address of an array

[stmfc]

int myArray[10];

it is said that array names are address information.
what i want to ask is
does myArray mean exactly the same thing with &myArray [0]
if so, why sizeof(myArray ) is not equal to sizeof(&myArray [0]).

1 -what is the relation and difference between myArray and &myArray [0]
2- what is the relation and difference between (*myArray ) and myArray [0]


thanks in advance...

[JosAH]

int myArray[10];

it is said that array names are address information.
what i want to ask is
does myArray mean exactly the same thing with &myArray [0]
if so, why sizeof(myArray ) is not equal to sizeof(&myArray [0]).

1 -what is the relation and difference between myArray and &myArray [0]
2- what is the relation and difference between (*myArray ) and myArray [0]


thanks in advance...

In C (and C++) the name of the array boils down to the address of the first
element of that array. The sizeof() operator is a bit funny, i.e. it is evaluated
at compile time: when you do 'sizeof(array)' the compiler knows what you're
talking about, i.e. the size in bytes taken by the entire array. OTOH when you
do this: 'sizeof(&array[0])' all the compiler knows is that you've taken the
address of the first element of that array, which is a simple pointer. So the
result of both sizeof() examples will most likely be different: the size of your
array measured in bytes versus the size of a simple pointer.

kind regards,

Jos

[AdrianH]

int myArray[10];

it is said that array names are address information.
what i want to ask is
does myArray mean exactly the same thing with &myArray [0]
if so, why sizeof(myArray ) is not equal to sizeof(&myArray [0]).

1 -what is the relation and difference between myArray and &myArray [0]
2- what is the relation and difference between (*myArray ) and myArray [0]


thanks in advance...

An array can degrad to a pointer, but it is not one.

If you pass an array to something that takes a pointer (of the same type as in the array) it will degrade into a pointer. At that point a sizeof will show the size of the pointernot of the array (because it is checked at compile time, not runtime). Example:

void f(int *array)
{
  cout << "second element = " << *(array+1) << endl;
  cout << "second element = " << array[1] << endl;
  ++array;
  cout << "second element = " << *array << endl;
  cout << "second element = " << array[0] << endl;
}
 
int main()
{
  int array[4] = { 1,2,3,4 };
  f(array);
}
 

If you pass an array to something that looks like it is supposed to take an array using the empty brackets ('[]'), beware. This is actually considered a pointer, and a non const one at that. It is for readability only. The sizeof operator works the same way as if it were a pointer. Example:

void f(int array[])
{
  cout << "second element = " << *(array+1) << endl;
  cout << "second element = " << array[1] << endl;
  ++array;
  cout << "second element = " << *array << endl;
  cout << "second element = " << array[0] << endl;
}
 
int main()
{
  int array[4] = { 1,2,3,4 };
  f(array);
}
 

If you pass an array to something that takes a reference to an array, it needs to be the same size.

void f(int (&array)[4])
{
  cout << "second element = " << *(array+1) << endl;
  cout << "second element = " << array[1] << endl;
  // ++array;  // This is invalid
}
 
int main()
{
  int array[4] = { 1,2,3,4 };
  f(array);
}
 

Adrian