pointers arrays and functions

Question

0.00/5 (No votes)

See more:

This question requires a bit of theory that i don't have.
First off, an array is basically the pointer to the first element in the array definition, and depending on the index specified, it can reference all of the data sequentially after it. Correct?

C++

int array[5];

Which is why we can just declare a pointer, and then allocate an array of values to it afterwards like so:

C++

int * array;
array = new int[5];

So now, why is it that when I pass an int pointer like to a function, allocate the space there and then after the function has run try to access those values, I get an access violation error?

Here's an example:

C++

#include <iostream>

void function( int * Array)
{
     Array = new int[5];
     for(int i = 0; i < 5; i++)
         Array[i] = 0;           // works fine here
}

int main(int argc, char * argv[])
{
     int * Array = NULL;
     function(Array);
     Array[3] = 2;               // access violation here
     system("PAUSE");
     return 0;

}

Posted 1-Oct-11 11:57am

FatalCatharsis

Updated 2-Oct-11 8:15am

André Kraak

v4

Add a Solution

Comments

Philippe Mori 1-Oct-11 19:30pm

A much easier way is to uses STL vector instead.

André Kraak 2-Oct-11 14:16pm

Edited question:
Added pre tags
Formatted text/code
Spelling/Grammar

2 solutions

Solution 2

Lets write your code with a little bit different style:

#include <iostream>

void function( int* Array)
{
     Array = new int[5];
     for(int i = 0; i < 5; i++)
         Array[i] = 0;           // works fine here
}
 

int main(int argc, char* argv[])
{
     int* Array = NULL;
     function(Array);
     Array[3] = 2;               // access violation here
     system("PAUSE");
     return 0;
 
}

(just put the * close to int, so that it is clear that int* is a type name)
and now compare it to this one:

#include <iostream>

void function( int val)
{
   val = 5;
}
 

int main(int argc, char * argv[])
{
     int val =0;
     function(val);
     std::cout << val << std::endl;
     system("PAUSE");
     return 0;
 
}

Are you surprised in seeing "0" as a result, instead of 5?

Now, your function, int* array is a value of type "pointer to int", that -on function call- points to the same location as the main Array (NULL in this case) but, inside function you changed the pointer to point to something else (new int[5], in your case), than you assign values to the pointed array.
But inside your function there is nothing that makes main to now about the change you did: Array, in main is still pointing to NULL, and you try to write int NULL+2.

To solve the problem you should allocate in main, than call function, and inside the function, use the pointer to assign the values to the elements.
Or, pass a reference to the pointer (so that if you change it, the "original" is also changed:

#include <iostream>

void function( int*& Array) //note the &
{
     Array = new int[5];
     for(int i = 0; i < 5; i++)
         Array[i] = 0;           
}
 

int main(int argc, char * argv[])
{
     int * Array = NULL;
     function(Array);
     Array[3] = 2;               
     system("PAUSE");
     delete[]Array;      //required, or will result in a memory leak.        
     return 0;
 
}

Posted 2-Oct-11 6:29am

Emilio Garavaglia

Updated 2-Oct-11 6:30am

v2

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Simon Bang Terkildsen · Accepted Answer · 2011-10-01T12:04:00

That's because a pointer e.g. int* is essentially an integer (a location in memory) when you pass this pointer to your function you pass the integer of the pointer and not the address of the pointer.
So inorder to change the value of a pointer in your function you'll have to pass the address of the pointer and not the value it would look like this:

C++

#include <iostream>
void function(int ** Array)
{
     *Array = new int[5];
     for(int i = 0; i < 5; i++)
         Array[i] = 0;           
}


int main(int argc, char * argv[])
{
     int * Array = NULL;
     function(&Array);
     Array[3] = 2;               
     system("PAUSE");
     return 0;
}

I find it hard to explain this, so instead of sitting here and rewrite this 5 time, I'll just let you ask any questions you have regarding my explanation or the updated source code I posted, and I'll try to go into more detail.

Just found this article Pointer to Pointer and Reference to Pointer[^] I haven't read it myself but it seems to go over double pointers in more detail. In any case it doesn't hurt to have multiple sources of information :)