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Passing an array of chars to a function

jr
Sorry for this very dumb question, but I've clearly got a long way to go!
Can someone please help me pass an array into a function. Here's a starting
point.

void TheMainFunc()
{
// Body of code...

TCHAR myArray[512];
DoStuff(myArray );
}

void DoStuff(TCHAR theArray)
{
...
}
I can't quite get my head around what the variable "myArray" is - is it a
pointer to a memory address that would hold a TCHAR (a wide char if
UNICODE)?

Thanks

John

--
---
If you need to reply personally, append "text" to the domain name in my
email adr.
Thanks
Jul 22 '05 #1
58 10152
jr wrote:
Sorry for this very dumb question, but I've clearly got a long way to
go! Can someone please help me pass an array into a function.
You cannot pass arrays into functions. Usually, a pointer to the array's
first element is passed instead.
Here's a starting point.

void TheMainFunc()
{
// Body of code...

TCHAR myArray[512];
DoStuff(myArray );
}

void DoStuff(TCHAR theArray)
{
...
}
I can't quite get my head around what the variable "myArray" is - is
it a pointer to a memory address that would hold a TCHAR (a wide char
if UNICODE)?
myArray is an array of TCHAR. Your DoStuff function takes a parameter of
type TCHAR, i.e. _one_ character.

Try:
void DoStuff(TCHAR* theArray)


When you pass the array to DoStuff, it will be automatically converted
into a pointer to its first element.

Jul 22 '05 #2
On Mon, 19 Jan 2004 23:54:58 -0000 in comp.lang.c++, "jr"
<jo***@tele.co. uk> was alleged to have written:
TCHAR myArray[512];
DoStuff(myArray );
}

void DoStuff(TCHAR theArray)
{
void DoStuff(TCHAR * theArray)
{
I can't quite get my head around what the variable "myArray" is - is it a
pointer to a memory address that would hold a TCHAR (a wide char if
UNICODE)?


No, it's not a pointer, it's the actual storage for 256 instances of
TCHAR. But you often may refer to it as if it was a pointer, and C is
very ambivalent about that, converting the array variable name to a
pointer with little provocation where non-array types would require
operator& to get the pointer. C++ would probably prefer to consistently
require the operator&, but goes along with C for compatibility.

Jul 22 '05 #3
David Harmon wrote:
TCHAR myArray[512]; I can't quite get my head around what the variable "myArray" is - is
it a pointer to a memory address that would hold a TCHAR (a wide char
if UNICODE)?


No, it's not a pointer, it's the actual storage for 256 instances of
TCHAR.


Sure about that? ;-)

Jul 22 '05 #4

"Rolf Magnus" <ra******@t-online.de> wrote in message
news:bu******** *****@news.t-online.com...
jr wrote:
Sorry for this very dumb question, but I've clearly got a long way to
go! Can someone please help me pass an array into a function.


You cannot pass arrays into functions. Usually, a pointer to the array's
first element is passed instead.

Of course you pass arrays into functions:
#include <iostream>

void foo(char arr[]){
std::cout<< arr<<'\n';
arr[5]='!';
}

int main( )
{
char arr[10] = "hello";
foo(arr);
std::cout<< arr<<'\n';
}

Also if you alter the array inside the function the original is also
altered.

<snip>

HTH.
Jul 22 '05 #5
On Tue, 20 Jan 2004 01:55:11 +0100 in comp.lang.c++, Rolf Magnus
<ra******@t-online.de> was alleged to have written:
David Harmon wrote:
TCHAR myArray[512];I can't quite get my head around what the variable "myArray" is - is
it a pointer to a memory address that would hold a TCHAR (a wide char
if UNICODE)?


No, it's not a pointer, it's the actual storage for 256 instances of
TCHAR.


Sure about that? ;-)


Oopsie. I didn't think he really needed the full 512.

Jul 22 '05 #6
On Tue, 20 Jan 2004 01:34:21 -0000, "Jumbo"
<pcr1000011<nos pam>@uko2.co.uk > wrote in comp.lang.c++:

"Rolf Magnus" <ra******@t-online.de> wrote in message
news:bu******** *****@news.t-online.com...
jr wrote:
Sorry for this very dumb question, but I've clearly got a long way to
go! Can someone please help me pass an array into a function.


You cannot pass arrays into functions. Usually, a pointer to the array's
first element is passed instead.

Of course you pass arrays into functions:


No, you can't.

From the ISO C++ standard, "5.2.2 Function call", paragraph 7:

"The lvalue tor value (4.1), array to pointer (4.2), and function to
pointer (4.3) standard conversions are performed on the argument
expression."

The name of an array used as a function call argument is always
converted to a pointer to its first element.

But you can pass an array to a function if it is enclosed in a struct
or class and you pass an instance of the struct or class by value.

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.l earn.c-c++
http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html
Jul 22 '05 #7
Hi jr,
when u declare an array in c or c++, the variable name is a
pointer to the first element of the array. So u can treat the variable
name as a pointer. In your case myArray is a pointer to a TCHAR. (
similar to TCHAR * myArray ).
as far as your function signature goes, in order for it to compile
successfully it must be either of these 2 i guess..

void DoStuff( TCHAR * theArray )
{
}

or

void DoStuff( TCHAR theArray[] )
{
}

Hi All,
This is my first post and reply. So plz bear with my mistakes, if
any. But for jr, i hope there are no mistakes.

Thanks and Regards,
Arun Prakash. B

"jr" <jo***@tele.co. uk> wrote in message news:<bu******* ***@newsg3.svr. pol.co.uk>...
Sorry for this very dumb question, but I've clearly got a long way to go!
Can someone please help me pass an array into a function. Here's a starting
point.

void TheMainFunc()
{
// Body of code...

TCHAR myArray[512];
DoStuff(myArray );
}

void DoStuff(TCHAR theArray)
{
...
}
I can't quite get my head around what the variable "myArray" is - is it a
pointer to a memory address that would hold a TCHAR (a wide char if
UNICODE)?

Thanks

John

Jul 22 '05 #8

"Jack Klein" <ja*******@spam cop.net> wrote in message
news:kr******** *************** *********@4ax.c om...
On Tue, 20 Jan 2004 01:34:21 -0000, "Jumbo"
<pcr1000011<nos pam>@uko2.co.uk > wrote in comp.lang.c++:

"Rolf Magnus" <ra******@t-online.de> wrote in message
news:bu******** *****@news.t-online.com...
jr wrote:

> Sorry for this very dumb question, but I've clearly got a long way to > go! Can someone please help me pass an array into a function.

You cannot pass arrays into functions. Usually, a pointer to the array's first element is passed instead.

Of course you pass arrays into functions:


No, you can't.

From the ISO C++ standard, "5.2.2 Function call", paragraph 7:

"The lvalue tor value (4.1), array to pointer (4.2), and function to
pointer (4.3) standard conversions are performed on the argument
expression."

The name of an array used as a function call argument is always
converted to a pointer to its first element.

But you can pass an array to a function if it is enclosed in a struct
or class and you pass an instance of the struct or class by value.

Yip :-)
Jul 22 '05 #9
Jumbo wrote:

"Rolf Magnus" <ra******@t-online.de> wrote in message
news:bu******** *****@news.t-online.com...
jr wrote:
> Sorry for this very dumb question, but I've clearly got a long way
> to go! Can someone please help me pass an array into a function.
You cannot pass arrays into functions. Usually, a pointer to the
array's first element is passed instead.

Of course you pass arrays into functions:
#include <iostream>

void foo(char arr[]){


arr is not an array, but a pointer to its first element. You can see
that it's not an array, because it doesn't have a size between its [],
and arrays _always_ have a size. The only place where this is allowed
is if you define an array and initialize it, so the size can be
determined from the initializer, like:

char c[] = "Hello, world\n";
std::cout<< arr<<'\n';
arr[5]='!';
}

int main( )
{
char arr[10] = "hello";
foo(arr);
std::cout<< arr<<'\n';
}

Also if you alter the array inside the function the original is also
altered.


Yes, that's because you didn't pass the array. If you had done that, the
function would have a copy of it and the original couldn't be modified
by that function.

Jul 22 '05 #10

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