New keyword and function scope

**maxsthekat** · 10-23-2010

Hi guys,

In the code below, the new keyword creates 5 integers for a pointer. However, whenever I am outside of the createmem() function, I cannot access this memory-- visual studio says that integers's memory address is 0x00000000). Why is this happening? Is there a good workaround for it?

Thanks!

-Max

Code:

#include <iostream>

using namespace std;

void createmem(int *myints);

void createmem(int *myints)
{
	myints = new int[5];
}


int main()
{
	int *integers = NULL;
	createmem(integers);

	delete[] integers;
}

**QuadraticFighte** · 10-23-2010

Code:

#include <iostream>

using namespace std;

void createmem(int *myints);

void createmem(int *myints)
{
        // 5 ints allocated, but main won't see it because myInts
        // is a copy of the int* in main.
	myints = new int[5];
}


int main()
{
	int *integers = NULL; // No space allocated
        // integers ptr copied in createmem, but integers won't actually point to valid memory after the call returns.
        // furthermore, you will leak memory in createmem because the copy myInts is lost when the calll returns
	createmem(integers); 


	delete[] integers;
}

That's not how it works. What you do is say

Code:

int* myArrayOfInts = new int[5];

If you want a function to return some ints, then you should have either

Code:

int* createmem() // Returning the array
{
	return new int[5];
}

or

Code:

void createmem(int** ptrToArrayOfInts) // Passing by reference from main
{
       *myints = new int[5];
}

Are you new to pointers?

**maxsthekat** · 10-23-2010

No, I'm not new to pointers. This is actually a much simpler version of what I'm doing... In another program, I have a function which needs to create a 2D array, a single array, and perform some operations on them (like clearing them out).

This doesn't really address my question, though. Why is this illegal?

Here's a more full code. inLine is a string pulled from a file, which represents a number of nodes and their edges, ex: 5 (1,2) (2,3) (3,4) represents 5 nodes, with 1<->2, 2<->3, and 3<->4.

The code then does other things (in other functions) with matrix, and nodeCode.

Code:

void fillMatrix(string inLine, int numNodes, int **matrix, int *nodeCode)
{
	stringstream tempStream;  //used for extracting formatted text from file
	int edgeStart;		//node where an edge begins
	int edgeEnd;		//node where an edge ends
	int edgeCount = 0;	//number of edges specified by user

	//begin by removing parenthesis and commas from data
	size_t found;		//used for finding unwanted formatting
	found = inLine.find_first_of("(),");

	while (found != string::npos)
	{
		if (inLine[found] == '(')
		{
			edgeCount++;		//count the number of edges we have from the user
		}
		inLine[found] = ' '; //replace with a space
		found = inLine.find_first_of("(),");
	}

	tempStream << inLine; //put the string onto the temporary stream
	
	//Data in tempStream is just numbers now, the first is the 
	//number of nodes
	tempStream >> numNodes;

	//create the matricies according to the size
	nodeCode = new int[numNodes];
	matrix = new int* [numNodes];

	for (int i = 0; i < numNodes; i++)
	{
		matrix[i] = new int[numNodes];
	}
	
	//clear matricies
	for (int i=0; i < numNodes; i++)
	{
		nodeCode[i] = UNVISITED;
		for (int j=0; j< numNodes; j++)
		{
			matrix[i][j] = UNCONNECTED;
		}
	}
	//Everything else specifies an edge
	
	for (int i=0; i < edgeCount; i++)
	{
		//get the edge node points
		tempStream >> edgeStart;
		tempStream >> edgeEnd;

		//Correct for offset (matrix starts at 0, nodes start at 1)
		edgeStart--; 
		edgeEnd--;

		//place them in the adjacency matrix
		matrix[edgeStart][edgeEnd] = CONNECTED;
		matrix[edgeEnd][edgeStart] = CONNECTED; //since this is an undirected graph
	}
	
}

EDIT: CONNECTED, UNCONNECTED, VISITED, and UNVISITED are all just enum values for better readability

**QuadraticFighte** · 10-23-2010

Because it's not the right way of actually initializing the memory. You lost a pointer to allocated memory in createmem. When the call returned, the pointer to the actual memory was not returned in main because the int* myInt is a value copied from main. The int* integers in main is still NULL. C++ is a call by value language. If you want to change the value of integers in main, you have to pass it by reference. Anyway, here's your leak.

Code:

==6483== HEAP SUMMARY:
==6483==     in use at exit: 20 bytes in 1 blocks
==6483==   total heap usage: 1 allocs, 0 frees, 20 bytes allocated
==6483== 
==6483== 20 bytes in 1 blocks are definitely lost in loss record 1 of 1
==6483==    at 0x4C285EB: operator new[](unsigned long) (vg_replace_malloc.c:305)
==6483==    by 0x40082D: main (foo.cpp:9)
==6483== 
==6483== LEAK SUMMARY:
==6483==    definitely lost: 20 bytes in 1 blocks
==6483==    indirectly lost: 0 bytes in 0 blocks
==6483==      possibly lost: 0 bytes in 0 blocks
==6483==    still reachable: 0 bytes in 0 blocks
==6483==         suppressed: 0 bytes in 0 blocks
==6483== 
==6483== For counts of detected and suppressed errors, rerun with: -v
==6483== ERROR SUMMARY: 1 errors from 1 contexts (suppressed: 4 from 4)

That's the line: myints = new int[5];

Also, if I try to dereference your pointer in main, we can see it clearly doesn't work:

Code:

#include <iostream>

using namespace std;

void createmem(int *myints);

void createmem(int *myints)
{
	myints = new int[5];
}


int main()
{
	int *integers = NULL;
	createmem(integers);
	integers[0] = 5; // Line I added

	delete[] integers;
}

Valgrind output:

Code:

==6506== Invalid write of size 4
==6506==    at 0x40082E: main (foo.cpp:17)
==6506==  Address 0x0 is not stack'd, malloc'd or (recently) free'd
==6506== 
==6506== 
==6506== Process terminating with default action of signal 11 (SIGSEGV)
==6506==  Access not within mapped region at address 0x0
==6506==    at 0x40082E: main (foo.cpp:17)
[...]
Segmentation fault

**maxsthekat** · 10-23-2010

Great. It doesn't work. I fully understand that. I even understand it's not the proper way of doing this. I still don't see what is the "proper" way of doing it.

Let me break down the problem for you:

1. You have two pointers, one is a pointer to a pointer.
2. You need to create memory for both pointers, inside of a function, and then perform operations on these inside of the main body of the program.

How would you go about that?

**QuadraticFighte** · 10-23-2010

Originally Posted by maxsthekat

Great. It doesn't work. I fully understand that. I even understand it's not the proper way of doing this. I still don't see what is the "proper" way of doing it.

Let me break down the problem for you:

1. You have two pointers, one is a pointer to a pointer.
2. You need to create memory for both pointers, inside of a function, and then perform operations on these inside of the main body of the program.

How would you go about that?

Look, I am not giving you attitude. I am taking time out to explain exactly why it didn't work. There is value in that, and if you can't see value in it... Oh dear...

I will tell you how to fix the basic example. You can extrapolate to your application

Code:

#include <iostream>

using namespace std;

void createmem(int **myints);

void createmem(int **myints) // We accept pointer to int*
{
	*myints = new int[5]; // Allocate the array and make it "visible" to main
}


int main()
{
	int *integers = NULL;
	createmem(&integers); // We correctly pass the reference of integers*
	integers[0] = 5; // Voila, now we have valid memory to use

	cout << integers[0] << endl;

	delete[] integers;
}

**Elysia** · 10-23-2010

How about:

Code:

#include <iostream>

void createmem(int* & myints);

void createmem(int* & myints) // We accept reference to int*
{
	myints = new int[5]; // Allocate the array and make it "visible" to main
}


int main()
{
	int* integers = nullptr;
	createmem(integers); // We correctly pass the reference of integers*
	integers[0] = 5; // Voila, now we have valid memory to use

	cout << integers[0] << endl;

	delete[] integers;
}

Also, QuadraticFighte, be careful with your terminology. You are not passing a reference to a pointer. You are passing the address of a pointer.
In C, they happen to be the same (although you can argue that it is incorrect), but in C++, they are not the same thing.

Furthermore, note that you do not have to do manual memory allocation for 2D array. You can simply use a std::vector for that.

Code:

std::vector<std::vector<int>> v(size_of_first_dimension, std::vector<int>(size_of_second_dimension));

Boost.Multiarray also provides some tools for easier creating of 2D arrays. You might check it out if you want.

**QuadraticFighte** · 10-23-2010

Originally Posted by Elysia

Code:

void createmem(int* & myints) // We accept pointer to int*
{
	myints = new int[5]; // Allocate the array and make it "visible" to main
}

Wow, that's super nifty. I never quite understood how the & worked in that context. So that assures that int*'s are passed by reference?

**Elysia** · 10-23-2010

Yes, everything to the left on the "&" is the type that is passed by reference, just like a pointer is a pointer to the type that's left of the last *.
int** -> pointer to int*.
int*& -> reference to int*.

**maxsthekat** · 10-23-2010

Thanks guys. I knew that if you wanted to change variables you could always do a pass by reference. I didn't realize there was a comparable situation for pointers

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