How do I access an individual character from an array of strings in c?

Question

Just trying to understand how to address a single character in an array of strings. Also, this of course will allow me to understand pointers to pointers subscripting in general. If I have char **a and I want to reach the 3rd character of the 2nd string, does this work: **((a+1)+2)? Seems like it should...

Answer 1

Almost, but not quite. The correct answer is:

*((*(a+1))+2)

because you need to first de-reference to one of the actual string pointers and then you to de-reference that selected string pointer down to the desired character. (Note that I added extra parenthesis for clarity in the order of operations there).

Alternatively, this expression:

a[1][2]

will also work!....and perhaps would be preferred because the intent of what you are trying to do is more self evident and the notation itself is more succinct. This form may not be immediately obvious to people new to the language, but understand that the reason the array notation works is because in C, an array indexing operation is really just shorthand for the equivalent pointer operation. ie: *(a+x) is same as a[x]. So, by extending that logic to the original question, there are two separate pointer de-referencing operations cascaded together whereby the expression a[x][y] is equivalent to the general form of *((*(a+x))+y).

Answer 2

Try a[1][2]. Or *(*(a+1)+2).

Basically, array references are syntactic sugar for pointer dereferencing. a[2] is the same as a+2, and also the same as 2[a] (if you really want unreadable code). An array of strings is the same as a double pointer. So you can extract the second string using either a[1] or *(a+1). You can then find the third character in that string (call it 'b' for now) with either b[2] or *(b + 2). Substituting the original second string for 'b', we end up with either a[1][2] or *(*(a+1)+2).

Answer 3

Iirc, a string is actually an array of chars, so this should work:

a[1][2]

Answer 4

You don't have to use pointers.

int main(int argc, char **argv){

printf("The third character of argv[1] is [%c].\n", argv[1][2]);

}

Then:

$ ./main hello The third character of argv[1] is [l].

That's a one and an l.

You could use pointers if you want...

*(argv[1] +2)

or even

*((*(a+1))+2)

As someone pointed out above.

This is because array names are pointers.

Answer 5

Quote from the wikipedia article on C pointers -

In C, array indexing is formally defined in terms of pointer arithmetic; that is, the language specification requires that array[i] be equivalent to *(array + i). Thus in C, arrays can be thought of as pointers to consecutive areas of memory (with no gaps), and the syntax for accessing arrays is identical for that which can be used to dereference pointers. For example, an array array can be declared and used in the following manner:

int array[5];      /* Declares 5 contiguous (per Plauger Standard C 1992) integers */
int *ptr = array;  /* Arrays can be used as pointers */
ptr[0] = 1;        /* Pointers can be indexed with array syntax */
*(array + 1) = 2;  /* Arrays can be dereferenced with pointer syntax */

So, in response to your question - yes, pointers to pointers can be used as an array without any kind of other declaration at all!