Question

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?

Answer 1

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]);