Help with sizeof and pointers

[Metzen]

hello,
ok, I want to find the length of an int array that is being passed to a function:

main()
{
int array[]={1,2,3,4,5,6,7 ,8};
function(array) ;
}
function(int *array)
{
int arraylen = sizeof(*array)/sizeof(int);
}
arraylen should be 8 I get 1.

What am I doing Wrong?

Thanx

[Kevin Goodsell]

Metzen wrote:

hello,
ok, I want to find the length of an int array that is being passed to a function:
You can't pass an array to a function, at least not directly.

main()
int main()

There is no "implicit int" rule in C++, and there hasn't been for many
years.
{
int array[]={1,2,3,4,5,6,7 ,8};
function(array) ;
This passes a pointer to the function.
}
function(int *array)
'array' is a somewhat misleading name for a pointer.
{
int arraylen = sizeof(*array)/sizeof(int);
sizeof(*array) == sizeof(int), so of course you get 1 here. But no
matter what you do, you won't get the size of the memory pointed to by
'array' using sizeof.
}
arraylen should be 8 I get 1.

What am I doing Wrong?

You need to pass the size into the function.

-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.

[Steven C.]

You have two choices either end your array with a unque int or pass the
size.

main()
{
int array[]={1,2,3,4,5,6,7 ,8};
function (array, sizeof(array)/sizeof(int))
}

function(int *array, int size)
{
}

"Metzen" <an****@yahoo.c om> wrote in message
news:4e******** *************** ***@posting.goo gle.com...
hello,
ok, I want to find the length of an int array that is being passed to a
function:

main()
{
int array[]={1,2,3,4,5,6,7 ,8};
function(array) ;
}
function(int *array)
{
int arraylen = sizeof(*array)/sizeof(int);
}
arraylen should be 8 I get 1.

What am I doing Wrong?

Thanx

[Kevin Goodsell]

Steven C. wrote:

You have two choices

Please don't top-post. Re-read section 5 of the FAQ for posting guidelines.

http://www.parashift.com/c++-faq-lite/

-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.

[Jon Bell]

In article <4e************ **************@ posting.google. com>,
Metzen <an****@yahoo.c om> wrote:

hello,
ok, I want to find the length of an int array that is being passed to a
function:
You can't. You have to pass in the length as a separate
argument/parameter.
main()
{
int array[]={1,2,3,4,5,6,7 ,8};
function(array );
}
function(int *array)
{
int arraylen = sizeof(*array)/sizeof(int);
}
arraylen should be 8 I get 1.

What am I doing Wrong?

You're using arrays instead of vectors.

#include <vector>

using namespace std;

void function (vector<int>& array)) // functions in C++ must have
// return types
{
int arraylen = array.size();
}

int main () // according to the C++ standard, main() must return an int.
{
vector<int> array(8);
array[0] = 1;
array[1] = 2;
// etc.
function (array);
return 0;
}

--
Jon Bell <jt*******@pres by.edu> Presbyterian College
Dept. of Physics and Computer Science Clinton, South Carolina USA

[Julián Albo]

Metzen escribió:

ok, I want to find the length of an int array that is being passed to afunction:

main()
{
int array[]={1,2,3,4,5,6,7 ,8};
function(array) ;
}
function(int *array)
{
int arraylen = sizeof(*array)/sizeof(int);
}

arraylen should be 8 I get 1.

What am I doing Wrong?

You are using sizeof (*array), * array is equivalent in this case to
array [0], that is, an int, then you are dividing sizeof (int) by sizeof
(int).

You need something like sizeof (array) / sizeof (int), if what you
intend is obtain the number of elements in the array. Or better yet,
sizeof (array) / sizeof (* array), that will no require change if you
later change your mind and use an array of double, for example.

Regards.

[Big Brian]

> You need to pass the size into the function.
Or, if you can make some assumptions about the array, like what's the
last value in it ( like '\0' terminated c style char arrays ), you can
determine the length of the array.

-Brian

[Julián Albo]

Julián Albo escribió:

main()
{
int array[]={1,2,3,4,5,6,7 ,8};
function(array) ;
}
function(int *array)
{
int arraylen = sizeof(*array)/sizeof(int);
}

arraylen should be 8 I get 1.

What am I doing Wrong?

You are using sizeof (*array), * array is equivalent in this case to
array [0], that is, an int, then you are dividing sizeof (int) by sizeof
(int).

You need something like sizeof (array) / sizeof (int), if what you
intend is obtain the number of elements in the array. Or better yet,
sizeof (array) / sizeof (* array), that will no require change if you
later change your mind and use an array of double, for example.

Ops, I don't see you passed array to a function. In that case sizeof
array will give you the size of a pointer, not the size of the array.

Regards.

[Buster]

"Metzen" <an****@yahoo.c om> wrote

hello,
ok, I want to find the length of an int array that is being passed to a

function:

The following might work. I can't work out whether the standard says it does.

// Begin listing
#include <iostream>
#include <ostream>

#define MACRO(a) (sizeof (a) / sizeof (int))

template <typename T, unsigned N>
int function (T (&) [N]) { return N; }

int test (int a [])
{
// std::cout << function (a) << ", "; // compile-time error
std::cout << MACRO (a) << std::endl; // no error signaled
}

int main ()
{
int array [] = { 1, 2, 3, 4, 5, 6, 7, 8 };

std::cout << function (array) << ", ";
std::cout << MACRO (array) << std::endl;

test (array);
}
// End listing

I get what I expected on mingw32-g++ 3.2 and on bcb32:
the output is "8, 8"; "1". Both compilers report the compile
time error marked if I uncomment that line.

So the template version is slightly superior in that, although
it only works in a small set of situations, it will not compile
unless it can give you the correct answer.

It's probably only rarely that this can be used instead of the
techniques mentioned in the other replies, and even more
rarely that it will provide any significant advantage in speed
or code size. But maybe sometimes.

Yours,
Buster.

[Shane Beasley]

an****@yahoo.co m (Metzen) wrote in message news:<4e******* *************** ****@posting.go ogle.com>...

function(int *array)
{
int arraylen = sizeof(*array)/sizeof(int);
That's the size of a pointer divided by the size of the object to
which it points, the result of which is meaningless.
}
arraylen should be 8 I get 1.

What am I doing Wrong?

You're passing a pointer, not an array. Technically, in modern C++,
you can fix the problem by passing an array by reference like this:

#include <cstddef>

// the size of an array is most properly represented by std::size_t
template <std::size_t N>
void function (int (&array) [N]) {
std::size_t arraylen = N;
}

However, that most people don't write functions like this. Rather,
they just pass in the size themselves:

void function (int *array, std::size_t arraylen) { }
function(array, sizeof(array) / sizeof(array[0]));

Perhaps you can combine the approaches:

template <typename T, std::size_t N>
std::size_t size (T (&) [N]) { return N; }
template <typename T, std::size_t N>
std::size_t size (const T (&) [N]) { return N; }
function(array, size(array));

- Shane