done with programs feeding for now

Author: Anku9053

Pointer.c

@@ -1,10 +1,14 @@
 #include <stdio.h>
 #include <conio.h>
-int valuebyusingpointer(int x, int y, int z);
-int pointerdiscussion(int x, int y, int z);
-int fun(int x);
-int basicdeclaration();
-int ProblemOnPointer();
+// int valuebyusingpointer(int x, int y, int z);
+// int pointerdiscussion(int x, int y, int z);
+// int fun(int x);
+// int basicdeclaration();
+// int ProblemOnPointer();
+// int twolevelpointer();
+// int HardpointerValuChange();
+// int easyPointerChange();
+int pointerToArray();
 int main()
 {
  int x, y, z;
@@ -20,7 +24,11 @@ int main()
  // pointerdiscussion(x,y,z);
  // valuebyusingpointer(x, y, z);
  // printf("%x",d);
- ProblemOnPointer();
+ // ProblemOnPointer();
+ // twolevelpointer();
+ // HardpointerValuChange();
+ // easyPointerChange();
+ pointerToArray();
  printf("\nhurray i learnt pointer");
 }
 
@@ -59,39 +67,111 @@ int fun(int x)
 {
  // x = 30;
  int *ptr = &x;
-
 }
 // Write a program in C to show the basic declaration of a pointer Expected Output :Pointer : Show the basic declaration of pointer:Here is m=10, n and o are two integer variable and *z is an nteger z stores the address of m = 0x7ffd40630d44 *z stores the value of m = 10 &m is the address of m =x7ffd40630d44 &n stores the address of n = 0x7ffd40630d48 &o stores the address of o =0x7ffd40630d4c &z stores the address of z= 0x7ffd40630d50
 
-int basicdeclaration(){
+int basicdeclaration()
+{
  int m = 1000;
- int n,o;
- int* z;
+ int n, o;
+ int *z;
  z = &m;
- printf("\nthe value of m is %d",m);
- printf("\nThe value of m through address is %x",*z);
-
- printf("\nThe adress of n is %x",&n);
- printf("\nThe adress of o is %x",&o);
+ printf("\nthe value of m is %d", m);
+ printf("\nThe value of m through address is %x", *z);
 
+ printf("\nThe adress of n is %x", &n);
+ printf("\nThe adress of o is %x", &o);
 }
 
-
-int valuechangethroughfunction(int x){
+int valuechangethroughfunction(int x)
+{
  x = 400;
  return 0;
 }
 
-
 // some basic problems on pointer
 
-int ProblemOnPointer(){
- int x=10,y=18;
- int *ptr1,*ptr2;
+int ProblemOnPointer()
+{
+ int x = 10, y = 18;
+ int *ptr1, *ptr2;
  ptr1 = &x;
  // ptr2 = &y;
  ptr2 = &y;
- *ptr2= *ptr1;
- printf("a = %d %d %d",x,*ptr1,*ptr2);
+ *ptr2 = *ptr1;
+ printf("a = %d %d %d", x, *ptr1, *ptr2);
+}
+
+// Two levelpointer
+
+int twolevelpointer()
+{
+ int s = 12111230;
+ int *ptr1, **ptr2;
+ ptr1 = &s;
+ ptr2 = &ptr1;
+
+ printf("The address of s is %x", &s);
+ printf("\n");
+ printf("The value of s is %d", s);
+ printf("\n");
+ printf("The address of ptr1 is %x", ptr1);
+ printf("\n");
+ printf("The value of ptr1 is %d", *ptr1);
+ printf("\n");
+ printf("The address of ptr2 is %x", ptr2);
+ printf("\n");
+ printf("The value of ptr2 is %p", **ptr2);
 }
 
+int HardpointerValuChange()
+{
+ int *pc, c;
+ c = 22;
+ printf("Address of c: %p\n", &c);
+ printf("Value of c: %d\n\n", c);
+ pc = &c;
+ printf("Address of pointer pc: %p\n", pc);
+ printf("Content of pointer pc: %d\n\n", *pc);
+ c = 11;
+ printf("Address of pointer pc: %p\n", pc);
+ printf("Content of pointer pc: %d\n\n", *pc);
+ *pc = 2;
+ printf("Address of c: %p\n", &c);
+ printf("Value of c: %d\n\n", c); // 2
+ return 0;
+}
+
+int easyPointerChange()
+{
+ int *pc, c;
+ c = 5;
+ pc = &c;
+ *pc = 1;
+ printf("%d\n", *pc);
+ printf("%d", c);
+}
+
+
+// Pointer To Array
+// a program where we got to know the base adress of the array is the array first element address aactually
+int pointerToArray(){
+ int arr[5];
+ int len = sizeof(arr)/sizeof(arr[0]);
+ // printf("lenggth is %d",len);
+ for(int i=0;i<len;i++){
+ printf("%x\n",&arr[i]);
+ }
+ printf("%x",&arr);
+ printf("\n");
+ printf("%x",&arr[1]);
+ printf("\n");
+ printf("%x",&arr[2]);
+ printf("\n");
+ printf("%x",&arr[3]);
+ printf("\n");
+ printf("%x",&arr[4]);
+}
+
+//By taking value and printing the answer
+

Pointer.exe

@@ -165,4 +165,5 @@ struct date baisicstructure(){
  printf("Enter the date : ");
  scanf("%d/%d/%d",&Somedate.dates,&Somedate.months,&Somedate.year);
  return Somedate;
-};
+};
+