A keynote on Problem Solving using Computers

1. Pointers - Continued
V. Kamakoti

2. Let us revise
#include <stdio.h>
main() {
What is the output?
int u1, u2;
int v = 3;
int *pv;
u1 = 2* (v + 5);
pv = &v;
u2 = 2*(*pv + 5);
printf(“n u1=%d u2=%d”,u1,u2);
}

3. Let us revise
#include <stdio.h>
main() {
u1 = 16 u2 = 16
int u1, u2;
int v = 3;
int *pv;
u1 = 2* (v + 5);
pv = &v;
u2 = 2*(*pv + 5);
printf(“n u1=%d u2=%d”,u1,u2);
}

4. Let us revise
char my_word[50];
int a[100];
scanf(“%s %d”,my_word,&a[50]);
Note: No “&” for arrays, but “&” required for
reading an element of an array

5. Passing partial arrays
#include<stdio.h>
main() {
int z[100];
void process(int z[]);
process(&z[50]);
print(“%d %d”,z[50],z[51]);
}
What is the output?
void process(int a[]) {
a[0] = 5;
a[1] = 6;
}

6. Passing partial arrays
#include<stdio.h>
main() {
int z[100];
void process(int z[]);
process(&z[50]);
print(“%d %d”,z[50],z[51]);
}
56
void process(int a[]) {
a[0] = 5;
a[1] = 6;
}

7. Functions can return pointers
main() {
int z[100];
int *a;
int *scan(int b[]);
a = scan(z);
printf(“%d %d”,a[0],a[1]);
}
What is the output?
int *scan(int b[]) {
int *c;
b[0] = 5;
b[1] = 6;
c = b;
return( c );
}

8. Functions can return pointers
main() {
int z[100];
int *a;
int *scan(int b[]);
a = scan(z);
printf(“%d %d”,a[0],a[1]);
}
int *scan(int b[]) {
int *c;
b[0] = 5;
b[1] = 6;
c = b;
return( c );
}
56

9. Functions can return pointers
main() {
int z[100];
int *a;
int *scan(int b[]);
a = scan(z);
z[3] = 8;
printf(“Output: %d”,a[3]);
}
int *scan(int b[]) {
int *c;
b[0] = 5;
b[1] = 6;
c = b;
return( c );
}
What is the Output?

10. Functions can return pointers
main() {
int z[100];
int *a;
int *scan(int b[]);
a = scan(z);
z[3] = 8;
printf(“Output: %d”,a[3]);
}
int *scan(int b[]) {
int *c;
b[0] = 5;
b[1] = 6;
c = b;
return( c );
}
Output: 8

11. Pointers and 1-D arrays
•
•
•
•
•
int x[100];
x : pointer to x[0] - say address 1002
x + 1: pointer to x[1] - address 1006
x + 2: pointer to x[2] - address 1010
x + i: pointer to x[i]
= 1000 + i * sizeof(int);
= Base Address + i * sizeof(int);

12. What happens
main() {
int z[100];
scanf(“%d”, z+5);
printf(“%d %d”,z[5],*(z+5));
}
If you enter 10, the code outputs
10 10

13. What is missing?
• You should know the size of the array
while writing the code
• Else space allocation cannot be done
• It is a limitation to the user of the code
• Any solution?
– Yes - through Dynamic Memory Allocation

14. These are Equivalent
• int x[10];
• Another way
– #define SIZE 10
– int x[SIZE]
• int x[10] = {1,2,3,5,6,7,8,9,10,11};
• int x[] = {1,2,3,4,5,6,7,8,9,10,11};
• Another way
– int *x;
– x = (int *) malloc(10 * sizeof(int));

15. These are Equivalent
– int *x;
– x = malloc(10 * sizeof(int));
– This shall return a pointer to character inconsistent with definition of “x” - which is an
integer pointer.
– x = (int *) malloc(10* sizeof(int));
– This is called “Pointer Casting”.

16. Look at this
main() {
int p;
int *q;
printf(“Enter size of arrayn”);
scanf(“%d”,&p);
q = (int *) malloc(p * sizeof(int));
}
Creates an array of size as desired by user.

17. So What?
int line[80];
int *pl;
line[2] = line[1]; //is equivalent to
*(line + 2) = line[1];
*(line + 2) = *(line + 1);
line[2] = *(line + 1);
p1 = &line[1]; //Is equivalent to
p1 = line + 1;