array name and pointer name confusion

array name and pointer name confusion - c

In C language, an Array is a collection of same data type, that is when we are defining an array like this :
int my_list[50];
we are allocating 50 consecutive location in the computer memory to store 50 integer.
when somebody says my array is "my_list", what does it mean? does it collectively denotes all the 50 spaces that I have defined in the memory?
if that is the case , when I'm trying to learn pointers, they say, no no, there is a pointer whose name is "my_list" and it points to my_list[0];
my conclusions are like this:
"my_list" does denote all the 50 memory location collectively, but when we are defining the array , the machine is generating a pointer whose name is "accidentally" "my_list" and it points to my_list[0];
array my_list denotes all the 50 location collectively and pointer my_list only points to my_list[0], though their name overlaps they are totally different things.
please correct me if i'm wrong.

does it collectively denotes all the 50 spaces that I have defined in the memory?
Yes, my_list is the array.
… they say, no no, there is a pointer whose name is "my_list" and it points to my_list[0];
“They” are wrong. There is no pointer named my_list.
However, when an array is used in an expression, it is automatically converted to a pointer to its first element except when it is the operand of sizeof, is the operand of unary &, or is a string literal used to initialize an array.
Thus, if you, for example, use my_list like a pointer, as in int *p = my_list + 5;, or you pass it to a function, as in foo(50, my_list);, then it is automatically converted to a pointer, as if you had written int *p = &my_list[0] + 5; or foo(50, &my_list[0]);.
For the exceptions: With sizeof, sizeof my_list gives the size of the whole array, the size of 50 int. There is no automatic conversion to a pointer, so sizeof my_list does not give the size of a pointer. And with unary &, &my_list is a pointer with type “pointer to array of 50 int” rather than a pointer with type “pointer to pointer to int”.
(An array you declare with a name, like my_list, is of course not a string literal. A string literal can be used to initialize an array, as in char MyString[] = "Hello";. In this case, the array formed by the string literal is used to initialize buffer; it is not converted to a pointer.)

Related

Different Types Of Initializing Character Arrays in C

I was learning about pointers and strings.
I understood that,
Pointers and Arrays/Strings have similar behaviours.
array[] , *array , &array[0]. They all are one and the same.
Why does the three statements in this code work, and char * help one does not ?
#include <stdio.h>
void display(char*help){
for(int i=0; help[i]!='\0'; i++){
printf("%c", help[i]);
}
}
int main(){
// char help[] = "Help_Me"; //Works
// char help[] = {'H','e','l','p','_','M','e','\0'}; //Works
// char *help = "Help_Me"; //Works
char *help = {'H','e','l','p','_','M','e','\0'}; //Error
display(help);
}
Error Messages :
warning: initialization of 'char *' from 'int' makes pointer from integer without a cast
warning: excess elements in scalar initializer

Pointers and Arrays/Strings have similar behaviours.
Actually, no, I wouldn't agree with that. It is an oversimplification that hides important details. The true situation is that arrays have almost no behaviors of their own, but in most contexts, an lvalue designating an array is automatically converted to pointer to the first array element. The resulting pointer behaves like a pointer, of course, which is what may present the appearance that pointers and arrays have similar behaviors.
Additionally, arrays are objects, whereas strings are certain configurations of data that char arrays can contain. Although people sometimes conflate strings with the arrays containing them or with pointers to their first elements, that is not formally correct.
array[] , *array , &array[0]. They all are one and the same.
No, not at all, though the differences depend on the context in which those appear:
In a declaration of array (other than in a function prototype),
type array[] declares array as an array of type whose size will be determined from its initializer;
type *array declares array as a pointer to type; and
&array[0] is not part of any valid declaration of array.
In a function prototype,
type array[] is "adjusted" automatically as if it were type *array, and it therefore declares array as a pointer to type;
type *array declares array as a pointer to type; and
&array[0] is not part of any valid declaration of array.
In an expression,
array[] is invalid;
*array is equivalent to array[0], which designates the first element of array; and
&array[0] is a pointer to array[0].
Now, you ask,
Why does the three statements in this code work, and char * help one does not ?
"Help_Me" is a string literal. It designates a statically-allocated array just large enough to contain the specified characters plus a string terminator. As an array-valued expression, in most contexts it is converted to a pointer to its first element, and such a pointer is of the correct type for use in ...
// char *help = "Help_Me"; //Works
But the appearance of a string literal as the initializer of a char array ...
// char help[] = "Help_Me"; //Works
... is one of the few contexts where an array value is not automatically converted to a pointer. In that context, the elements of the array designated by the string literal are used to initialize the the array being declared, very much like ...
// char help[] = {'H','e','l','p','_','M','e','\0'}; //Works
. There, {'H','e','l','p','_','M','e','\0'} is an array initializer specifying values for 8 array elements. Note well that taken as a whole, it is not itself a value, just a syntactic container for eight values of type int (in C) or char (in C++).
And that's why this ...
char *help = {'H','e','l','p','_','M','e','\0'}; //Error
... does not make sense. There, help is a scalar object, not an array or a structure, so it takes only one value. And that value is of type char *. The warnings delivered by your compiler are telling you that eight values have been presented instead of one, and they have, or at least the one used for the initialization has, type int instead of type char *.

array[] , *array , &array[0]. They all are one and the same.
No. Presuming array names some array, array[] cannot be used in an expression (except where it might appear in some type description, such as a cast).
array by itself in an expression is automatically converted to a pointer to its first element except when it is the operand of sizeof or the operand of unary &. (Also, a string literal, such as "abc", denotes an array, and this array has another exception to when it is converted: When it is used to initialize an array.)
In *array, array will be automatically converted to a pointer, and then * refers to the element it points to. Thus *array refers to an element in an array; it is not a pointer to the array or its elements.
In &array[0], array[0] refers to the first element of the array, and then & takes its address, so &array[0] is a pointer to the first element of the array. This makes it equivalent to array in expressions, with the exceptions noted above. For example, void *p = array; and void *p = &array[0]; will initialize p to the same thing, a pointer to the first element of the array, because of the automatic conversion. However, size_t s = sizeof array; and size_t s = sizeof &array[0]; may initialize s to different values—the first to the size of the entire array and the second to the size of a pointer.
// char help[] = "Help_Me"; //Works
help is an array of char, and character arrays can be initialized with a string literal. This is a special rule for initializations.
// char help[] = {'H','e','l','p','_','M','e','\0'}; //Works
help is an array, and the initializer is a list of values for the elements of the array.
// char *help = "Help_Me"; //Works
help is a pointer, and "Help_Me" is a string literal. Because it is not in one of the exceptions—operand of sizeof, operand of unary &, or used to initialize an array—it is automatically converted to a pointer to its first element. Then help is initialized with that pointer value.
char *help = {'H','e','l','p','_','M','e','\0'}; //Error
help is a pointer, but the initializer is a list of values. There is only one thing to be initialized, a pointer, but there are multiple values listed for it, so that is an error. Also, a pointer should be initialized with a pointer value (an address or a null pointer constant), but the items in that list are integers. (Character literals are integers; their values are the codes for the characters.)
{'H','e','l','p','_','M','e','\0'} is not a syntax that creates a string or an array. It is a syntax that can be used to provide a list of values when initializing an object. So the compiler does not recognize it as a string or array and does not use it to initialize the pointer help.

Pointer is not the array and it cant be initialized like an array. You need to create an object, then you can assign its reference to the pointer.
char *help = (char[]){'H','e','l','p','_','M','e','\0'};

Difference between &ar[0][0] and ar [duplicate]

I am having a tough time understanding the type and use of the name of the array in C. It might seems a long post but please bear with me.
I understand that the following statement declares a to be of type int [] i.e array of integers.
int a[30];
While a also points the first element of array and things like *(a+2) are valid. Thus, making a look like a pointer to an integer. But actually the types int [] and int* are different; while the former is an array type and later is a pointer to an integer.
Also a variable of type int [] gets converted into a variable of type int* when passing it to functions; as in C arrays are passed by reference (with the exception of the sizeof operator).
Here comes the point which makes me dangle. Have a look at the following piece of code:
int main()
{
int (*p)[3];
int a[3] = { 5, 4, 6 };
p = &a;
printf("a:%d\t&a:%d\n",a,&a);
printf("%d",*(*p + 2));
}
OUTPUT:
a:2686720 &a:2686720
6
So, how does the above code work? I have two questions:
a and &a have the same values. Why?
What exactly does int (*p)[3]; do? It declares a pointer to an array, I know this. But how is a pointer to an array different from the pointer to the first element of the array and name of the array?
Can anyone clarify things up? I am having a hell of a lot of confusions.
I know that I should use %p as a placeholder instead of using %d for printing the value of pointer variables. As using the integer placeholder might print truncated addresses. But I just want to keep things simple.

Other answers already explained the issue. I am trying to explain it with some diagram. Hope this will help.
When you declare an array
int a[3] = {5, 4, 6}
the memory arrangement looks like
Now answering your question:
a and &a have the same values.How?
As you already know that a is of array type and array name a becomes a pointer to first element of array a (after decay),i.e it points to the address 0x100. Note that 0x100 also is the starting address of the memory block (array a). And you should know that, in general, the address of the first byte is said to be the address of the variable. That is, if a variable is of 100 bytes, then its address is equal to the address of its first byte.
&a is address of the entire memory block, i.e it is an address of array a. See the diagram:
Now you can understand why a and &a both have same address value although both are of different type.
What exactly it does int (*p)[3]; Declares a pointer to an array,i know this.But,how a pointer to an array is different from the pointer to the first element of the array and name of the array?
See the above figure, it is explained clearly how pointer to an array is different from the pointer to an array element.
When you assign &a to p, then p points to the entire array having starting address 0x100.
NOTE: Regarding to the line
... as in C arrays are passed by references (with exception of sizeof function).
In C, arguments are passed by value. No pass by reference in C. When an ordinary variable is passed to a function, its value is copied; any changes to corresponding parameter do not affect the variable.
Arrays are also passed by value, but difference is that the array name decays to pointer to first element and this pointer assigned to the parameter (here, pointer value is copied) of the function; the array itself isn't copied.
In contrast to ordinary variable, an array used as an argument is not protected against any change, since no copy is made of the array itself, instead copy of pointer to first element is made.
You should also note that sizeof is not a function and array name does not act as an argument in this case. sizeof is an operator and array name serves as an operand. Same holds true when array name is an operand of the unary & operator.

a and &a have the same values.How?
They have the same value but different types. Array objects have no padding between elements (before or after) so the address of the array and the address of the first element of the array are the same.
That is:
(void *) a == (void *) &a
What exactly it does int (*p)[3]; Declares a pointer to an array,i know this.But,how a pointer to an array is different from the pointer to the first element of the array and name of the array?
These are two different pointer types. Take for example, pointer arithmetic:
a + 1 /* address of the second element of the array */
&a + 1 /* address one past the last element of the array */
EDIT: due to popular demand I added below some information about conversion of arrays.
With three exceptions, in an expression an object of type array of T is converted to a value of type pointer to T pointing to the first element of the array. The exceptions are if the object is the operand of sizeof or & unary operator or if the object is a string literal initializing an array.
For example this statement:
printf("a:%d\t&a:%d\n", a, &a);
is actually equivalent to:
printf("a:%d\t&a:%d\n", &a[0], &a);
Also please note that d conversion specifier can only be use to print a signed integer; to print a pointer value you have to use p specifier (and the argument must be void *). So to do things correctly use:
printf("a:%p\t&a:%p\n", (void *) a, (void *) &a);
respectively:
printf("a:%p\t&a:%p\n", (void *) &a[0], (void *) &a);

a corresponds to the pointer pointing at 0th element of the array. Whereas,the same is the case with &a.It just gives the starting address of the array.
As,a --> pointer pointing to starting element of array a[],it does not know about other element's location..
&a --->address location for storing array a[] which stores first element location,but knows every element's location.
Similarly,other elements location will be (a+2),(a+4) and so upto the end of the array.
Hence,you got such result.
int (*p)[3] is a pointer to the array. had it been int *p[3],it would been meant entirely different. It'd have meant an array of pointers which would have been totally different from this context.
Pointer to an array will automatically take care of all the other
elements in the array.In this case,your's is (p);
Whereas,the pointer to the first element of the array,i.e., a will
only know about first element of the array.You'll have to manually
give pointer arithmetic directions to access next elements.See,in this
case---we can get second element from a by adding 2 to a,i.e.
a+2,third element by adding 4 to a,i.e., a+4 and so on. // mind the
difference of two as it is an integer array!

In answer to question 1, this is simply an aspect of the C language as designed, unlike most other modern languages C/C++ allows direct manipulation of addresses in memory and has built in facilities to 'understand' that. There are many articles online that explain this better than I could in this small space. Here is one and I am sure there are many others: http://www.cprogramming.com/tutorial/c/lesson8.html

From C99 Standard n1124 6.3.2.1 p3
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.

a and &a have the same value because a long time ago you were required to use the address operator & on arrays to get the array's address, but it is no longer necessary. The name of the array (a in this case) these days just represents the memory address of the array itself, which is also what you get from &a. It's a shorthand that the compiler handles for you.

Can a dereferenced pointer point to the same address as the pointer itself? [duplicate]

I am having a tough time understanding the type and use of the name of the array in C. It might seems a long post but please bear with me.
I understand that the following statement declares a to be of type int [] i.e array of integers.
int a[30];
While a also points the first element of array and things like *(a+2) are valid. Thus, making a look like a pointer to an integer. But actually the types int [] and int* are different; while the former is an array type and later is a pointer to an integer.
Also a variable of type int [] gets converted into a variable of type int* when passing it to functions; as in C arrays are passed by reference (with the exception of the sizeof operator).
Here comes the point which makes me dangle. Have a look at the following piece of code:
int main()
{
int (*p)[3];
int a[3] = { 5, 4, 6 };
p = &a;
printf("a:%d\t&a:%d\n",a,&a);
printf("%d",*(*p + 2));
}
OUTPUT:
a:2686720 &a:2686720
6
So, how does the above code work? I have two questions:
a and &a have the same values. Why?
What exactly does int (*p)[3]; do? It declares a pointer to an array, I know this. But how is a pointer to an array different from the pointer to the first element of the array and name of the array?
Can anyone clarify things up? I am having a hell of a lot of confusions.
I know that I should use %p as a placeholder instead of using %d for printing the value of pointer variables. As using the integer placeholder might print truncated addresses. But I just want to keep things simple.

a and &a have the same values.How?
They have the same value but different types. Array objects have no padding between elements (before or after) so the address of the array and the address of the first element of the array are the same.
That is:
(void *) a == (void *) &a
What exactly it does int (*p)[3]; Declares a pointer to an array,i know this.But,how a pointer to an array is different from the pointer to the first element of the array and name of the array?
These are two different pointer types. Take for example, pointer arithmetic:
a + 1 /* address of the second element of the array */
&a + 1 /* address one past the last element of the array */
EDIT: due to popular demand I added below some information about conversion of arrays.
With three exceptions, in an expression an object of type array of T is converted to a value of type pointer to T pointing to the first element of the array. The exceptions are if the object is the operand of sizeof or & unary operator or if the object is a string literal initializing an array.
For example this statement:
printf("a:%d\t&a:%d\n", a, &a);
is actually equivalent to:
printf("a:%d\t&a:%d\n", &a[0], &a);
Also please note that d conversion specifier can only be use to print a signed integer; to print a pointer value you have to use p specifier (and the argument must be void *). So to do things correctly use:
printf("a:%p\t&a:%p\n", (void *) a, (void *) &a);
respectively:
printf("a:%p\t&a:%p\n", (void *) &a[0], (void *) &a);

a corresponds to the pointer pointing at 0th element of the array. Whereas,the same is the case with &a.It just gives the starting address of the array.
As,a --> pointer pointing to starting element of array a[],it does not know about other element's location..
&a --->address location for storing array a[] which stores first element location,but knows every element's location.
Similarly,other elements location will be (a+2),(a+4) and so upto the end of the array.
Hence,you got such result.
int (*p)[3] is a pointer to the array. had it been int *p[3],it would been meant entirely different. It'd have meant an array of pointers which would have been totally different from this context.
Pointer to an array will automatically take care of all the other
elements in the array.In this case,your's is (p);
Whereas,the pointer to the first element of the array,i.e., a will
only know about first element of the array.You'll have to manually
give pointer arithmetic directions to access next elements.See,in this
case---we can get second element from a by adding 2 to a,i.e.
a+2,third element by adding 4 to a,i.e., a+4 and so on. // mind the
difference of two as it is an integer array!

In answer to question 1, this is simply an aspect of the C language as designed, unlike most other modern languages C/C++ allows direct manipulation of addresses in memory and has built in facilities to 'understand' that. There are many articles online that explain this better than I could in this small space. Here is one and I am sure there are many others: http://www.cprogramming.com/tutorial/c/lesson8.html

From C99 Standard n1124 6.3.2.1 p3
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.

a and &a have the same value because a long time ago you were required to use the address operator & on arrays to get the array's address, but it is no longer necessary. The name of the array (a in this case) these days just represents the memory address of the array itself, which is also what you get from &a. It's a shorthand that the compiler handles for you.

Pointer and array 'a' and '&a' giving same output? [duplicate]

I am having a tough time understanding the type and use of the name of the array in C. It might seems a long post but please bear with me.
I understand that the following statement declares a to be of type int [] i.e array of integers.
int a[30];
While a also points the first element of array and things like *(a+2) are valid. Thus, making a look like a pointer to an integer. But actually the types int [] and int* are different; while the former is an array type and later is a pointer to an integer.
Also a variable of type int [] gets converted into a variable of type int* when passing it to functions; as in C arrays are passed by reference (with the exception of the sizeof operator).
Here comes the point which makes me dangle. Have a look at the following piece of code:
int main()
{
int (*p)[3];
int a[3] = { 5, 4, 6 };
p = &a;
printf("a:%d\t&a:%d\n",a,&a);
printf("%d",*(*p + 2));
}
OUTPUT:
a:2686720 &a:2686720
6
So, how does the above code work? I have two questions:
a and &a have the same values. Why?
What exactly does int (*p)[3]; do? It declares a pointer to an array, I know this. But how is a pointer to an array different from the pointer to the first element of the array and name of the array?
Can anyone clarify things up? I am having a hell of a lot of confusions.
I know that I should use %p as a placeholder instead of using %d for printing the value of pointer variables. As using the integer placeholder might print truncated addresses. But I just want to keep things simple.

a and &a have the same values.How?
They have the same value but different types. Array objects have no padding between elements (before or after) so the address of the array and the address of the first element of the array are the same.
That is:
(void *) a == (void *) &a
What exactly it does int (*p)[3]; Declares a pointer to an array,i know this.But,how a pointer to an array is different from the pointer to the first element of the array and name of the array?
These are two different pointer types. Take for example, pointer arithmetic:
a + 1 /* address of the second element of the array */
&a + 1 /* address one past the last element of the array */
EDIT: due to popular demand I added below some information about conversion of arrays.
With three exceptions, in an expression an object of type array of T is converted to a value of type pointer to T pointing to the first element of the array. The exceptions are if the object is the operand of sizeof or & unary operator or if the object is a string literal initializing an array.
For example this statement:
printf("a:%d\t&a:%d\n", a, &a);
is actually equivalent to:
printf("a:%d\t&a:%d\n", &a[0], &a);
Also please note that d conversion specifier can only be use to print a signed integer; to print a pointer value you have to use p specifier (and the argument must be void *). So to do things correctly use:
printf("a:%p\t&a:%p\n", (void *) a, (void *) &a);
respectively:
printf("a:%p\t&a:%p\n", (void *) &a[0], (void *) &a);

a corresponds to the pointer pointing at 0th element of the array. Whereas,the same is the case with &a.It just gives the starting address of the array.
As,a --> pointer pointing to starting element of array a[],it does not know about other element's location..
&a --->address location for storing array a[] which stores first element location,but knows every element's location.
Similarly,other elements location will be (a+2),(a+4) and so upto the end of the array.
Hence,you got such result.
int (*p)[3] is a pointer to the array. had it been int *p[3],it would been meant entirely different. It'd have meant an array of pointers which would have been totally different from this context.
Pointer to an array will automatically take care of all the other
elements in the array.In this case,your's is (p);
Whereas,the pointer to the first element of the array,i.e., a will
only know about first element of the array.You'll have to manually
give pointer arithmetic directions to access next elements.See,in this
case---we can get second element from a by adding 2 to a,i.e.
a+2,third element by adding 4 to a,i.e., a+4 and so on. // mind the
difference of two as it is an integer array!

In answer to question 1, this is simply an aspect of the C language as designed, unlike most other modern languages C/C++ allows direct manipulation of addresses in memory and has built in facilities to 'understand' that. There are many articles online that explain this better than I could in this small space. Here is one and I am sure there are many others: http://www.cprogramming.com/tutorial/c/lesson8.html

From C99 Standard n1124 6.3.2.1 p3
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.

a and &a have the same value because a long time ago you were required to use the address operator & on arrays to get the array's address, but it is no longer necessary. The name of the array (a in this case) these days just represents the memory address of the array itself, which is also what you get from &a. It's a shorthand that the compiler handles for you.

What exactly is the array name in c?

I am having a tough time understanding the type and use of the name of the array in C. It might seems a long post but please bear with me.
I understand that the following statement declares a to be of type int [] i.e array of integers.
int a[30];
While a also points the first element of array and things like *(a+2) are valid. Thus, making a look like a pointer to an integer. But actually the types int [] and int* are different; while the former is an array type and later is a pointer to an integer.
Also a variable of type int [] gets converted into a variable of type int* when passing it to functions; as in C arrays are passed by reference (with the exception of the sizeof operator).
Here comes the point which makes me dangle. Have a look at the following piece of code:
int main()
{
int (*p)[3];
int a[3] = { 5, 4, 6 };
p = &a;
printf("a:%d\t&a:%d\n",a,&a);
printf("%d",*(*p + 2));
}
OUTPUT:
a:2686720 &a:2686720
6
So, how does the above code work? I have two questions:
a and &a have the same values. Why?
What exactly does int (*p)[3]; do? It declares a pointer to an array, I know this. But how is a pointer to an array different from the pointer to the first element of the array and name of the array?
Can anyone clarify things up? I am having a hell of a lot of confusions.
I know that I should use %p as a placeholder instead of using %d for printing the value of pointer variables. As using the integer placeholder might print truncated addresses. But I just want to keep things simple.

a and &a have the same values.How?
They have the same value but different types. Array objects have no padding between elements (before or after) so the address of the array and the address of the first element of the array are the same.
That is:
(void *) a == (void *) &a
What exactly it does int (*p)[3]; Declares a pointer to an array,i know this.But,how a pointer to an array is different from the pointer to the first element of the array and name of the array?
These are two different pointer types. Take for example, pointer arithmetic:
a + 1 /* address of the second element of the array */
&a + 1 /* address one past the last element of the array */
EDIT: due to popular demand I added below some information about conversion of arrays.
With three exceptions, in an expression an object of type array of T is converted to a value of type pointer to T pointing to the first element of the array. The exceptions are if the object is the operand of sizeof or & unary operator or if the object is a string literal initializing an array.
For example this statement:
printf("a:%d\t&a:%d\n", a, &a);
is actually equivalent to:
printf("a:%d\t&a:%d\n", &a[0], &a);
Also please note that d conversion specifier can only be use to print a signed integer; to print a pointer value you have to use p specifier (and the argument must be void *). So to do things correctly use:
printf("a:%p\t&a:%p\n", (void *) a, (void *) &a);
respectively:
printf("a:%p\t&a:%p\n", (void *) &a[0], (void *) &a);

a corresponds to the pointer pointing at 0th element of the array. Whereas,the same is the case with &a.It just gives the starting address of the array.
As,a --> pointer pointing to starting element of array a[],it does not know about other element's location..
&a --->address location for storing array a[] which stores first element location,but knows every element's location.
Similarly,other elements location will be (a+2),(a+4) and so upto the end of the array.
Hence,you got such result.
int (*p)[3] is a pointer to the array. had it been int *p[3],it would been meant entirely different. It'd have meant an array of pointers which would have been totally different from this context.
Pointer to an array will automatically take care of all the other
elements in the array.In this case,your's is (p);
Whereas,the pointer to the first element of the array,i.e., a will
only know about first element of the array.You'll have to manually
give pointer arithmetic directions to access next elements.See,in this
case---we can get second element from a by adding 2 to a,i.e.
a+2,third element by adding 4 to a,i.e., a+4 and so on. // mind the
difference of two as it is an integer array!

In answer to question 1, this is simply an aspect of the C language as designed, unlike most other modern languages C/C++ allows direct manipulation of addresses in memory and has built in facilities to 'understand' that. There are many articles online that explain this better than I could in this small space. Here is one and I am sure there are many others: http://www.cprogramming.com/tutorial/c/lesson8.html

From C99 Standard n1124 6.3.2.1 p3
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.

a and &a have the same value because a long time ago you were required to use the address operator & on arrays to get the array's address, but it is no longer necessary. The name of the array (a in this case) these days just represents the memory address of the array itself, which is also what you get from &a. It's a shorthand that the compiler handles for you.