;
}
/* Search for proper location */
tmp = oldList;
while ((tmp->next != NULL) && (tmp->next->value < item->value))
tmp = tmp->next;
item->next = tmp->next;
tmp->next = item;
return oldList;
}
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Linked List
Inserting Items into a Linked List
struct element* insert(struct element* item, struct element* oldList)
int main()
{
/* Insert some items into list */
list = insert(newElement(65), list);
list = insert(newElement(2), list);
list = insert(newElement(97), list);
list = insert(newElement(3), list);
list = insert(newElement(300), list);
printList(list);
}
list
2
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NULL
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Linked List
Freeing Items in a Linked List


Elements in a linked list need to be “freed” or you
will have a memory leak.
When “freeing” items in a linked list, be careful
not to “saw off the limb you’re standing on”.
int main()
{
/* Create a linked list */
list = newElement(50);
list->next = newElement(75);
list->next->next = newElement(99);
free(list->next->next);
free(list->next);
free(list);
}
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Linked List
Printing Items in a Linked List

Use a temporary pointer to walk down the list


print values as you go
end up with a newline
void printList(struct element *ptr)
{
struct element *tmp;
tmp = ptr;
while (tmp != NULL)
{
printf(“ %d”, tmp->value);
tmp = tmp->next;
}
printf("\n");
}
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void printList(struct element *ptr)
{
if (ptr == NULL) printf("\n");
else
{
printf(" %d", ptr->value);
printList(ptr->next);
}
}
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Example 2
Example 2

Thermostat temperature entry:
typedef unsigned int u16;
typedef unsigned char u8;
enum { SUN, MON, TUE, WED, THUR, FRI, SAT };
typedef struct Setting_struct
{ struct Setting_struct* link;
union
{ u16 time;
// day:hour/minute
struct
{ u8 day:3;
// day of week (0-6)
u8 hour:5;
// hour (0-23)
u8 minute;
// minute (0-59)
} times;
} date;
u8 high;
u8 low;
} Setting;
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link
date
high
low
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41
Example 2
Create New Node
// create a new entry
Setting* newSetting(u8 day, u8 hour, u8 minute, u8 high, u8
low)
{
// malloc a new node
Setting* temp = (Setting*)malloc(sizeof(Setting));
// store entries
temp->date.times.day = day;
temp->date.times.hour = hour;
temp->date.times.minute = minute;
temp->high = high;
temp->low = low;
// null link
temp->link = NULL;
return temp;
} // end newSetting
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Example 2
main
enum { SUN=0, MON, TUE, WED, THUR, FRI, SAT };
int main()
{
Setting *list = NULL;
// Create linked list
list = newSetting(MON, 6, 30, 22<<1, 20<<1);
list->link = newSetting(WED, 20, 30, 17<<1, 15<<1);
list->link->link = newSetting(FRI, 8, 30, 22<<1, 20<<1);
list->link->link->link = newSetting(SAT, 18, 30, 20<<1, 18<<1);
}
0256
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025e
0266
026e
0000
1e31
1ea3
1e45
1e96
2c
22
2c
28
28
1e
28
24
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Example 2
Insert Setting
Setting* insertSetting(Setting* setting, Setting* oldList)
{
Setting* temp;
if (oldList == NULL) return setting;
// oldList is empty
// The new item goes first (pre-pend)
if (setting->date.time < oldList->date.time)
{ setting->link = oldList;
return setting;
}
// Search for proper location
temp = oldList;
while (temp->link != NULL)
{ if (temp->date.time == setting->date.time)
{ // replace
temp->high = setting->high;
temp->low = setting->low;
free(setting);
return oldList;
}
if (temp->link->date.time > setting->date.time) break;
temp = temp->link;
// next
}
setting->link = temp->link;
// insert
temp->link = setting;
return oldList;
} // end insertSetting
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Example 2
Print Setting
char* days[] = { "Sun", "Mon", "Tue",
"Wed", "Thu", "Fri", "Sat" };
void printList(Setting* ptr)
{
Setting* temp;
temp = ptr;
while (temp != NULL)
{
printf("\n%s %02d:%02d %4.1f-%4.1f",
days[temp->date.times.day],
temp->date.times.hour,
temp->date.times.minute,
(float)temp->high / 2.0,
(float)temp->low / 2.0 );
temp = temp->link;
}
printf("\n");
} // end printList
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Example 2
Final Test
void main(void)
{
Setting *list = NULL;
eZ430X_init(CALDCO_1MHZ);
lcd_init();
lcd_backlight(ON);
// init board
// init LCD
// Create linked list
list = newSetting(MON, 6, 30, 22<<1, 20<<1);
list->link = newSetting(WED, 20, 30, 17<<1, 15<<1);
list->link->link = newSetting(FRI, 8, 30, 22<<1, 20<<1);
list->link->link->link = newSetting(SAT, 18, 30, 20<<1, 18<<1);
// Insert some items into list
list = insertSetting(newSetting(MON,
list = insertSetting(newSetting(SUN,
list = insertSetting(newSetting(TUE,
list = insertSetting(newSetting(MON,
list = insertSetting(newSetting(SAT,
}
6, 30, 24<<1, 18<<1), list);
6, 30, 22<<1, 20<<1), list);
6, 30, 22<<1, 20<<1), list);
6, 30, 20<<1, 10<<1), list);
20, 30, 22<<1, 20<<1), list);
printList(list);
return;
BYU CS/ECEn 124
Structures
46
BYU CS/ECEn 124
Structures
47