Is array name equivalent to a pointer??

[me_ansh]

Hello every one,
I have a confusion regarding arrays and pointers, My question is: "Is array name a pointer (rather a constant pointer)"?? I feel the answer is 'yes' because we can use the de-referencing operator(*) with the array name, Also, because array name is a constant pointer so we cannot assign a different address to it.... but the link http://c-faq.com/aryptr/aryptr2.html clearly says that arrays are not pointers....

[Narue]

>My question is: "Is array name a pointer (rather a constant pointer)"??
No, but the reality is similar. An array name is converted to a pointer to the first element most of the time. The official standard terminology is that an array name is converted to a pointer to the first element when used in a value context as opposed to an object context. A value context is when you're retrieving and subsequently using the value of an object, such as the right hand side of x = y; . An object context is when you're modifying or querying the object itself, not the value it contains. The left hand side of x = y; is one example.

People try to make sense of the array-to-pointer conversion rule by saying that the array is converted to a constant pointer, and that's why you can't assign to it or perform modifying arithmetic such as ++a . What really happens is that operators such as ++ see the operand in an object context, and those operators aren't defined for array types. The conversion is not made at all, so pretending that you're working with a constant pointer rather than an array is a fundamental misunderstanding.

There are a few places where an array can legally be used in an object context, which is why saying an array name is a pointer is wrong. Two of the common places an array is used in object context are:

1. As an operand to the address-of operator, converting to a pointer won't work because then the type would be all muxed up even though the actual address would likely be the same. When you say &a (where a is an array type), you want the result to be a pointer to an array, not a pointer to a pointer. If the conversion happened, this wouldn't work as expected.
2. As an operand to the sizeof operator, converting to a pointer would produce entirely the wrong result. You'd get the size of a pointer, not the size of the array, which is nearly always not what you expected. A common mistake is using sizeof on an "array" function parameter:

   void foo ( int a[] )
   {
     printf ( "%zd\n", sizeof a );
   }

   This will print the size of a pointer, not the size of the array, because the function arguments are passed in a value context. The array argument is converted to a pointer and the function expects a pointer parameter. This is also why the first dimension for an array parameter doesn't need to be specified (ie. int a[] vs. int a) and why the array notation and pointer notation are equivalent (eg. int a[] == int *a) as function parameters.
   

[me_ansh]

Suppose we take the example of following code:

char arr[] = "daniweb";
arr[2] = 'u';

In the second line above, neither we are retrieving and using the value (value context) nor we are modifying or querying the object itself(object context). Also we know that arr[2] = *(arr+2), that means the array is being converted to pointer here.... So which context is this...

[Narue]

>So which context is this...
You answered your own question, it's value context. Why? Because the subscript operator arr[2] is syntactic sugar for *(arr + 2). Clearly (arr + 2) is a subexpression which results in a value: the address offset by two elements. The indirection operator then takes that address value and retrieves the object that's required by the assignment operator.