240-222 CPT: Arrays of Pointers/12 1

240-222 Computer Programming Techniques240-222 Computer Programming TechniquesSemester 1, 1998Semester 1, 1998

Objectives of these slides:– to illustrate the use of arrays of pointers and

contrast them with 2D arrays

12. Arrays ofPointers

240-222 CPT: Arrays of Pointers/12 2

Overview:Overview:

1. An Array of Pointers

2. A 2D Array of Characters

3. An Array of char*

4. Sorting Words

5. Dynamic Memory Allocation

6. A Dynamic Array

7. Static versus Dynamic Memory

8. main() Arguments

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1. An Array of Pointers1. An Array of Pointers

int *b[10];

int x[3], y[35];::

b[0] = x; b[1] = y;

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2. A 2D Array of Characters2. A 2D Array of Characters

char names[][5] ={ "Bob", "Jo", "Ann", "Fred"};

printf("%s", names[0]);

‘B’ ‘o’ ‘b’ ‘\0’

‘J’ ‘o’ ‘\0’

‘A’ ‘n’ ‘n’ ‘\0’

‘F’ ‘r’ ‘e’ ‘d’ ‘\0’

names

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3. An Array of char * Sec 7.9 / 7.83. An Array of char * Sec 7.9 / 7.8

char *names[] ={"Augustin", "Ann", "Bob", "Freddy"}

printf("%s", names[0]);

A u g u s t i n \0

A n n \0

B o b \0

F r e d d y \0

names[0]

names[1]

names[2]

names[3]

names

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4. Sorting Words4. Sorting Words test.dat:

– A is for apple or alphabet pie which all get a slice of, come taste it and try.

Sort the words:$ sort_words < test.dat

Output:Aaallalphabet...which

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sort_words.csort_words.c

#include <stdio.h>#include <stdlib.h>#include <string.h>

#define MAXWORD 50 /* max word length */#define SIZE 1000 /* array size */

void sort_words(char [][MAXWORD], int);void string_swap(char **, char **);

:

continued

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int main(){ char w[SIZE][MAXWORD]; /* array of fixed length strings */ int num, i;

for (i=0; scanf("%s", w[i]) == 1; i++){ if (i >= SIZE) { printf("Too many Words!"); exit(1); } }

::

continued

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: num = i; sort_words(w, num); for (i=0; i < num; i++) printf("%s\n", w[i]);

return 0;}

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5. Dynamic Memory Allocation5. Dynamic Memory Allocation

sort_words.c can save space by defining the size of each w[i] dynamically at run time.

New data structure:char *w[SIZE];

Make a pointer to a block of memory using:calloc(<num of elems>, <elem size>)

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New Top LevelNew Top Level

#include <stdio.h>#include <stdlib.h>#include <string.h>

#define MAXWORD 50 /* max word length */#define SIZE 1000 /* array size */

void sort_words(char *[], int);void string_swap(char **, char **);

:

continued

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int main(){ char *w[SIZE]; /* array of pointers */ char word[MAXWORD]; /* work space */ int num, i;

for (i=0; scanf("%s", word) == 1; i++){ if (i >= SIZE) { printf("Too many Words!"); exit(1); } w[i] = calloc(strlen(word)+1, sizeof(char)); strcpy(w[i], word); }

:continued

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: num = i; sort_words(w, num); for (i=0; i < num; i++) printf("%s\n", w[i]);

return 0;}

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Some CommentsSome Comments

calloc(strlen(word)+1, sizeof(char))

cannot just use strcpy(w[i], word)

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void sort_words(char *w[], int n)/* n elements are to be sorted *//* similar to bubble sort */{ int i, j;

for(i = 0; i < n; i++) for (j = i+1; j < n; j++) if (strcmp(w[i], w[j]) > 0) string_swap(&w[i], &w[j]);}

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In picture form:In picture form:

A u g u s t i n \0

A n n \0

B o b \0

F r e d d y \0

w[0]

w[1]

w[2]

w[3]

w

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void string_swap(char **p, char **q)/* swap the strings using pointers */{ char *temp;

temp = *p; *p = *q; *q = temp;}

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In picture form:In picture form:

A u g u s t i n \0

A n n \0

w[0]

w[1]

A u g u s t i n \0

A n n \0

w[0]

w[1]

Before

After

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6. A Dynamic Array6. A Dynamic Array

#include <stdio.h>#include <stdlib.h>

int main(){ int *a; /* will point to the array */ int i, size, sum = 0;

printf("An array will be created dynamically. Input an array size followed by values: ");

:

continued

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scanf("%d", &size);

/* allocate space in a for size int's */ a = (char *)malloc(size * sizeof(int)); for (i=0; i < size; i++) scanf("%d", &a[i]);

for(i=0; i < size; i++) sum += a[i]; printf("Sum is %d\n", sum);

free(a); return 0;}

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7. Static versus Dynamic Memory7. Static versus Dynamic Memory

The choice between static or dynamic memory is usually based on if you know how much data will be stored in your program.

If you know that an array of 100 integers is required then declare:

int data[100];

If you do not know, then consider creating dynamic memory.

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8. 8. main()main() Arguments Arguments

/* my_echo.c: Echo command line input */#include <stdio.h>

int main(int argc, char *argv[]){ int i;

printf("argc = %d\n", argc); for (i = 0; i <= argc-1; ++1) printf("argv[%d] = %s\n", i, argv[i]); return 0;}

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Compilation and Execution:Compilation and Execution:

$ gcc -Wall -o my_echo my_echo.c

$ my_echo a is for apple

argc = 5argv[0] = my_echoargv[1] = aargv[2] = isargv[3] = forargv[4] = apple

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PictoriallyPictorially

$ my_echo hello, world

produces:

m y _ e c h o \0

w o r l d \0

argv[0]

argv[1]

argv[2]

argv[3]

h e l l o , \0

NULL

argv