character pointer question (should be easy, but I'm new to this)

**Dynesclan** · 06-10-2012

I have a character pointer question.

Given a char pointer called ptr (i.e. char *ptr) and the alphabet array (const char alphabet[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ"), how do I make ptr point to the letter 'D' in the alphabet array (i.e. how to set ptr to memory address where 'D' is stored)?

The constant alphabet array I initialized to the string below looks like this (as shown above):

const char alphabet[] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

I assume that this is an easy question, but I am still a little lost at the moment.

**laserlight** · 06-10-2012

How would you get ptr to point to alphabet[0]?

**Dynesclan** · 06-10-2012

Oops...

const char alphabet [26] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

**laserlight** · 06-10-2012

Um, please answer my question. In this case, specifying the 26 makes no difference.

**Matticus** · 06-10-2012

While we wait for an answer to the question laserlight asked, I would like to point out one thing.

Putting the alphabet in double quotes makes it a string constant - meaning that the compiler will automatically append the string with a null character (\0). Therefore, the array size would need to be the number of characters in the string plus one for the null character.

Code:

char alphabet[27] = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";

You could get away with using 26 if you didn't want a string, but simply an array of chars. In that case, you would have to define each array element independently; something like:

Code:

char alphabet[26] = {'A','B','C', ... 'Z'};
/* just an example; each element would have to be assigned, I only used the
ellipsis because I didn't feel like typing out each letter */

**laserlight** · 06-10-2012

Originally Posted by Matticus

You could get away with using 26 if you didn't want a string, but simply an array of chars. In that case, you would have to define each array element independently; something like:

No, Dynesclan's line of code in post #3 initialises the array named alphabet such that it does not contain a null terminated string: the null character from the string literal is just ignored. So, you don't need to "define each array element independently", unless you want to be compatible with C++.

**Matticus** · 06-10-2012

I stand corrected. Thank you.

[edit] Considering the level of the OPs experience, based on previous posts, I thought this advice might help future learning pitfalls in using strings

**Dynesclan** · 06-10-2012

Sorry to keep you waiting for an answer. Honestly, I'm not sure how to set ptr to point to an individual letter in the array. At first I thought it would have to do with converting the letter to their ASCII values, but dismissed the idea because I thought it didn't relate to the memory addresses.

**laserlight** · 06-10-2012

Well, you would take the address of that character, so it could be:

Code:

ptr = &alphabet[0];

Alternatively, since an array is converted to a pointer to its first element, you could write:

Code:

ptr = alphabet;

**Dynesclan** · 06-10-2012

Are you saying that the [0] points to 'A', at the start of the array? Does that mean that since 'D' is 3 characters away from 'A', the ptr would use [3]?

For example:
ptr = &alphabet [3];

**laserlight** · 06-10-2012

Originally Posted by Dynesclan

Are you saying that the [0] points to 'A', at the start of the array?

Not quite: the index of the first element of the array is 0.

Originally Posted by Dynesclan

Does that mean that since 'D' is 3 characters away from 'A', the ptr would use [3]?

The index of the element of the array with the value of 'D' is 3.

This distinction is necessary when you see that I use the address of operator & to get the pointer.

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