> But, to pass an array is ok in C++. Can you try the following code
please?
It is OK, but the language defines such passing is implemented in such a
way the pointer to the array is passed instead, not the array value,
which is what Uli described in his first reply, and which is evidenced
by your sample. You can see Foo1 modified the input, meaning the array
was not passed by its value, but rather by a pointer to its members -
actually instead of char input[12] the parameter is passed as char *input.
I think you could perhaps consider reading some good book about C++
language, where basics like this would be described?
Ondrej
George napsal(a):
Thanks for your clarification, Tim!
But, to pass an array is ok in C++. Can you try the following code please?
#include <iostream>
using namespace std;
void Foo1 (char input[12])
{
std::cout << typeid(input).name() << endl; // output char*
input [0] = '1';
return;
}
void Foo2 (char (&input) [12])
{
std::cout << typeid(input).name() << endl; // output char [12]
input [0] = '2';
return;
}
void Foo3 (char* input)
{
input [0] = '3';
return;
}
int main()
{
char buf[] = "Hello World";
Foo1 (buf);
cout << buf << endl; //output "1ello World"
Foo2 (buf);
cout << buf << endl; //output "2ello World"
Foo3 (buf);
cout << buf << endl; //output "3ello World"
return 0;
}
regards,
George
"Tim Roberts" wrote:
George <George@discussions.microsoft.com> wrote:
Why GetArrayLength(const T(&arr)[size]) works, but
GetArrayLength(const T(arr)[size]) -- I removed & does not work?
Ulrich already answered that question. You can't pass an array to a
function. You can pass is the address of an array, either by reference or
by pointer.
--
Tim Roberts, timr@probo.com
Providenza & Boekelheide, Inc.