Passing array of int is like passing reference?

[Assassinbeast]

Why is it passing by reference and not making copies?

void hmm(int x[])
{
   for(int i=0;i<4;++i)
	x[i]=20;
}

int main()
{
   int g[4];

   hmm(g);

   for(int i=0;i<4;++i)
	cout << g[i] << endl;

}

the result is 20, 20, 20, 20.... :S

[Disch]

Built-in arrays cannot be passed by value because they cannot be copied. Therefore they are always passed by reference.

If you want to pass by value, you will need to use a container class, like vector, or valarray.

[Peter87]

What is being passed is a pointer to the first element in the array. x is a pointer so you could have written line 1 as void hmm(int* x) with the exact same result.

[Assassinbeast]

ahh ok ty for reply

[vlad from moscow]

You are wrong! In your example the array is passed by value. To pass an array by reference you have to declare the function as

void hmm(int ( &x )[4])
{
   for(int i=0;i<4;++i)
	x[i]=20;
}

As in your original example the array is passed by value then a copy of the address of the first element is created.

It is semantically equivalent to the statement

int *x = new int[4];

that is a copy of the address of the first element of unnamed array is assigned to x.

Take into account that if you have an array as for example

int a[4];

then using the name of an array as for example

*a = 10;

means that name a is implicitly converted to the address of the first element of the array. The same mechanism works then the name is passed to a function but in this last case a copy of the address of the first element of the array is created. So changing this copy does not change the original value of the array name.