Asad Abbas <cyberstuden...@...> wrote:
>
> While dealing with functions in C++.
> Generally, the C++ convention is the call-by-value.
> An exception is where arrays are passed as arguments,
> Arrays are called by reference by default.
No, they aren't.
It's possible to pass by reference in C++ by using the
reference operator.
It is hence possible to pass an array by reference, but
only explicitly...
void foo(int (&array)[10])
If you try...
void foo(int array[10])
...the parameter will 'decay' and be treated as...
void foo(int *array)
This is _not_ the same thing as pass by reference. The array
will be treated as a pointer to an element. The pointer is
passed by value.
Even though it _looks_ like pass by reference, a further
difference can be seen in the loss of dimension. Observe...
#include <iostream>
using namespace std;
void foo(int a[5])
{
for (int i = 0; i < 3; i++)
cout << a[i] << '\n';
}
int main()
{
int a[3] = { 1, 2, 4 };
foo(a);
}
The program above should compile and run just fine because
the declaration of foo is treated exactly as if it were...
void foo(int *a)
But change the declaration to use a reference...
void foo(int (&a)[5])
...and you should see an error.
> Is ther any way to pass the arrays to the functions by
> value rather than reference?
People have mentioned wrapping in structs, but that is
not passing an array, that is passing a struct. There is
no way to pass an array by value in C++ (or C.)
--
Peter
