CODEWITHSUNDEEP
arrayscpointers
CaptainMagmelaton Cap
CaptainMagmelaton Cap

Reputation: 71

difference between an array name and a pointer

is the array name is actually like typing the address of the first element of the array ?

based on my understanding of the array name

the array name variable isn't a pointer that has it's own memory address and it's content is the address of the first element of the array

but it's a way of typing the address of first element in a plain english name

i will try now to explain it now visually

consider the following code it will be related to the image i'm about to show you

char[] text = "Hey";
char *ptr = text;

and now look at the image

enter image description here

the *ptr variable has it's own memory address and the content of the pointer is the memory address of H which is 0101

now the text variable doesn't have it's own memory address when we type text C interprets it to the memory address 0101

so basically text is 0101

am i right ?

Upvotes: 1

Views: 345

Answers (2)

Vlad from Moscow
Vlad from Moscow

Reputation: 310980

For starters this declaration

char[] text = "Hey";

is not a valid C construction.:) It seems you mean

char text[] = "Hey";

An array type and a pointer type are two different types. Consider the following demonstrative program.

#include <stdio.h>

int main(void) 
{
    char text[] = "Hello World!";
    char *ptr   = "Hello World!";
    
    printf( "sizeof( text ) = %zu\n", sizeof( text ) );
    printf( "sizeof( ptr ) = %zu\n", sizeof( ptr ) );
    
    return 0;
}

Its output might look like

sizeof( text ) = 13
sizeof( ptr ) = 8

Or if you will apply the address of operator & to the variable text like &text then the type of the expression will be char ( * )[13]. While applying the operator to the variable ptr like &ptr then the type of the expression will be char **.

Indeed according to the C Standard an array designator is implicitly converted in expressions (except the expressions mentioned above) to a pointer to its firs element.

So you may write

char text[] = "Hey";
char *ptr = text;

The values of the expressions &text and ptr (though they have different types) will be equal because they both point to the initial address of the extent of the memory occupied by the array.

The fact that an array designator is converted to a pointer to its first element is related to how the compiler adjusts a function parameter having an array type to the type of pointer to the array element type. That is for example these two function declarations declare the same one function

void f( char s[] );

and

void f( char *s );

Upvotes: 1

Sourav Ghosh
Sourav Ghosh

Reputation: 134326

Well, almost but not quite.

text is a variable of type char [4], it's an array of chars. Now, as per the spec, chapter 6.3.2.1/P3

Except when it is the operand of the sizeof operator, the _Alignof 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.

That means,

char * ptr = text;

is the same as writing

char * ptr = &text[0];

That said,

now the text variable doesn't have it's own memory address when we type text C interprets it to the memory address 0101

No, this is wrong. The variable text has an address (try doing &text, it'll be of type char (*) [4]). Now, since an array is a collection of same objects (member object type) in contiguous memory locations, the address of the array is the address of the first element of the array, so while printing the address of the array and the address of the first element of the array will turn out to be same - but remember, they certainly do differ in type.

Upvotes: 2

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