What is the difference between functions, which have reference to an array:
// reference to array
void f_(char (&t)[5]) {
auto t2 = t;
}
and simply array:
// just array
void f__(char t[5]) {
auto t2 = t;
}
as a parameters?
The calling code is:
char cArray[] = "TEST";
f_(cArray);
f__(cArray);
char (&rcArr)[5] = cArray;
f_(rcArr);
f__(rcArr);
In both cases t2 is char*, but in first function my VS2019 is showing that t inside function has type char(&t)[] and t inside second function has type char*.
So after all, is there any practical difference between those functions?
Arrays can be passed by reference OR by degrading to a pointer. For example, using char arr[1]; foo(char arr[]). , arr degrades to a pointer; while using char arr[1]; foo(char (&arr)[1]) , arr is passed as a reference. It's notable that the former form is often regarded as ill-formed since the dimension is lost.
In this tutorial, we will learn how to pass a single-dimensional and multidimensional array as a function parameter in C++ with the help of examples. In C++, we can pass arrays as an argument to a function. And, also we can return arrays from a function.
In the Call by Value method, there is no modification in the original value. In the Call by Reference method, there is a modification in the original value. In the case of Call by Value, when we pass the value of the parameter during the calling of the function, it copies them to the function's actual local argument.
No difference. At all. In a function parameter list, int arr[] is nothing but an "alternative" writing for int* arr .
Reference to an array means aliasing an array while retaining its identity. Reference to an array will not be an int* but an int []. Let us discuss this in detail by discussing the difference between these two.
The object can be extended and the changes will be reflected in the whole prototype chain for other objects. Arrays are objects only in javascript. The major difference is that they store the data in an ordered collection in which the data can be accessed using a numerical index. They are also mutable and data can be modified at any index.
Reference to an array will not be an int* but an int []. Let us discuss this in detail by discussing the difference between these two. This is quite weird that int [] is the same as int* but still compiler perspective on both is entirely different.
For compiler, a and b are the same data type i.e int* here for compiler they are just int* pointing to address. But for compiler type of as an array is int [2] and type of b as an array is int [3] which are completely different from each other.
You can specify a complete array type parameter as for example
void f( int ( &a )[N] );
and within the function you will know the number of elements in the passed array.
When the function is declared like
void f( int a[] );
then the compiler adjusts the function declaration like
void f( int *a );
and you are unable to determine the number of elements in the passed array. So you need to specify a second parameter like
void f( int *a, size_t n );
Also functions with a referenced array parameter type may be overloaded. For example these two declarations
void f( int ( &a )[] );
and
void f( int ( &a )[2] );
declare two different functions.
And functions with a referenced array parameter type may be called with a braced list (provided that the corresponding parameter has the qualifier const) like for example
f( { 1, 2, 3 } );
Here is a demonstration program
#include <iostream>
void f( const int ( &a )[] )
{
std::cout << "void f( const int ( & )[] ) called.\n";
}
void f( const int ( &a )[2] )
{
std::cout << "void f( const int ( & )[2] ) called.\n";
}
void f( const int a[] )
{
std::cout << "void f( const int [] ) called.\n";
}
int main()
{
f( { 1, 2, 3 } );
}
The program output is
void f( const int ( & )[] ) called.
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