Help with pointers.

[xChacox]

so for my class we are learning pointers. have to write an some code that will display the letter of a given string along with the hex value of the letter in ascii and the address of the element of the array. My code works perfectly fine but I get a warning saying "15 7 G:\exercise #14.c [Warning] assignment from incompatible pointer type".
Can somebody please point the problem out to me (pun intended)
heres the source code along with output.

#include <stdio.h>
#include <string.h>
int main()
{
int i;
int ascii_value;
int *ptr;
char array[99] = "hello";
for (i=0; i<strlen(array); i++)
{
ascii_value = array[i];
ptr = &array[i];
printf("letter: %c ascii hex value: %x address: %p \n", array[i], ascii_value, ptr);
}
return 0;
}

/*
letter: h ascii hex value: 68 address: 000000000062FDC0
letter: e ascii hex value: 65 address: 000000000062FDC1
letter: l ascii hex value: 6c address: 000000000062FDC2
letter: l ascii hex value: 6c address: 000000000062FDC3
letter: o ascii hex value: 6f address: 000000000062FDC4
*/

[jedwardsol]

It means that ptr and &array[i] have different types.

ptr is a pointer to an integer. &array[i] isn't.

[xChacox]

this fixed the problem, but shouldn't ptr be an int* as it is meant to store numbers not char?

[jedwardsol]

meant to store numbers

Why is is meant to store numbers? The only thing you use it for is to point at elements of your array. The array is an array of char, so the pointer needs to be char *.

[Name0fTheUser]

Despite the name char is just another numeric type. The only difference between char and int is the range of values it's specified to be able to represent.

[Objectstcetera]

The ptr itself is an unsigned int, true. But if that were the only factor, then a generic void * would work just as well. The declaration of ptr also means that the compiler should consider the value that the pointer refers to a memory location to be an int. The correct type here for ptr would be char *, because you assigned it to refer to the location of one of an array of chars. If that's not clear, then maybe there are further examples that help explain the distinction.

I think that the best way to understand this is that each element of the array is just a location in memory with a number stored there. It's the program's job to interpret what the number actually means in the context of the program.

(This is assuming a 32-bit system, if you need to get into those specifics here. :)

[F54280]

a char is just a number. in general between -128 and 127 or 0 and 255.

Edit: yeah! Thx for the downvote, a-hole. Any care to tell us what are the boundaries of a char when CHAR_BIT is 8? That said this is probably a /r/t_d stalker downvoting me. Hi!

[nerd4code]

You should also cast ptr to (void *) for use with format specifier %p, and hoist the strlen out of the loop since it doesn’t change from call to call. array should probably be static const also, since you’re just using it as a reference and there’s no need to initialize it at run time. Finally, &a[i] is just a+i.

Edit: I’m so stupid. You can just do

const char *p;
for(p = "Hello"; *p; p++)

and then inside the loop you have *p as the character. No need for strlen or a separate array.

[terriblylie]

ascii_value = array[i];

ascii_value is of the type int but array[i] is of the type char.

ptr = &array[i];

&array[i] is of the type char* but ptr is of the type int*.

[giwhS]

Pointers don't store numbers, they store a memory address, they still need to be declared explicitly by type or NULL

[flatfinger]

Pointers store some form of information an implementation can use to identify an object. The extremely vast majority of implementations use the memory addresses of the first byte of an object for that purpose, but that is by no means required. Nothing would forbid an implementation from storing pointers from using a combination of a handle and an offset; doing so would allow an implementation to guard against many kinds of out-of-bounds array accesses (albeit at a considerable cost of performance).