Pointer in C.txt

Fact 1

int x, y, *z; // we declare 3 interger variables and z is a pointer variable

x = 40;
y = 44;
z = &x; // here "z" is holding address of x , so " *z " has the value of "x"

printf("value of x = %d", x); // 40
printf("value of *z = %d", *z); // 40

Fact 2:
**&y = ?

It will be simplied as : *(*(&y))

Example :

int x = 5, *y = &x;

(&y) is own address of y; which is ofcourse different the x's address not a big deal !
*(&y) => describes what has in y?, as y has x's address means i.e. &x . *(&y) prints address of x, (while *y prints value of x).
*(*(&y)) As *(&y) is address of x, *(address of x) gives the value of x , i.e. 5


Fact 3:

(1) if i = &j then *i  has to be true i.e i is a pointer variable.

(2) *(i = &j) is equivalent of *(&j)

(3) 
int *i, j = 5;
i = &j;

printing **i or *(*i) will give error (because you have not declared as double pointer variable), you can write *(&j) or *&j  i.e value of j (5) or *(&i) i.e. value of i, 
which the address no. of j.

Fact 4: Pointer and Array 

#include <stdio.h>

int main(){
	  int i,*p, odd[5] = {1,3,5,7,9};
	  p = odd; // p = odd is equivalent of p = &odd[0] (address of first element of array).
			         // *p hold value of odd[0] i.e. 1 .
	 for (i = 0; i < 5; i++){
		  printf("%d \t", *p); // 1,3,5,7,9
		  p++;  // with each iteration pointer points to next element of array.
	 }
}

Fact 5 :
int x = 0;
printf("%d", x++); // 0
//printf("%d", x);
because x++ evaluates like this first printed out current value of x, then increment value of x. so if we uncomment 3rd line
1 will be printed.

========================

int x = 0;
x++; // here x's value incremented by 1, so x is now 1
printf("%d", x++); // like before printf() will execute x's current value which is 1, then increment it by 1, so x will be 2 but
		we will not see that on screen!

=========================

int x = 0;
x++;
printf("%d", x); // 1 but x will not get any increment.

==========================

int x = 0;
printf("%d", ++x); // here x will be incremented first than print on screen which is 1. 

==========================

int x = 0;
++x;
printf("%d",++x); //here x is incremented in line 2 , now x is 1 then in line 3 , x will be incremented again by 1 then print which is 2

===========================
int x=0;
++x;
printf("%d \n", x++); // here x will be 1 for line 2 but in line 3, x will be printed as 1 , then increment by 1 , which we will can't see.

Fact 6:

int x,*p, odd[5] = {1,3,5,7,9};
	
p = odd;
x = *p++;  // Order of evaluation  *(p++) , 

printf("%d \n", x); 	// x has *p++ , x will be printed equal to the value of *(p) which is 1
			// then p will be increment by 1, that means p's address will be updated
			// by 1, p will point the next element of array. so *p has value 3 now!
			// but we can't see that
			// if you uncomment next line it will print 3

//printf("%d \n", *p);	

===============================

int x,*p, odd[5] = {1,3,5,7,9};
	
p = odd;
x = *++p;  // Order of evaluation  *(++p) , 

printf("%d \n", x); 	// x has *++p , in line 3 p will be incremented by 1, i.e. it will point 
			// next elment of array which is address of odd[1], then *(address of odd[1])
			// will give value 3, which will be saved in x, and it will be printed !
			// Here no increment will happen on p after the printf().
			
===============================

int x,*p, odd[5] = {1,3,5,7,9};
	
p = odd;
x = ++*p;  // Order of evaluation  ++(*p) , 

printf("%d \n", x); 	// x has ++*p , in line 3 *p i.e. 1 will be increment first by 1 so
			// x will save 2 and it will be printed !