Basic and Practice in Programming Lab7

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Basic and Practice in Programming Lab7

Variable and Its Address (1/2) What is the variable? Abstracted representation of allocated memory Having address & value Memory address 10 0x00000010 a int a = 10; value 0x00000004 0x00000000

Variable and Its Address (2/2) /* Practice 1 : Variable and Address */ #include <stdio.h> int main(void) int a = 10; printf( value: %d address: %p\n, a, &a); return 0;

A Consideration Let s consider follow function form int add (int x, int y) x += y; return x;

A Consideration Let s consider follow function form int add (int x, int y) x += y; return x; int add (_x, _y) int x, y; x = _x; y = _y; x += y; return x; Realistic executable function form The function uses arguments as form of local variables

A Consideration Let s consider follow function form int add (int x, int y) x += y; return x; int main (void) int x = 10, y = 10, sum; sum = add(x, y); return 0; What is the value of x? int add (_x, _y) int x, y; x = _x; y = _y; x += y; return x; Realistic executable function form The function uses arguments as form of local variables

Call by Value /* Practice 2 : Call by value */ #include <stdio.h> int add (int x, int y) x += y; return x; int main(void) printf( %d + %d = %d\n, 10, 20, add(10, 20)); return 0;

Call by Value /* Practice 3 : Call by value 2 */ #include <stdio.h> int add (int x, int y) x += y; printf ( x: %p y: %p, &x, &y); return x; int main(void) int x = 10, y = 20; printf( x: %p y: %p, &x, &y); printf( %d + %d = %d\n, x, y, add(x, y)); return 0;

A Question How to use same variable Across two functions The answers are Using global variable but, Using global variable is NOT recommended Security, consistency, reliability problems Using variable s address instead of variable itself Pointer

Pointer It is not a concept Just a method There is no serious philosophy in the pointer Pointer is just variable for containing address int a = 10; int *p = &a; //declaration form of pointer //p == &a, *p == a

Pointer Memory address 10 0x00000010 int a = 10; int *p; p = &a; value 0x00000010 value 0x00000004 *p 0x00000000

Functions Two types of function call Call by value Call by reference void func1(int a) Call by value Call by value Argument is just value Call by reference Argument is given by variable s address void func2(int *a) Call by reference

Call by Reference (1/3) Swap function Exchange two variable s contents int a = 10, b = 20; a 10 b 20 swap(a, b); a 20 b 10

Call by Reference (2/3) /* Practice 4 : Call by reference example1*/ #include <stdio.h> int main(void) int a = 10, b = 20; void swap (int x, int y) int temp; temp = y; y = x; x = temp; printf( a : %d b : %d, a, b); swap(a, b); printf( a : %d b : %d, a, b); return 0;

Call by Reference (3/3) /* Practice 5 : Call by reference example2*/ #include <stdio.h> int main(void) int a = 10, b = 20; void swap (int *x, int *y) int temp; temp = *y; *y = *x; *x = temp; printf( a : %d b : %d, a, b); swap(&a, &b); printf( a : %d b : %d, a, b); return 0;

Reference? swap Call by value x 10 main y 20 a b 10 0x1000 20 0x1004

Reference? swap Call by value x 20 main y 10 a b 10 0x1000 20 0x1004

Reference? swap Call by value x 20 main y 10 a b 10 0x1000 20 0x1004 swap *x 0x1000 Call by reference *y 0x1004

Reference? swap Call by value x 20 main y 10 a 10 0x1000 b 20 0x1004 swap *x 0x1000 0x1000 10 Call by reference *y 0x1004 0x1004 20

Reference? swap Call by value x 20 main y 10 a b 20 0x1000 10 0x1004 Reference swap *x 0x1000 0x1000 20 Call by reference *y 0x1004 0x1004 10

Type Pointer has type Integer, character, floating point Pointer must be used to point to same type variable ip = &int_val int *ip int int_val = 10; char *cp cp = &char_val char char_val = a ; fp = &f_val float *fp float f_val = 3.14;

Pointer of Pointer Pointer is just a variable Of course, it has its address /* Practice 4 : Pointer of pointer */ #include <stdio.h> int main (void) int x = 10, *p, **pp; p = &x; pp = &p; printf ( %d %d %d\n, x, *p, **pp); printf ( %p %p %p\n, &x, p, *pp); printf ( %p %p\n, &p, pp); return 0;

Exercise 1 Left shifter Input is given as form of sequence of 5 distinct charac ters Last input is the number of shift Skeleton code is given Check the homepage You must not print last input A B C D E 3 D E A B C

Array (1/8) Pointer Container of memory address Accessing variable s value by using * int *p; p = &a; Array Reference of continuous memory addresses Not a variable int arr[10] arr address of arr[0] arr[2] arr[4] arr[6] arr[8] arr[1] arr[3] arr[5] arr[7] arr[9] arr == &arr[0] *arr?

Array (2/8) /* Practice 1 : Array 1*/ #include <stdio.h> int main (void) int arr[10] = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9; printf( %p %p\n, &arr[0], arr); printf( %d %d\n, arr[0], *arr); return 0;

Array (3/8) Array in memory Address is linearly increasing By size of type int arr[5] arr[0] arr[1] arr[2] arr[3] arr[4] 0x1000 0x1004 0x1008 0x100c 0x1010 sizeof(int) Memory

Array (4/8) /* Practice 2 : Array 2*/ #include <stdio.h> int main (void) int arr[10] = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9; int i; for (i=0;i<10;i++) printf( %p %p\n, &arr[i], arr+i); arr[i] == arr+i == &arr[0] + i*sizeof(int) for (i=0;i<10;i++) printf( %d %d\n, arr[i], *(arr+i)); return 0;

Array (5/8) /* Practice 3 : Array 3*/ #include <stdio.h> int main (void) char arr[10] = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9; int i; for (i=0;i<10;i++) printf( %p %p\n, &arr[i], arr+i); arr[i] == arr+i == &arr[0] + i*sizeof(char) for (i=0;i<10;i++) printf( %d %d\n, arr[i], *(arr+i)); return 0;

Array (6/8) 2D array int arr[2][5] arr[0][0] arr[0][1] arr[0][2] arr[0][3] arr[0][4] arr[1][0] arr[1][1] 0x1000 0x1004 0x1008 0x100c 0x1010 0x1014 0x1018 sizeof(int) sizeof(int) * 5 sizeof(int) Memory

Array (7/8) /* Practice 4 : 2D Array 1*/ #include <stdio.h> int main (void) int arr[2][5] = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9; printf( %p %p %p\n, arr, arr[0], &arr[0][0]); printf( %p %p %p\n, arr+1, arr[1], &arr[1][0]); printf( %p %p %p\n, *(arr+1)+2, arr[1]+2, &arr[1][2]); return 0;

Array (8/8) /* Practice 5 : 2D Array 2*/ #include <stdio.h> int main (void) int arr[2][5] = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9; printf( %p %p %p\n, arr, arr[0], &arr[0][0]); printf( %p %p %p\n, arr+1, arr[1], &arr[1][0]); printf( %p %p %p\n, *(arr+1)+2, arr[1]+2, &arr[1][2]); printf("%d %d %d\n", **arr, *arr[0], arr[0][0]); printf("%d %d %d\n", **(arr+1), *arr[1], arr[1][0]); printf("%d %d %d\n", *(*(arr+1)+2), *(arr[1]+2), arr[1][2]); return 0;

Pointer (1/6) Pointer can be used as array int arr[5]; int *p; p = &arr[0]; //same as p = arr; *p 0x1000 arr[0] arr[1] arr[2] arr[3] arr[4] 0x1000 Memory

Pointer (2/6) /* Practice 6: Pointer 1*/ #include <stdio.h> int main (void) int arr[10] = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9; int i; int *p; p = &arr[0]; //same as p = arr for (i=0;i<10;i++) printf( %p %p %p\n, &p[i], p+i, &arr[i]); for (i=0;i<10;i++) printf( %d %d %d\n, p[i], *(p+i), arr[i]); return 0;

Pointer (3/6) Pointer must be carefully used int arr[5]; int *p; printf( %d\n, p[1]); *p p[1] == *(p + sizeof(int)) garbage arr[0] arr[1] arr[2] arr[3] arr[4] 0x1000 Memory?

Pointer (4/6) /* Practice 7 : Pointer misuse*/ #include <stdio.h> int main (void) int arr[10] = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9; int i; int *p; //p = &arr[0]; for (i=0;i<10;i++) printf( %p %p %p\n, &p[i], p+i, &arr[i]); for (i=0;i<10;i++) printf( %d %d %d\n, p[i], *(p+i), arr[i]); return 0;

Pointer (5/6) /* Practice 8 : Pointer 2*/ #include <stdio.h> int string_length(char *s) int n; for (n = 0;*s!= \0 ;s++) n++; return n; int main (void) char str[10]; scanf( %s, &str[0]); printf( %s\nlength : %d\n, str, string_lenth(str)); return 0;

Pointer (6/6) String copy strcpy() is defined in standard C library

Pointer Example /* Practice 9 : String copy 2*/ #include <stdio.h> void mystrcpy(char s[], char d[]) int i; for (i = 0; s[i]!= \0 ; i++) d[i] = s[i]; d[i] = s[i]; int main (void) char str[10], dest[10]; scanf( %s, &str[0]); printf( %s\nlength : %d\n, str, string_lenth(str)); return 0;

Pointer Array Pointer can be declared as array Example: string array char *str_array[5] str_array[0] str_array[1] str_array[2] str_array[3] str_array[4] All work and no play makes Jack a dull boy

Pointer Array /* Practice 10 : Pointer array */ int main (void) char *str_array[5] = All work, and, no play, makes, Jack a dull boy ; printf( %s %s %s %s %s\n, str_array[0], str_array[1], str_array[2], str_array[3], str_array[4]); return 0;

Pointer Array /* Practice 11 : Pointer array 2 */ int main (void) char *str_array[5] = All work, and, no play, makes, Jack a dull boy ; printf( %s %s %s %s %s\n, str_array[0], str_array[1], str_array[2], str_array[3], str_array[4]); str_swap(&str_array[0], &str_array[2]); printf( %s %s %s %s %s\n, str_array[0], str_array[1], str_array[2], str_array[3], str_array[4]); return 0; void str_swap (char **s, char **d) char *temp; temp = *s; *s = *d; *d = *s;

Exercise 1 Modifying strcpy() Two arguments Source string Destination string Don t use local variable Using pointer operations Modifying practice 9 s code

Exercise 2 Word changer Input is composed of three strings A sentence not including white space Target word Changing word You find target words in a sentence, and change by c hanging word Two words have same length AllworkandnoplaymakesJackadullboy work play AllplayandnoplaymakesJackadullboy

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