Pointers, Arrays, and Recursion 1

Original Course link: Pointers, Arrays, and Recursion

Week 1, Read pointer 1

C pointer

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

void g(int x, int * y) {
printf("In g, x = %d, *y = %d\n", x, *y);
x++;
*y = *y - x;
y = &x;
}

void f(int * a, int b) {
printf("In f, *a = %d, b = %d\n", *a, b);
*a += b;
b *= 2; // * multiply, not pointer identifier here;
g(*a, &b);
printf("Back in f, *a = %d, b = %d\n", *a, b);
}

int main(void) {
int x = 3;
int y = 4;
f(&x, y);
printf("In main: x = %d, y = %d\n", x, y);
return EXIT_SUCCESS;
}

Output

In f, *a = 3, b = 4 In g, x = 7, *y = 8 Back in f, *a = 7, b = 0 In main: x = 7, y = 4

execute the code by hand and write the output.

Why? Back in f, *a = 7, b = 0

g(*a, &b) = (7, 8)

&b means, the first address of the value 8, so b =0;

Swap with Hardware

how pointers work?

int *y, y = &x;

what’s y, x, *y, &x difference in the above program?

Pointers symbols: *, &

dereference the arrow?

the * operator, dereferences a pointer (follows an arrow),

the & operator, takes the address of something (gives an arrow pointing at that thing).

shorthand for dereferencing a pointer to a struct and selecting a field

(**q*)*.x = q->x*

*(*(*(*q).r).s).t. = q->r->s->t*

Output#

Why? Back in f, *a = 7, b = 0#

Output

Why? Back in f, *a = 7, b = 0