25

I created ptr as pointer to an array of 5 chars.

char (*ptr)[5];

assigned it the address of a char array.

char arr[5] = {'a','b','c','d','e'};
ptr = &arr;

using pointer ptr can I access the char values in this array?

printf("\nvalue:%c", *(ptr+0));

It does not print the value.

In my understanding ptr will contain the base address of array but it actually point to the memory required for the complete array (i.e 5 chars). Thus when ptr is incremented it moves ahead by sizeof(char)*5 bytes. So is it not possible to access values of the array using this pointer to array?

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  • 1
    Please remember to accept the correct answer to your question which helped you the most. You accept questions with the tick mark next to it. Commented Sep 2, 2011 at 8:00
  • Everyone, pretty please read this c-faq.com/aryptr/ptrtoarray.html before posting an answer. Commented Sep 2, 2011 at 8:23

4 Answers 4

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25

When you want to access an element, you have to first dereference your pointer, and then index the element you want (which is also dereferncing). i.e. you need to do:

printf("\nvalue:%c", (*ptr)[0]); , which is the same as *((*ptr)+0)

Note that working with pointer to arrays are not very common in C. instead, one just use a pointer to the first element in an array, and either deal with the length as a separate element, or place a senitel value at the end of the array, so one can learn when the array ends, e.g.

char arr[5] = {'a','b','c','d','e',0}; 
char *ptr = arr; //same as char *ptr = &arr[0]

printf("\nvalue:%c", ptr[0]);
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1 Comment

Array pointers have their (limited) uses, for example they are useful when declaring multi-dimensional arrays dynamically and you want the memory declared adjacently. So I wouldn't dismiss them entirely. But as we can see from this thread, most C programmers are clueless about them, so that alone is a reason to stay away from then.
10

Most people responding don't even seem to know what an array pointer is...

The problem is that you do pointer arithmetics with an array pointer: ptr + 1 will mean "jump 5 bytes ahead since ptr points at a 5 byte array".

Do like this instead:

#include <stdio.h>

int main()
{
  char (*ptr)[5];
  char arr[5] = {'a','b','c','d','e'};
  int i;

  ptr = &arr;
  for(i=0; i<5; i++)
  {
    printf("\nvalue: %c", (*ptr)[i]);
  }
}

Take the contents of what the array pointer points at and you get an array. So they work just like any pointer in C.

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0

Your should create ptr as follows:

char *ptr;

You have created ptr as an array of pointers to chars. The above creates a single pointer to a char.

Edit: complete code should be:

char *ptr;
char arr[5] = {'a','b','c','d','e'};
ptr = arr;
printf("\nvalue:%c", *(ptr+0));
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2 Comments

char (*ptr)[5] is a pointer to array, and char *ptr[5] is array of pointers.
The OP asked about array pointers specifically. So I suspect that they know how to use plain pointers and arrays already.
0

Use of pointer before character array

Normally, Character array is used to store single elements in it i.e 1 byte each

eg:

char a[]={'a','b','c'};

we can't store multiple value in it.

by using pointer before the character array we can store the multi dimensional array elements in the array

i.e.

char *a[]={"one","two","three"};
printf("%s\n%s\n%s",a[0],a[1],a[2]);
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