Thread: can u help me with pointers?plzzzzzzzz

I cant answer it cuz
Iam studing for my mid semester exams
so I dont have the time
and if u want the truth
I didnt under stand the pointers
so i cant answer it
if u gona explain to me
I will be thankful

no, kurz7 is right.

either pick a question and ask it(one you have, not homework), or show some code.

First, I wanna say nobody wanna do the homework for somebody else, when this person is lazy to try to do them.

So, I won't do your work. I wanna just to try to explain the work with pointers.....
For Further information, read the www.cprogramming.com tutorials.....


pointers
===================

declaration syntax:

Datatype *pointervariable

Example:
+++++++++++++

int x=5;

address: 0000:1234
name: x
value: 5

code:

Code:

#include <stdio.h>

int main()
{
   int x=5;
   printf("The address of x is %p \n",&x);
   return 0;
}

Something about addresses.......

code:

Code:

#include <stdio.h>

int main()
{
 int check;
 int chapter1 = 5;
 int chapter2 = 60;
 int chapter3 = 166;
 int *data;

 do{
     printf("\tChapter overview\n");
     printf("\t*******************************\n\n");
     printf("\t-1- Chapter 1\n");
     printf("\t-2- Chapter 2\n");
     printf("\t-3- Chapter 3\n");
     printf("\t-4- End\n");
     printf("\n");
     printf("\tOption: ");
     scanf("%d",&check);
     printf("\tYou find chapter %d on ",check);

     switch(check)
      {
       case 1 :  data=&chapter1;
                 printf("page %d\n",*data);
                 break;
       case 2 :  data=&chapter2;
                 printf("page %d\n",*data);
                 break;
       case 3 :  data=&chapter3;
                 printf("page %d\n",*data);
                 break;
       default:  printf("page ???\n");
                 break;
     }
   }
   while(check<4);
 return 0;
}

This is another example:

Code:

#include <stdio.h>

int main()
{
 int x=5;
 int *y;

 printf("address x=%p, value x=%d\n",&x,x);
 printf("address *y=%p, value *y=%d(crap)\n",&y,*y);

 printf("\ny=&x;\n\n");
 y=&x;              /*y has now the address of x */

 printf("address x=%p, value x=%d\n",&x,x);
 printf("address *y=%p, value *y=%d\n",&y,*y);

 printf("\ny points to this address %p\n",y);
 printf("this is the address of x = %p\n",&x);

 printf("\nWATCH OUT!!!\n\n");

 *y=10;
 printf("*y=10\n\n");
 printf("address x=%p, value x=%d\n",&x,x);
 printf("address *y=%p, value *y=%d\n",&y,*y);

 printf("\nypoints to this address %p\n",y);
 printf("it is still the address of x (%p)\n",&x);

 return 0;
}

The next example is imporatant. And never use such crap, it coul be very dangerous:

Code:

#include <stdio.h>

int main()
{
int *y;
*y=10;
printf("the value of *y is %d\n",*y);
return 0;
}

This is dangerous, because y points to a not defined address..........

you could try to prevent such mistakes, when u use:

Code:

#define NULL (void *)0

example:

Code:

#include <stdio.h>

int main()
{
int *y=NULL;

if(y==NULL)
{
printf("The pointer has an illegal/not defined address\n");
}

else
{
*y=10;
}

return 0;
}

Some other things to know:

Code:

int *ptr;    /* declaration */
int var;

ptr=&var;    /* initialistion : ptr get the address of var */
*ptr=100;   /* dereferenciation: var get the value 100 */
*ptr+=100; /* dereferenciation: we add 100 to var */
*ptr++;    /* dereferenciation: var has now the value of 201 */
*ptr--;    /* var has now the value of 200 again */
printf("%d",*ptr);    /* outputs the value of var */
printf("%p",&ptr);    /* outputs the address of ptr */
printf("%p",ptr)      /* outputs the address of var */

And the second part........


pointers, which points onto other pointers

example:

Code:

int *ptr1;
int *ptr2;
ptr1=ptr2;

code:

Code:

#include <stdio.h>

int main()
{
 int value=10;
 int *ptr1;
 int *ptr2;

 ptr1=&value;
 ptr2=ptr1;

 printf("value=%d address in *ptr1=%p\n",*ptr1,ptr1);
 printf("address of *ptr1=%p\n",&ptr1);

 printf("value=%d address in *ptr2=%p\n",*ptr2,ptr2);
 printf("address of *ptr2=%p\n",&ptr2);
 printf("address of the int value=%p\n",&value);
 return 0;
}

Pointers in C/c++ have always to have the same datatypes as variables...........

example, this code returns crap..... and good compilers like gcc would not compile the code and return an error, like cannot convert double* to int*:

Code:

#include <stdio.h>

int main()
{
  int *int_ptr;
  double double_value=999.999;

  int_ptr=&double_value;
  printf("*int_ptr=%d double=%f\n",*int_ptr, double_value);
  return 0;
}

Now, I gonna explain the usage of pointers as function parameters "Call-By-Reference"
There is another method "call-by-value", which has often disadvanteges, because the compiler has always to copy the parameters, cause the call-by-value parameters are used as local variables in the functions.......

example:

Code:

#include <stdio.h>
#define PI 3.141592

float circlearea(float value)
{
   return (value = value * value * PI);
}

int main()
{
   float r, area;
   printf("We calculate a circle area!!\n\n");
   printf("Input r : ");
   scanf("%f",&r);

   area = circlearea(r);
   printf("\nThe circle area: %f\n",area);
   return 0;
}

In such a case, we can easily use the address of r and pass the address of r as argument to the function.......

call-by-reference example:

Code:

#include <stdio.h>
#define PI 3.141592

void circlearea(float *value)
{
 *value = ( (*value) * (*value) * PI );
}

int main()
{
   float r;
   printf("We calculate the circle area\n\n");
   printf("Input r: ");
   scanf("%f",&r);
   circlearea(&r);
   printf("\nThe circlearea : %f\n",r);
   return 0;
}

It think, that this part will be very interesting for you:

ARRAYS and POINTERS

First....
Arrays and pointers have nothing in common. A pointer is the address of an address and an array name is only an address.

Here is an example, how to access arrays with pointers:

Code:

#include <stdio.h>

int main()
{
 int element[8]= {1,2,4,8,16,32,64,128};
 int *ptr;
 int i;

 ptr=element;

 printf("*ptr points onto the value of %d\n",*ptr);
 printf("*ptr+1 points now onto the value of %d\n",*(ptr+1));
 printf("*(ptr+3) = %d\n",*(ptr+3));

 printf("\nAnd now a for loop: \n");

 for(i=0; i<8; i++)
 {
      printf("element[%d]=%d \n",i,*(ptr+i));
 }
 return 0;
}

When you use this:
ptr=element
You send the address of element to the pointer.
This works without the address operator, cause the ANSI standard says, that the Array name is always like a pointer, which points onto the first array element.

Here the example:

Code:

#include <stdio.h>

int main()
{
 int element[8] = {1,2,4,8,16,32,64,128};
 int i;

 printf("*element     = %d\n",*element);
 printf("*(element+1) = %d\n",*(element+1));
 printf("*(element+3) = %d\n",*(element+3));

 printf("\nNow a for loop: \n");

 for(i=0; i<8; i++)
 {
      printf("*(element+%d) = %d \n",i,*(element+i));
 }
 return 0;
}

When u wanna use the address operator, it's ok too.
ptr=&element[0]; /* identical with ptr=element */

Now another example, call-by-reference with arrays and pointers.....

Code:

#include <stdio.h>

void function(int *array, int n_array)
{
 int i;
 for(i=0; i < n_array; i++)
  printf("%d ",array[i]);
 printf("\n");
}

int main()
{
 int values[] = { 1,2,3,5,8,13,21 };

 function(values, sizeof(values)/sizeof(int));

 return 0;
}

I hope, I have helped you.......
~codingmaster