Chapter 5 Array, Pointers and Strings

Chapter 5 Array, Pointers and Strings

By C. Shing

ITEC Dept

Radford University

Slide 2

Objectives Understand how to use arrays (one and

multiple dimensions) Understand how to use pointers Understand the relationship between pointers

and arrays Understand how to call function using

pass-by-reference and pass-by-value Understand how to create dynamic memory

Slide 3

Objectives (Cont.) Understand how to pass function as argument Understand how to use strings Understand the relationship between pointers

and strings Understand array of pointers Understand how to use command line

arguments Understand how to use Library function: scanf

Slide 4

Array A contiguous memory with the same data type Declared as

1-dimension

type array_name [array_size]; 2-dimension

type array_name [row_size][coulmn_size];…

Slide 5

Array (Cont.) Has index (or indices) starting from 0

to access each memory 1-dimension: specify the content at index

array_name[index]Example:

type array_name [array_size];declare memory: array_name[0], array_name[1], …,

array_name[array_size-1] 2-dimension: specify the content

at row_index and column_indexarray_name[row_index][column_index]

Slide 6

Array (Cont.) Programmer must make sure that

indices are not out of bound.

Otherwise, buffer overflow security problem

will occur.

Slide 7

Array (Cont.) Initialization:

row-wise (starts fill in 1st row first) Fill in 0/NULL if not enough initial integer (pointer)

values specified Form:

1-dimension:type array_name[array_size] = {initial

values}; 2-dimension:

type array_name[row_size][column_size]= {{row 0 initial values}, …, {row row_size-1 initial values}}

Slide 8

Array (Cont.) Initialization: (Cont.)

Example: 1-dimension:

int a [4] = {4,3,2,1};orint a[]={4,3,2,1};

Then a[0]=4, a[1]=3, a[2]=2, a[3]=1 2-dimension:

int a[3][4]= {{10,11,12,13}, {20,21,22,23}, {30,31,32,33}}orint a[][]= {{10,11,12,13}, {20,21,22,23}, {30,31,32,33}}orint a[][4]= {10,11,12,13, 20,21,22,23, 30,31,32,33}

Then a[0][0]=10, a[0][1]=11, a[0][2]=12, a[0][3]=13, …, , a[2][0]=30, a[2][1]=31, a[2][2]=32, a[2][3]=33

Slide 9


Example: (Cont.) 1-dimension:

int a [4] = {4,3};

Then a[0]=4, a[1]=3, a[2]=0, a[3]=0 2-dimension:

int a[3][4]

= {{10}, {20,21,22},

{30,31}}

Then a[0][0]=10, a[0][1]=a[0][2]=a[0][3]=0, a[2][0]=30,

a[2][1]=31, a[2][2]=a[2][3]=0

Slide 10


Example: Initialize the whole array 1-dimension:

int a [4] = {0};Then a[0]=0, a[1]=0, a[2]=0, a[3]=0 2-dimension:

int a[3][4]={0}

Then a[0][0]=a[0][1]=a[0][2]=a[0][3]=…= a[2][0]=a[2][1]=a[2][2]=a[2][3]=0

Slide 11

Pointers Store address using a word More efficient than using index Declare as:

type *pointer_variable;This request a word size pointer_variable that

stores an address of the typeNote: do not use pointer to a register type variable

Use in a statement: (Referencing/Addressing operator &)pointer_variable = & type_variable;

The pointer_variable stores the address of the type_variable

Note: do not use & expression

Slide 12

Dereferencing Operator * Find the content that pointed by an address Use in a statement:

variable = * pointer_variable;Example:int i , *p;i= 10;p= &i;

j= *p;Then j also contains 10. (Note: & and * cancel each other)

Example

Slide 13

Top-Down Design Example - Use Pointer

Example: write a program to convert a series of Fahrenheit degrees into Celsius degree (use sentinel -1 to stop)The Output will look like

Fahrenheit Celsius_________ ______212.00 100.0032.00 0.00

Slide 14


Functions:1st level: main(void)

2nd level: PrintHeading: this function prints header and __ in screen

input: nothingoutput: nothing

GetInput: this function reads data from keyboard and stores fahrenheit degreeinput: fahrenheit address (double*)- due to pass by valueoutput: nothing

ProcessData: this function converts the fahrenheit to celsius degreeinput : fahrenheit (double) and celsius address (double *) from

mainoutput : nothing

OutputResult: this function prints out fahrenheit and celsius degrees in screeninput: fahrenheit (double), celsius (double) from mainoutput : nothing

Slide 15


program

data

Top-Down Design Example - Use Array Program (sentinel: -1)

data

Slide 16

Array and Pointer Relation Array name stores address of the array

1-dim: type array_name [array_size];

The base address is &array_name[0]The array_name stores &array_name[0]The (array_name+i) stores &array_name[i]Example:int a[N], *p;p = a; // p = &a[0] p= a+i; // p=&a[i]Note: if the address of the array a is 1000, then p contains the address= 1000+ i*sizeof(int), which is not 1000+i

Example

Slide 17

Array and Pointer Relation (Cont.) Do not attempt to change the address of the arrayExample:

int a[100], *p;

& 100 is not allowed&a is not allowed++a is not alloweda = (int *) 100 is not allowed

p= (int *) 100 is allowed

Slide 18

Array and Pointer Relation (cont.) Array name stores address of the array (Cont.)

2-dim:

type array_name [row_size][column_size];

The base address is &array_name[0][0]

The array_name stores &array_name[0]:

beginning address of 1st row

(different from &array_name[0][0])

The (array_name+i) stores &array_name[i]

Slide 19

Array and Pointer Relation (cont.)Example:

int a[M][N], *p;

p = a; // p = &a[0], where a[0] is an array of N elements

// *p is the array a[0], which contains N elements

p= a+i; // p=&a[i]

Class Example

Slide 20


2-dim:

Question:

int a[M][N], *p;

Can you use pointer p (where base address p=&a[0][0])

to find a[i][j]?

Slide 21


2-dim: (Cont.)int a[M][N], *p;p=&a[0][0]; [Note: (int *)a also gives bas address]

Answer:a[i][j]= *(p+i*N+j) or a[i][j]= *((int *)a+i*N+j)

Slide 22


2-dim:

Question:

int a[M][N], *p;

p=a;

Can you use pointer p to find a[i][j]?

Hint: a[i]=*(a+i)

Slide 23

Array and Pointer Relation (cont.)Idea:1. Find address of 1-dim array a[i] by (a+i)2. Find the 1 dim array a[i] by *(a+i).Then if you use index now, the cell a[i][j] content is either (*(a+i))[j] or3. Find the address of the cell j of a[i] by *(a+i)+j4. The cell a[i][j] content is *(*(a+i)+j)

Note: when you cast (int *) on a, then it specifies the base address &a[0][0]See Example.

Slide 24


2-dim: (Cont.)int a[M][N], *p;p=a;

Answer:1. a[i][j]= (*(p+i))[j] or2. a[i][j]= *(*(p+i)+j)Is the following correct?a[i][j]= *(p+i+j)Why?

Slide 25

Array and Pointer Relation (cont.)See Example1

or

Example 2

Summary Example

Slide 26

Operators Priority

Operator Same Priority Rule

(),postfix++/--,[],.,-> Left to right

Unary +/-, prefix++/--, !, ~,&, * Right to left

*, /, % Left to right

+, - Left to right

<<, >> Left to right

<, >, <=, >= Left to right

==, != Left to right

Slide 27

Operators Priority (Cont.)

Operator Same Priority Rule

&

^

|

&& Left to right

|| Left to right

?: Right to left

, Left to right

Slide 28

Passing Array Element to Functions Pass-by-value: pass a copy of only some or an

array element The array element will not be changed after

function call

Syntax: functionname(a[n]);

Example

Slide 29

Passing Entire Array to Functions Pass-by-reference: pass the address of an entire

array Only a word size is passed, very efficient ANSI C uses pass by reference for entire array only The array content can be changed however the

array address is not changed ANSI C provides const type to prevent the whole array is accidentally changed

Syntax: functionname(a);Example

Slide 30

Passing Array Element to Functions- by Value

Example: 1-dimint a[100];a[0]=0;

byValue(a[0]);printf(“a[0]=%d”,a[0]); // a[0]=0 is printed

void byValue(int b) { b=100; }

Slide 31

Passing Array to Functions- by Reference

Example: 1-dimint a[100], i;

for (i=0;i<100;++i)a[i]=i;

byReference(a); for (i=0;i<100;++i)

printf(“a[%d]=%d”,i,a[i]); // a[0]=100, a[1]=99, … are printed void byReference(int *a) // a is a pointer { int i;

for (i=0;i<100;++i) a[i]=100-i; // use pointer[index] to specify the cell element

} See byReference.c

Slide 32

Use Array1. Read in data to function inputData and returns count of number of data: array1.c

syntax: count = functionname(…);2. Read in data to function inputData and store count of number of data in a variable count: array3.c

syntax: functionname(…, &count , …);Example: using 1-dim int array (read in character),

data

Slide 33

Use Array (Cont.) Linear Search: array2.c, Data: array2_data.txt Run: (Use a sentinel -1)

a.out < array2_data.txt

Slide 34

Class Example Example. 2D array vendorWrite a program to use the following input and produce the output. Input(keyboard): For each vendor, prompt for and enter the following data:(by storing them in a file and use redirect input from the file) Vendor Invoice Invoice DiscountNumber Number Amount Rate217 1239 2309.12 0.10349 9823 670.00 0.09

Slide 35

Class Example (Cont.) Example. 2D array vendor (Cont.)Output(screen): For each vendor, print the following report: Author ACCOUNTS PAYABLEInvoice Number: XXXXVendor Number: XXXInvoice Amount: $99999.99Discount Amount: $ 999.99Amount Due: $99999.99

Example

Slide 36

Class Example (Cont.) Example. 2D array vendor (Cont.)

Example 1: program for input and output, data

a.out < 2d.txt

Example2 : 1 row data only Example: 2D array input & output

Hint to HW #2, test data

Slide 37

Class Example (Cont.) Example: (Cont.)

The following segment has a run-time error on void input_data (double *vendor) : Why?

#define MAXSIZE 50void input_data (double *vendor);int main (void) { double vendor[MAXSIZE][4]; input_data (vendor);}void input_data (double *vendor) { int i, j;

for (i=0;i<MAXSIZE;i++) for (j=0;j<4;j++) scanf(“%lf”, &vendor[i][j]);

}

Slide 38

Class Example (Cont.) Example : (Cont.)

Answer:The bug is at void input_data (double *vendor);Andvoid input_data (double *vendor) {

…scanf(“%lf”, &vendor[i][j]);

}

Slide 39


Answer: (Cont.)

Reason: when you pass vendor into

function input_data

It is the address of the array: vendor[0]

There is no such thing as

vendor[i][j] used in scanf statement.

Slide 40


Answer: (Cont.)Correction:

#define MAXSIZE 50void input_data (double vendor[][4]);int main (void) { double vendor[MAXSIZE][4]; input_data (vendor);}void input_data (double vendor[][4]) { int i, j;

for (i=0;i<MAXSIZE;i++) for (j=0;j<4;j++) scanf(“%lf”, &vendor[i][j]);

}

Slide 41

Bubble Sort Sort an array of int: a[0], …, a[count-1] # of passes: count -1 Program, Data Another way of using array address:

also use const on formal parameters to prevent

accidental parameter values changed

Slide 42

Passing Array to Functions- by Reference (Cont.)

Example: 1-dimint a[100], i;

void bubble_sort (int a[], int n) {

int i,j; for (i=0;i<n-1;++i) for (j=n-1;j>i;--j)

… swap (a+j-1, a+j); // swap a[j-1] and a[j]

}

Slide 43


Example: (Another version of bubble sort) 1-dim

int a[100], i; void bubble_sort (int a[], int n) {

int i,j; for (i=0;i<n-1;++i) for (j=0;j<n-i-1;++j)

if (a[j]>a[j+1]) swap (a+j, a+j+1); // swap a[j] and a[j+1]

}

Slide 44


Example: 1-dim (Cont.)void swap (int *const a, int * const b)

// a and b are pointers {

int tmp; tmp=*a;

*a=*b; *b=tmp;}

Slide 45

Passing 2D Array to Functions- by Reference

Example: 2-dimint a[10][100], i, j;

for (i=0;i<10;++i) for (j=0;j<100;++j)

a[i][j]=i*j; byRef (a); for (i=0;i<10;++i) for (j=0;j<100;++j)

printf(“a[%d][%d]=%d”,i,j,a[i][j]); // a[0][0]=100, a[0][1]=99, … are printed void byRef (int a[][100]) { int i, j;

for (i=0;i<10;++i) for (j=0;j<100;++j) a[i][j]=100-i-j;

}

Slide 46

Use Pointers only instead of indexing

Example

Remember you must define spaces for array

before using pointers. Otherwise, runtime error

informs you due to no spaces can be assigned

beyond the pointer that you declared.

Slide 47

Strings Array of characters end with ‘\0’ String name is a pointer to the base address

(i.e. the address of the 1st character) Initialize by

char *s=“string constant”; // initialization specifies the length of s = 16 or

char s[] =“string constant”; or char s[] ={‘s’,’t’,…, ‘n’, ‘t’,’\0’};

Slide 48

String and Pointer Relation Similar to the array and pointer relation

Example:char *p=“constant”;

printf(“%s\n”, p+2); // nstant printed

printf(“%c\n”, p[5]); // a printed

printf(“%c\n”, *(p+4)); // t printed

Slide 49

String Functions Use string function to work on strings String functions are defined in string.h String functions (not secure- buffer overflow attack):

strings s and t strcat (char *s, const char *t): s=s concatenate t strcpy (char *s, const char *t):

s=copy of t until ‘\0’ of t is reached strcmp (char *s, const char *t):

<0 if s<t=0 if s=t>0 if s>t

strlen (const char *s): return string length before ‘\0’,an unsigned integer or long int (type size_t)

strerror (int errornum): returns pointer to system dependent error message

Slide 50

String Functions (Cont.) String functions are defined in string.h String functions (More secure): strings s and t

strncat (char *s, const char *t, size_t n): s=s concatenate at most n characters of t

strncpy (char *s, const char *t, size_t n): s=copy of at most n characters of t

strncmp (char *s, const char *t, size_t n): compare up to n characters of s with t

<0 if s<t=0 if s=t>0 if s>t

Slide 51

String Functions (Cont.) Example:

char * strcpy (char *s, register const char *t)

{

register char *p=s;

while (*p++=*t++);

return s;

}

Slide 52

Keyboard Input Function – gets (Not Secure: buffer overflow problem)

Form: gets (string_variable);Read from keyboard into string_variable until end of line or end of file

Example:char s[80]; // remember s is address, do not use char *sgets (s); // sample data:a 100 -1.23 This is a sample data// s=“a 100 -1.23 This is a sample data”

Slide 53

A Safer Keyboard Input Function - getline

A non-standard function Form: ssize_t getline (char **lineptr, size_t *n, FILE *stream) Read from keyboard into a string lineptr at most n characters until end of line or end of file

Note: A standard function called fgets does a similar job. See sscanf slide (Slide 64) Example link.

Example:

Slide 54

Keyboard Output Function - puts Form: puts (string_variable);Print the content of the address of string_variable

Example:void output_string (const char *s) { // remember s is address

puts (s); // print out s:a 100 -1.23 This is a sample data

}

Slide 55

Keyboard Input Format Function - scanf Form: scanf(“format”, variable_address);

Format: similar to those used in printf, see Chapter 3 If use %s format, it reads all until it reaches the first white space

Example:char c;int i;double db;char s[80]; // remember s is addressscanf(“%c%d%lf%s”, &c, &i, &db, s); // sample data:a 100 -1.23 This is a sample data// c=‘a’, i=100, db=-1.23, s=“This”

Slide 56

Example - scanf Program: read characters into an arrays

(lname and fname) and convert them to strings Data

Slide 57

Keyboard Input Format Function – scanf (Cont.)

Form: scanf(“pure_format”, variable_address); (Cont.) Scan set:

%[range]Reads every character which belongs to range, stop at the character (not read in) which does not belong to the range

%[^range]Reads every character which does not belongs to range, stop at the character (not read in) which belongs to the range

%10dRead in 10 digit number %10cRead in 10 characters %*cSkip any number of characters

Example: Program, DataHint to Quiz #5, Hint Data

Slide 58

Keyboard Output Format Function - printf

Form: printf(“format”, variable_address); Format: %s, see those used in printf, see Chapter 2

Slide 59

Search Strings Find the index of the array if found. Return -1 if not found

Example: read the following data into array of strings (max 8 characters)- no duplication

Data:ABC 123PLO4ME123456AL-QADA2PLO4ME123V456

Slide 60

Search Strings (Cont.)int i;char plate[9];char license[1000][9];int frequency[1000]={0};int count=0, index;// count: # of different licenses in license array

for (i=0;gets(plate);i++){

index = search(license, plate, count);…

}

Slide 61

Search Strings (Cont.)

if (index == -1)

{

strcpy(license[count], plate);

…

}

else

{ …}

Slide 62

Search Strings (Cont.)int search (char license[][9], char *plate, int count){

int i;

for (i=0; i<count; i++){

if (strcmp (license[i],plate)== 0)return i; // found plate in license array

}

return -1; // not found}

Slide 63

Input and Output String Format Function – sscanf and sprintf

Form: sscanf (instring, “format”, variable_address); Similar to scanf, it inputs data from string

instead of keyboard Format: similar to those used in printf, see Chapter 4 If use %s format, it reads all until it reaches

the first white space Form: sprintf (outstring, “format”, variable);

Similar to printf, it outputs results to string instead of screen

More secure one:Form: snprintf (outstring, n, “format”, variable);

Slide 64

Input and Output String Format Function – sscanf and sprintf (Cont.)

Example:char c;int i;double db;char s[80], outstring[80],

*instring=“a 100 -1.23 This is a sample data”; sscanf(instring,“%c%d%lf%s”, &c, &i, &db, s); //Then c=‘a’, i=100, db=-1.23, s=“This”sprintf(outstring,“%c%d%lf%s”, c, i, db, s); // print to outstring: a 100 -1.23 This Orsnprintf(outstring,“%c%d%lf%s”,80, c, i, db, s); // print to outstring: a 100 -1.23 This

Example, Data

Slide 65

Class Example sscanf and gets:

Example

Hint to HW Substring of “string” from location 2 to 3 is “tr”

Slide 66

String Search Utilities Need to include <string.h>

char* strchr(const char *s, int c): return pointer to the found character c in s. Returns NULL

otherwiseExample: strchr(“This is a sunny day.”, ‘u’): returns s+11 size_t strcspn(const char *s, const char *range):

returns length of initial s that are not in the rangeExample: strcspn(“This is a sunny day.”, “abcd”): returns 8

Slide 67

String Search Utilities (Cont.) size_t strspn (const char *s, const char *range):

returns length of initial s that are in rangeExample: strspn(“This is a sunny day.”, “hist”):

returns 4 since space is not in “hist” char *strpbrk (const char *s, const char *range):

returns pointer of the 1st character of s found in range

Example: strpbrk(“This is a sunny day.”, “abcd”):

returns pointer of ‘a’

Slide 68

String Search Utilities (Cont.) char * strrchr (const char *s, int c):

returns pointer of the last found character c in sExample:

strrchr(“This is a sunny day.”, “s”): returns position of s in sunny

char *strstr (const char *s, const char *range): returns pointer of the 1st occurrence of range found in s

Example: strstr(“This is a sunny day.”, “is”):

returns pointer of “is” in “This”

Slide 69

String Search Utilities (Cont.) char * strtok (const char *s, const char *delimiter):

returns pointer of the token in s using the delimiterand save the position of the next token

Example: char s[]=“This is a sunny day.”;

char *token;

token=strtok(s, “ “);

do {

printf(“token=%s\n”, token);token=strtok(NULL, “ “); // must use this to get next token

} while (token != NULL);

The tokens printed are:Thisisasunnyday.

Slide 70

Other General String Utilities void *memcpy

(void * destination, const void *source, size_t n):copies n items of source to destination returns pointer to the result

Note: this function cannot overwrite the destination object

Example: char s[] = “This is a sunny day.”; char t[80];

memcpy(t,s, 21); returns pointer of t= “This is a sunny day.”

Slide 71

Other General String Utilities (Cont.) void *memmove

(void * destination, const void *source, size_t n):copies n items of source to destination returns pointer to the result

Note: this function can overwrite the destination object

Example: char s[] = “This is a sunny day.”;

memmove(s,&s[10], 3); returns “suns is a sunny day.”

Slide 72

Other General String Utilities (Cont.) void *memcmp

(const void * s, const void *t, size_t n):compares the 1st n elements of s and t and returns 0 if s=t, -1 if s<t, 1 if s>t

void *memchr (const void * s, int c, size_t n):returns pointer of the 1st found c in the 1st n elements of sExample: char s[] = “This is a sunny day.”;

memchr (s, ‘u’,16); returns pointer of “unny day.”

Slide 73

Other General String Utilities (Cont.) void *memset (void * s, int c, size_t n):

copies c in the 1st n elements of s and returns pointer of s

Example: char s[] = “This is a sunny day.”;

memset (s, ‘c’,6); returns pointer of “sssssss a sunny day.”

Slide 74

Pass Function as Argument The name of the function is a pointer,

passing a function is just passing a pointer

Slide 75

Pass Function as Argument (Cont.) Example: Write a function that calculate

the average of any double function f(x), where x = m, …, n (Cont.)

To find the average of sin (5), .., sin (10), we useaverage (sin, 5, 10);

To find the average of sqrt(10),…, sqrt(100), we use

average (sqrt, 10, 100);

Slide 76

Pass Function as Argument (Cont.) Example: Write a function that calculate

the average of any double function f(x), where x = m, …, n

// The function header can also be// double average (double f(double), int m, int n)double average (double (*f)(double), int m, int n){

int i;double sum=0.0;

for (i=m; i<=n; ++i) sum += f(i);

return sum/(n-m);}

Slide 77

Pass Function as Argument (Cont.) Example: Find

Using the trapezoidal rule

b

adx

x

1

Slide 78

Array Out of Bound When array index is out of bound, you willThe undesired memory:

ExampleOutput

Buffer overflow Attack:http://www.youtube.com/watch?

v=CDCEBKnrMxc&fmt=22Heap Overflow Attack: https://www.youtube.com/watch?v=2W4tuxHcOnE

Slide 79

Dynamic Memory Allocation Allocate n memory with a type Use function defined in stdlib.h

malloc (n*sizeof(type)): more efficient, do not initialize all memory

calloc(n, sizeof(type)): less efficient,will initialize all memory

Always check request memory is successful. If successful, return pointer to the n memory base_pointer with the generic type void *

Need to free spaces after finish function call by free(base_pointer);

Slide 80

Dynamic Memory Allocation (Cont.) Example:

int *a;

if (a=calloc(5, sizeof(int))){ a[0]=10; a[1]=20; a[2]=30; a[3]=40; a[4]=50;}else printf(“Error, request memory are not available.\n”);

Slide 81

Array of Pointers (Array of Strings) Store array of strings: save space Example:Assume that array w has 5 cells, each stores an address of strings as follows:w[0]=100=address of string “this”w[1]=200=address of string “is”w[2]=300=address of string “a”w[3]=400=address of string “snow”w[4]=500=address of string “day”

Slide 82

Array of Pointers (Cont.) Example: (Cont.)

The addresses of w are as follows:

address of w[0]=1000





Slide 83

Array of Pointers (Cont.) Example: (Cont.)

There are 3 methods to do bubble sort for array w.

Method 1 is the most efficient one.

Slide 84

Array of Pointers – Method 1 (Cont.) Example: (Cont.)To use bubble sort to sort the array w:void sort_strings (char *w[], int n){ int i, j; for (i=0; i<n; ++i) for (j=i+1; j<n;++j)

if (strcmp(w[i],w[j])>0) swap (&w[i], &w[j]);

}

Slide 85

Array of Pointers – Method 1 (Cont.) Example: (Cont.)To use bubble sort to sort the array w: (Cont.)

for i=0 and j=1, if (strcmp(w[i],w[j])>0)

This means strcmp(100, 200) which is required by Function strcmp to compare the contents ofaddresses 100 and 200

swap (&w[i], &w[j]);This sends in addresses 1000 and 1004 to function swap}

Slide 86

Array of Pointers – Method 1 (Cont.) Example: The swap function is written as

(Cont.) void swap (char **s, char **t){ char *tmp; tmp=*s; *s=*t; *t=tmp;}

Slide 87

Array of Pointers – Method 1 (Cont.) Example: (Cont.) You may also write it asvoid swap (char *s[], char *t[]){ char *tmp; tmp=*s; *s=*t; *t=tmp;}

Slide 88

Array of Pointers – Method 1 (Cont.) Example: The detailed function call (Cont.) void swap (char **s, char **t){/*The line above has char **, which means address of address (or address of pointers).For the 2 slides before, s=1000, t=1004 */

char *tmp; // tmp contains address tmp=*s; // *s=*1000=100 *s=*t; // *t=*1004=200 *t=tmp;} // The swap here swaps addresses 100 and 200

Slide 89

Array of Pointers – Method 1 (Cont.) Example: (Cont.)Instead of swapping address of w[i] and w[j], we can swap the string directly (a less efficient way):void sort_strings (char *w[], int n){ int i, j; for (i=0; i<n; ++i) for (j=i+1; j<n;++j)

if (strcmp(w[i],w[j])>0) swap (w[i], w[j]);

}

Slide 90


(Cont.) void swap (char *s, char *t){ char *tmp; strcpy(tmp, s); strcpy(s, t); strcpy(t, tmp);}

Slide 91

Array of Pointers – Method 1 (Cont.) Example: (Cont.)After bubble sort, the array w contains

w[0]=300=address of string “a”w[1]=500=address of string “day”w[2]=200=address of string “is”w[3]=400=address of string “snow”w[4]=100=address of string “this”

Slide 92

Array of Pointers – Method 1 (Cont.) Example: (Bubble sort of words)

1. Swap (direct addressing), word: static array2. Swap (in-direct addressing), word: static

array: efficient

1. Swap (in-direct addressing), word: dynamic array: efficient Program, Data

Slide 93

Array of Pointers – Method 2 Another way to write sort_strings is as below:To use bubble sort to sort the array w:void sort_strings (char *w[], int n){ int i, j; for (i=0; i<n; ++i) for (j=i+1; j<n;++j)

if (*w[i]> *w[j]) swap (w[i], w[j]);

}

Slide 94

Array of Pointers – Method 2 (Cont.) Another way to write sort_strings is as below:For i=1 and j=2,

if (*w[i]> *w[j])This means the content of address 100 (The 1st character at address 100)> that of address 200

swap (w[i], w[j]);This sends addresses 100 and 200 into function swap

Slide 95

Array of Pointers – Method 2 (Cont.) Example: The swap function is (Cont.) void swap (char *s, char *t){ char tmp; tmp=*s; *s=*t; *t=tmp;}

Slide 96

Array of Pointers – Method 2 (Cont.) Example: The detailed function call (Cont.) void swap (char *s, char *t){/* s=100, t=200 */ char tmp; tmp=*s; // *s=*100=‘t’ *s=*t; // *t=200=‘i’ *t=tmp;} // This swaps only the 1st char ofstings “this” and “is”

Slide 97

Array of Pointers – Method 2 (Cont.) Example: (Cont.)After bubble sort, the array w contains

w[0]=100=address of string “ahis”w[1]=200=address of string “ds”w[2]=300=address of string “i”w[3]=400=address of string “snow”w[4]=500=address of string “tay”

Slide 98

Array of Pointers – Method 3 Example: (Cont.)To use bubble sort to sort the array w:void sort_strings (char *w[], int n){ int i, j; for (i=0; i<n; ++i) for (j=i+1; j<n;++j)

if (strcmp (w[i],w[j])>0) swap (&w[i], &w[j]);

}

Slide 99


(Cont.) void swap (char **s, char **t){ char tmp; tmp=**s; **s=**t; **t=tmp;}

Slide 100

Array of Pointers – Method 3 (Cont.) Example: It can also be written as (Cont.)void swap (char *s[], char *t[]){ char tmp; tmp=**s; **s=**t; **t=tmp;}

Slide 101

Array of Pointers – Method 3 (Cont.) Example: The detailed function call (Cont.) void swap (char **s, char **t){/*The line above has char **, which means address of address (or address of pointers).For the 2 slides before, s=1000, t=1004 */

char tmp; // tmp contains content of string tmp=**s; // **s=**1000=*100=‘t’ **s=**t; // **t=**1004=*200=‘i’ **t=tmp;} // The swap here also swaps the 1st char of contents of addresses 100 and 200

Slide 102

Sort Array of strings- Method 4 Problem: In the previous license example, we want to sort array of strings license[] according to the frquencies of each licenseint main (void){

… sort_strings (license, frequency, count); …}

Slide 103

Sort Array of strings- Method 4 (Cont.)void sort_strings (char w[][9], int f[], int n){

int i, j;

for (i=0; i<n; ++i)for (j=i+1; j<n;++j)

if (f[i]<f[j]){

swapString (&w[i], &w[j]); // swap license arrayswapInt(&f[i], &f[j]); // swap frequency array

}}

}

Slide 104

Sort Array of strings- Method 4void swapString (char s[][9], char t[][9]){

char tmp[9]="";

strcpy(tmp,*s);strcpy(*s,*t);strcpy(*t,tmp);

}

Slide 105

Sort Array of strings- Method 4void swapInt (int *a, int *b){

int tmp;

tmp=*a;*a=*b;*b=tmp;

} Complete Example: Sort Names and Grades by Grades,

Sort Names and Grades by Lastname, Data

Slide 106

Command Line Arguments 3 arguments for main() function

argc: total number of arguments including the

command argv[]: array of pointers

argv[0]: contains command argv[1]: contains 1st argument after command

… etc env[]: array of environment variables

Slide 107

Command Line Arguments (Cont.) Example:// use the following function header// int main(int argc, char ** argv, char **env)orint main (int argc, char * argv[], char *env[]){ int i; for (i=0;i<argc;++i) printf(“argv[%d]=%s\n”,i, argv[i]); for (i=0;env[i] != NULL;++i) printf(“env[%d]=%s\n”,i, env[i]); return 0;}

Slide 108

Command Line Arguments (Cont.) Class Example: Print All Arguments

Example1

Example2

Example3

Example4 (Hint to Extra HW #1):

use strtok function

Slide 109

Practice 1

Given int i=3, j=5,*p=&i,*q=&j,*r; double x;

Find the values of the following table:

Expression Value

________ _____

p == &i

** &p

r=&x

7**p/*q+7

*(r=&j)*=*p *r=

Slide 110

Practice 1 (Answer)

Given int i=3, j=5,*p=&i,*q=&j,*r; double x;


Expression Value

________ _____

p == &i 1

** &p 3

r=&x 1

7**p/*q+7 11

*(r=&j)*=*p *r= 15

Slide 111

Practice 2

Given char s1[]="beautiful big sky country";

s2[]="how now brown cow";


Expression Value

------------- ------

strlen(s1)

strlen(s2+8)

strcmp(s1,s2)

printf("%s",s1+10);

strcpy(s1+10, s2+8); strcat(s1,"s!"); printf("%s",s1);

Slide 112

Practice 2 (Answer)

Given char s1[]="beautiful big sky country"; s2[]="how now brown cow"; Find the values of the following table: Expression Value------------- ------strlen(s1) 25strlen(s2+8) 9strcmp(s1,s2) < 0printf("%s",s1+10); “big sky country”strcpy(s1+10, s2+8); strcat(s1,"s!");printf("%s",s1);

“beautiful brown cows!”

Slide 113

Practice 3

Given int a[3][5]

(1) Is a== &a[0]?

(2) Is a == &a[0][0]?

(3) Find which ones in the following are not the

same as a[i][j]?

*(a[i]+j)

*(&a[0][0]+5*i+j)

a[i] + j

(*(a+i))[j]

*((*(a+i))+j)

Slide 114

Practice 3 (Cont.)

Answer:

(1) Yes

(2) No

(3) a[i] + j is different, the rest are the same

Example

Slide 115

Practice 4Use command line to execute a program called mycp using

mycp indata outresult If your main function header is

int main(int argc, char *argv[ ]) What's the value of the following? argc = argv[1][2] argv[2][2] (*argv)[0]= *(argv+1)[0]=**argv=

Slide 116

Practice 4 (Answer)Use command line to execute a program called mycp using

mycp indata outresult If your main function header is

int main(int argc, char *argv[ ]) What's the value of the following? argc = 3argv[1][2] = ‘d’argv[2][2] = ‘t’(*argv)[0]= ‘m’*(argv+1)[0]= ‘i’**argv= ‘m’

Slide 117

Practice 4 (Explanation) Example: (Cont.)

The command line arguments are in the array

argv which contains

*argv=argv[0]=100=address of string “mycp”

*(argv+1)=argv[1]=200=address of string “indata”

(argv+2)=argv[2]=300=address of string “outresult”

Slide 118

Practice 5

Given char c='A', s[]="Blue moon!"; Show what to be printed? statement print out ________ _______ printf("%c",c) printf("%2c",c) printf("%-3c",c) printf("%s",s) printf("%3s",s) printf("%.6s",s) printf("%-11.8s",s)

Slide 119

Practice 5 (Answer)

Given char c='A', s[]="Blue moon!"; Show what to be printed? (^ represents a blank)statement print out ________ _______ printf("%c",c) Aprintf("%2c",c) ^Aprintf("%-3c",c) A^^ printf("%s",s) Blue moon!printf("%3s",s) Blue moon!printf("%.6s",s) Blue mprintf("%-11.8s",s) Blue moo^^^

Slide 120

Appendix(1). Binary Search/* Purpose: use binary search to search the key on a sorted score array. If found, return the index. * If not found, return -1* * Input: key, sorted score array and the size n•* Output: either its index if found •or -1 if not found•*/

Slide 121

Appendixint binarySearch(int key, int score[50], int n) { int left=0, right=n-1; int middle;

while (left < right) {middle=(left+right)/2;if (key == score[middle])

return middle; // found key else

if (key > score[middle])left=middle+1; // key in right part

elseright=middle; // key in leftpart

} return -1; // key not found

}

Slide 122

Appendix(2). Selection Sortvoid selectionSort (int ar[], int n) { int i, j, minat, mi, tempn;

for (i=0;i<n-1;i++){minat=i;min = a[i];for(j=i+1;j<n ;j++) {// select min in array if (min > a[j]) {//ascending order for descending reverse

minat=j; //the position of the min elementmin=a[j];

} // if} // for jtemp=a[i] ; //swapa[i]=a[minat]; a[minat]=temp;

} // for i} // selection sort

Slide 123

References Deitel & Deitel: C How to Program, 4th ed.,

Chapter 6, 7 & 8, Prentice Hall

Documents

Chapter 5 Array, Pointers and Strings