Dynamic Array Allocation function

P4R4N01D · 05-15-2009, 12:25 AM

I have written a function to read in a variable number of ints into an array, & after this is called I print out the values in the array that were read in in main(). When i print out the contents of the array however, most of f the elements are garbage and bear no resemblance to what values I entered in. I think the array it is modifing is not modifing the size for the calling function but I am not sure how to fix it as the function also returns the number of elements read into the array. The array has expected values inside the read function and returns the correct value for n. Here is code:

Code:

#include <iostream>

using namespace std;

void read(int *a, int &n);
// read the first n elements of array a

int SIZE = 5;

int main()
{

    int *list1 = new int[SIZE];
    int *list2 = new int[SIZE];
    int noeles1=0, noeles2, test1, test2;

    read(list1, noeles1);

    read(list2, noeles2);

    cout << "List 1: " << endl;
    for (test1=0; test1<noeles1; test1++) {
        cout << list1[test1];
        cout << " ";
    }

    cout << "\nList 2: " << endl;
    for (test2=0; test2<noeles2; test2++) {
        cout << list2[test2];
        cout << " ";
    }

    return 0;
}

// write the body of following functions

void read(int *a, int &n)
{
	SIZE = 5;
	int k;
	cout << "Enter values:" << endl;
		n=0;
		do {
			if (n<SIZE) cin >> a[n];
			else {
				SIZE *= 2; //double array
				int *temp = new int[SIZE];
				for (k = 0; k<SIZE; k++) temp[k] = a[k];
				delete [] a;
				a = temp;
				cin >> a[n];
			}
		} while (a[n++] != -1);
	    n--;
}

**laserlight** · 05-15-2009, 12:30 AM

It looks like you forgot to pass the pointer by reference, so the assignment to a in read() merely changes the local pointer.

P4R4N01D · 05-15-2009, 12:43 AM

Thanks for the quick reply, I thought passing a pointer to int was passing by reference. Would changing the function to void read(int **a, int &n) pass by reference (as int &*a didn't work and int a would just be an int). If I change a to be passed as above & change every occurance of a to *a I get a runtime error: "the memory couldn't be written".

**laserlight** · 05-15-2009, 12:46 AM

Quote:

Originally Posted by P4R4N01D

Would changing the function to void read(int **a, int &n) pass by reference (as int &*a didn't work and int a would just be an int).

It would simulate pass by reference, but what I had in mind:

Code:

void read(int *&a, int &n)

Actually, to emphasize type, I would write it as:

Code:

void read(int*& a, int& n)

By the way, why are you not using std::vector<int> instead?

iMalc · 05-15-2009, 12:54 AM

Quote:

Originally Posted by P4R4N01D

Thanks for the quick reply, I thought passing a pointer to int was passing by reference. Would changing the function to void read(int **a, int &n) pass by reference (as int &*a didn't work and int a would just be an int). If I change a to be passed as above & change every occurance of a to *a I get a runtime error: "the memory couldn't be written".

That's because you tried to declare a pointer to a reference which is illegal. You need a reference to a pointer instead.

Types are read right to left (well there's more to it than that when you get more complex examples, but that's true enough for you, for now).

P4R4N01D · 05-15-2009, 12:56 AM

Quote:

Originally Posted by laserlight

By the way, why are you not using std::vector<int> instead?

As I have not learnt about vectors yet, I didn't even think about using them. Breifly looking at a reference on them I can see how they would be useful as an alternate method. Another reason why I used this approach was to modify an existing program that read in a fixed number of integers. Thanks for your help.

matsp · 05-15-2009, 02:04 AM

Can I also suggest that you modify the

Code:

			if (n<SIZE) cin >> a[n];
			else {
				SIZE *= 2; //double array
				int *temp = new int[SIZE];
				for (k = 0; k<SIZE; k++) temp[k] = a[k];
				delete [] a;
				a = temp;
				cin >> a[n];
			}

so that you only have one cin >> a[n] - there is no need to do that twice. Just change the condition to test FIRST if there is enough space, and then allocate more space if needed.

Also, you should not copy the NEW size from a[k], as that is outside the size of the original allocation, so you probably need another variable to hold the original (or new) size until you have updated the content.

Finally, I would suggest that you do not use a global variable for size - there is a bug in your current code because of the size for the frist array being changed, so that if you enter 100 values in the first set, the second set will overwrite the end of your allocation after the 5 first elements have been entered. You need to keep track of the size of each array, not have one variable for both!

--
Mats

05-15-2009, 12:25 AM	#1
P4R4N01D HelpingYouHelpUsHelpUsAll Join Date: Dec 2007 Location: In your nightmares Posts: 223	Dynamic Array Allocation function I have written a function to read in a variable number of ints into an array, & after this is called I print out the values in the array that were read in in main(). When i print out the contents of the array however, most of f the elements are garbage and bear no resemblance to what values I entered in. I think the array it is modifing is not modifing the size for the calling function but I am not sure how to fix it as the function also returns the number of elements read into the array. The array has expected values inside the read function and returns the correct value for n. Here is code: Code: #include <iostream> using namespace std; void read(int a, int &n); // read the first n elements of array a int SIZE = 5; int main() { int list1 = new int[SIZE]; int list2 = new int[SIZE]; int noeles1=0, noeles2, test1, test2; read(list1, noeles1); read(list2, noeles2); cout << "List 1: " << endl; for (test1=0; test1<noeles1; test1++) { cout << list1[test1]; cout << " "; } cout << "\nList 2: " << endl; for (test2=0; test2<noeles2; test2++) { cout << list2[test2]; cout << " "; } return 0; } // write the body of following functions void read(int a, int &n) { SIZE = 5; int k; cout << "Enter values:" << endl; n=0; do { if (n<SIZE) cin >> a[n]; else { SIZE = 2; //double array int temp = new int[SIZE]; for (k = 0; k<SIZE; k++) temp[k] = a[k]; delete [] a; a = temp; cin >> a[n]; } } while (a[n++] != -1); n--; } __________________ long time no C; //seige You miss 100% of the people you don't C; Code: if (language != LANG_C && language != LANG_CPP) drown(language); Last edited by P4R4N01D; 05-15-2009 at 12:30 AM. Reason: remove some debugging stuff
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05-15-2009, 12:30 AM	#2
laserlight C++ Witch Join Date: Oct 2003 Location: Singapore Posts: 11,338	It looks like you forgot to pass the pointer by reference, so the assignment to a in read() merely changes the local pointer. __________________ C + C++ Compiler: MinGW port of GCC Build + Version Control System: SCons + Bazaar Look up a C/C++ Reference and learn How To Ask Questions The Smart Way
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05-15-2009, 12:43 AM	#3
P4R4N01D HelpingYouHelpUsHelpUsAll Join Date: Dec 2007 Location: In your nightmares Posts: 223	Thanks for the quick reply, I thought passing a pointer to int was passing by reference. Would changing the function to void read(int *a, int &n) pass by reference (as int &a didn't work and int a would just be an int). If I change a to be passed as above & change every occurance of a to a I get a runtime error: "the memory couldn't be written". __________________ long time no C; //seige You miss 100% of the people you don't C; Code: if (language != LANG_C && language != LANG_CPP) drown(language); Last edited by P4R4N01D; 05-15-2009 at 12:44 AM. Reason: read -> written*
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05-15-2009, 02:04 AM	#7
matsp Kernel hacker Join Date: Jul 2007 Location: Farncombe, Surrey, England Posts: 15,686	Can I also suggest that you modify the Code: if (n<SIZE) cin >> a[n]; else { SIZE = 2; //double array int temp = new int[SIZE]; for (k = 0; k<SIZE; k++) temp[k] = a[k]; delete [] a; a = temp; cin >> a[n]; } so that you only have one cin >> a[n] - there is no need to do that twice. Just change the condition to test FIRST if there is enough space, and then allocate more space if needed. Also, you should not copy the NEW size from a[k], as that is outside the size of the original allocation, so you probably need another variable to hold the original (or new) size until you have updated the content. Finally, I would suggest that you do not use a global variable for size - there is a bug in your current code because of the size for the frist array being changed, so that if you enter 100 values in the first set, the second set will overwrite the end of your allocation after the 5 first elements have been entered. You need to keep track of the size of each array, not have one variable for both! -- Mats __________________ Compilers can produce warnings - make the compiler programmers happy: Use them! Please don't PM me for help - and no, I don't do help over instant messengers.
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