pointers

Code:

#include<stdio.h>
main(){
	int a[5]={5,50,100,150};
	int *iPtr;
	iPtr=a;
	printf("\n%d", ++*iPtr); // this will increment a[0] but not temporary
printf("\n%d", *iPtr+4); // this will increment a[0] temporary

could you please show me how to increment a[0] for 4 (+4) but while using pointers as above?

As in you want to write the expression a[0] + 4 using pointer notation rather than array index notation?

The easiest way, provided your pointer is valid would be *ptr++ or ++*ptr.

You can also do math this way ... *ptr = 8 * 3;

When defining a pointer as in int *ptr The star says "I need an address here"
When manipulating values at a pointer as in *ptr The star says "Do this at the given address".

Take note the difference between (*ptr)++ and *ptr++.

Code:

printf("\n%d", *iPtr+4); // this will increment a[0] temporary

guys I am asking how to increment as I do above but this is temporary and I would like my incrementation to not be temporary!!!

btw. I understand this

Originally Posted by Bayint Naung

Take note the difference between (*ptr)++ and *ptr++.

Originally Posted by c_lady

I am asking how to increment as I do above but this is temporary and I would like my incrementation to not be temporary!!!

Oh, then you should write:

Code:

a[0] += 4;

or:

Code:

*iPtr += 4;

But then you should already know this before starting on arrays and pointers.

Code:

#include <stdio.h>
int main(){

  int *ptr, a[5] = {0,0,0,0,0}; 

  ptr = a; // ptr points to a
  *ptr += 4; // Increase the value which ptr points to by 4, in this case a[0]+4

  *(++ptr) += 5; // This should increment a[1] by 5...  notice the use of () and ++

  printf("a[0]= %d\n", a[0]);
  printf("a[1]= %d\n", a[1]);

  return 0;
}

Output:

Code:

$ gcc test.c
$ ~/a.out
a[0]= 4
a[1]= 5

edit:

May not be needed to use () but I like to because in longer expressions things can get very complicated and C's order of operation isn't always clear.