Thread: Program to demonstrate strchr function

Originally Posted by november1992

I was wondering why the find_c variable always changes when I run the program.

find_c is a pointer. You probably wanted to write *find_c for printing. To be on the safe side, you should check that find_c is not a null pointer.

Originally Posted by november1992

"i" is the difference of the location of the character it's looking for (find_c) and string2, but what what is the value of string2? I was thinking it is the first index of the array because I remember my professor saying that the name of an array points to the first element.

An array is converted to a pointer to its first element.

Originally Posted by laserlight

An array is converted to a pointer to its first element.

Um...while I understand you're trying to be brief, this is incorrect and I believe this lack of elaboration is a root of the misconception that arrays are, or the same as, pointers.

To the OP, an array does is not converted to a pointer. Rather, you get the address of the start of the array when it is unsubscripted. You can assign the address of an array to a pointer, but not vice versa. An array is still a right hand data type.

Code:

    char *find_c;
 
    char string1[] = {"red"};
    char string2[] = {"the color red is good"};
 
find_c = string1;  // << is legal.
string1 = find_c; // << is not legal.

Originally Posted by Cynic

Um...while I understand you're trying to be brief, this is incorrect and I believe this lack of elaboration is a root of the misconception that arrays are, or the same as, pointers.

Incorrect? Uh...

Originally Posted by C99 Clause 6.3.2.1 Paragraph 3

Except when it is the operand of the sizeof operator or the unary & operator, or is a string literal used to initialize an array, an expression that has type "array of type" is converted to an expression with type "pointer to type" that points to the initial element of the array object and is not an lvalue. If the array object has register storage class, the behavior is undefined.

Originally Posted by Cynic

an array does is not converted to a pointer. Rather, you get the address of the start of the array when it is unsubscripted.

An address is (the value of) a pointer.

Originally Posted by Cynic

Then by all means, please show an example where you can assign a new address to an array.

You cannot, because the pointer thus obtained is not an lvalue.

Originally Posted by Cynic

Because it's not a pointer.

I quoted the standard to you. You're wrong. Period.

In the given context of november1992's question, the array string2 is converted to a pointer to its first element. This pointer is not an lvalue, but that's fine, because it can still be used in the pointer arithmetic to obtain the index value that is assigned to i.

EDIT:
To soften this a bit, I'll clarify by stating that you're wrong to claim that "an array (...) is not converted to a pointer". Otherwise, these statements are correct, or at least can be reasonably interpreted in a way that makes them correct:

Originally Posted by Cynic

Rather, you get the address of the start of the array when it is unsubscripted. You can assign the address of an array to a pointer, but not vice versa. An array is still a right hand data type.

The "start of the array" is its first element. This address comes with a type, i.e., the type of a pointer to an element of the array, hence we get a pointer to the array's first element. "Get" means conversion, since an array is not an address. Denying the conversion from array type to pointer type thus contradicts your own statement.

It is certainly true that "you can assign the address of an array to a pointer". But here's where type comes into play: a pointer to the array and a pointer to the array's first element have the same value when we are only considering addresses, but they have different types. Since november1992's question is concerned with pointer arithmetic, the type of the pointer comes into play. Hence, just talking about addresses (as numeric values rather than pointers with the notion of type) is insufficient.

You said string2 is converted to a pointer to its first element, so what is that value?

The address of the first element will be the value of the pointer, of course. When you dereference you get the first element.

It might be helpful to imagine that memory is a long ruler. This is helpful in understanding how you get completely different addresses with pointer arithmetic, and it should help you visualize the difference between addresses (i.e. the value of any pointer).

And how does it find the position by subtracting from find_c?

One way that I think is easy to understand is in the standard.

Originally Posted by ISO/IEC 9899:TC2 6.5.6.9 footnote 89

Another way to approach pointer arithmetic is first to convert the pointer(s) to character pointer(s): In
this scheme the integer expression added to or subtracted from the converted pointer is first multiplied
by the size of the object originally pointed to, and the resulting pointer is converted back to the
original type. For pointer subtraction, the result of the difference between the character pointers is
similarly divided by the size of the object originally pointed to.

If you need to know the rules about such expressions, that is actually more what 6.5.6.9 is about.

Originally Posted by laserlight

An array is converted to a pointer to its first element.

To be fair, a clearer way of stating that might be "the name of an array decays to a pointer to it's first element" or some such. After all, the "concept" of a pointer is a variable that stores the address of another whereas the "concept" of an array is a contiguous sequence of variables, so saying things like "an array is converted to a pointer to its first element" will invariably lead to confusion, IMO.