Programming Forums > Application Development > C++

bl00dninja · Sep 21st, 2006, 3:21 AM

alright, so you've got the recursive solution...it seems like you're having a clerical problem rather than a real logic problem. try using the dereferencing operator (*) to ...

never mind, just looked at it again, think about sizeof.

sizeof(array[1024]) = sizeof(array[1]) or array[whatever], just the size of the data element in bytes. messing with that is not going to help you. also the "sizeof" the array itself is only going to be the # of bytes required to hold a pointer to the array. quit fucking with sizeof. it won't tell you how many elements are there.

this discrepancy has led me to believe that you may have even incorrectly used array bracket notation in your original post as you apparently are unclear on some stuff. (not trying to be mean, i'm unclear on a lot of stuff)

maybe restate the question in plain english rather than (pseudo)code?

good luck man!

RemoteC2 · Sep 22nd, 2006, 8:09 PM

okay. i will restate my question in plain english: how can i pass arrays of any size to a function?

The Dark · Sep 22nd, 2006, 8:25 PM

You can just declare the parameter as

(Toggle Plain Text)

function (int array[])

however this does not tell the function anything about the size of the array.
Your function either needs to know the size of the array, or you could use some sentinel value in the array to tell it when to stop - this doesn't apply in your case, as you can't have a sentinel value as your array elements can contain any integer.
So you could pass the size in, e.g.

(Toggle Plain Text)

function (int array[], int size)

which is pretty staightforward. Then you just call the function like

(Toggle Plain Text)

function(myArray, sizeof myArray / sizeof myArray[0])

If you don't want to do that, then you could use the std::vector class instead of an array. The std::vector class has member functions that you can use to find the number of elements or to iterate through all the elements.

grumpy · Sep 22nd, 2006, 11:11 PM

Quote:

Originally Posted by The Dark

So you could pass the size in, e.g.

(Toggle Plain Text)

function (int array[], int size)

which is pretty staightforward. Then you just call the function like

(Toggle Plain Text)

function(myArray, sizeof myArray / sizeof myArray[0])

Everything else in DArk's post is good, but I have a problem with this particular bit of advice as it doesn't always work.

For example;

(Toggle Plain Text)

#include <iostream>

int function(int array[], int size)
{
     // print out the size and then the elements
     std::cout << "Size = " << size << '\n';
     for (int i = 0; i < size; ++i)
         std::cout << i << ' ' << array[i] << '\n';
}

int intermediary(int array[])
{
     function(array, sizeof array/sizeof array[0]);
}

int main()
{
     int x[] = {11,12,13,14,15};   // an array with 5 elements
     
     std::cout << "First try\n";
     function(x, sizeof x/sizeof x[0]);
     std::cout << "Second try\n";
     intermediary(x);
}

A superficial look at this code would suggest the first and second tries printing the array will achieve the same output, because the call of function() in main() is of the same form as the call of function() within intermediary(). The problem is that the function intermediary() only receives a pointer to the first element of the array, and receives no information about the size of the array.

Within main(), the expression "sizeof x/sizeof x[0]" will give the value 5 (as the compiler knows x is an array of 5 ints). That code works as intended.

However, the expression "sizeof array/sizeof array[0]" in intermediary() will not yield a value of 5 because array is received as a pointer, but comes with no information about the number of elements it is associated with. "sizeof array" therefore computes the size of the pointer. Let's assume we're on a 32 bit OS. The size of a pointer is typically 4 bytes and the size of an integer is often (not always though) 4 bytes on a 32 bit system. So "sizeof array/sizeof array[0]", on a lot of 32 bit systems will yield a value of 1.

Dameon · Sep 22nd, 2006, 11:51 PM

An interesting behavior, grumpy. Perhaps a good canidate for use in the Underhanded C Contest.

DaWei · Sep 23rd, 2006, 3:23 AM

"What is NOT a pointer"

bl00dninja · Sep 24th, 2006, 3:18 AM

[,],{,},-,=,_,+,<,>,!,%,&,*,(,)

yeah, they'll act gay. so does the problem logic. for every minute you spend trying to figure out why something doesn't work, you'll save thousands of hours in the future.

keep plugging away man!

Sep 21st, 2006, 3:21 AM	#11
bl00dninja Programming Guru Join Date: Oct 2004 Location: namespace std Posts: 1,246 Rep Power: 6	alright, so you've got the recursive solution...it seems like you're having a clerical problem rather than a real logic problem. try using the dereferencing operator (*) to ... never mind, just looked at it again, think about sizeof. sizeof(array[1024]) = sizeof(array[1]) or array[whatever], just the size of the data element in bytes. messing with that is not going to help you. also the "sizeof" the array itself is only going to be the # of bytes required to hold a pointer to the array. quit fucking with sizeof. it won't tell you how many elements are there. this discrepancy has led me to believe that you may have even incorrectly used array bracket notation in your original post as you apparently are unclear on some stuff. (not trying to be mean, i'm unclear on a lot of stuff) maybe restate the question in plain english rather than (pseudo)code? good luck man! __________________ i put on my robe and wizard hat... Have you ever heard of Plato, Aristotle, Socrates?...Morons.

Sep 22nd, 2006, 8:25 PM	#13
The Dark Expert Programmer Join Date: Jun 2005 Posts: 880 Rep Power: 4	You can just declare the parameter as (Toggle Plain Text) function (int array[]) function (int array[]) however this does not tell the function anything about the size of the array. Your function either needs to know the size of the array, or you could use some sentinel value in the array to tell it when to stop - this doesn't apply in your case, as you can't have a sentinel value as your array elements can contain any integer. So you could pass the size in, e.g. (Toggle Plain Text) function (int array[], int size) function (int array[], int size) which is pretty staightforward. Then you just call the function like (Toggle Plain Text) function(myArray, sizeof myArray / sizeof myArray[0]) function(myArray, sizeof myArray / sizeof myArray[0]) If you don't want to do that, then you could use the std::vector class instead of an array. The std::vector class has member functions that you can use to find the number of elements or to iterate through all the elements.

Sep 22nd, 2006, 11:51 PM	#15
Dameon Troll Join Date: Apr 2005 Location: Texas Posts: 732 Rep Power: 4	An interesting behavior, grumpy. Perhaps a good canidate for use in the Underhanded C Contest. __________________ MD5(sig) = bcef75433db02e9ad9bf81d6f7c5c270

Sep 23rd, 2006, 3:23 AM	#16
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	"What is NOT a pointer" __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Sep 24th, 2006, 3:18 AM	#17
bl00dninja Programming Guru Join Date: Oct 2004 Location: namespace std Posts: 1,246 Rep Power: 6	[,],{,},-,=,_,+,<,>,!,%,&,*,(,) yeah, they'll act gay. so does the problem logic. for every minute you spend trying to figure out why something doesn't work, you'll save thousands of hours in the future. keep plugging away man! __________________ i put on my robe and wizard hat... Have you ever heard of Plato, Aristotle, Socrates?...Morons.

Sep 22nd, 2006, 8:09 PM	#12
RemoteC2 Programmer Join Date: Jan 2006 Posts: 55 Rep Power: 3	okay. i will restate my question in plain english: how can i pass arrays of any size to a function?

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Programming Forums > Application Development > C++

stumped...