[SOLVED] Using free() after malloc() in C.

amikoyan · 02-26-2023, 05:42 AM

The following function is from a program to add two polynomials.

Code:

int* add(int A[], int B[], int m, int n)
{
	
   
    int* sum = malloc(sizeof(int));  
 /* Initialize the sum polynomial */
    for (int i = 0; i < m; i++)
        sum[i] = A[i];
	
    /*Take every term of first polynomial */
    for (int i = 0; i < n; i++)
        sum[i] += B[i];
 
    return sum;
    */free() here??? */
}

I can not work out where to use free(). If I put free(sum) after return sum, and the program compiles and adds two polynomials. However splint tells me:

Code:

 
poly_add.c:26:10: Dead storage sum passed as out parameter to free: sum
  Memory is used after it has been released (either by passing as an only param
  or assigning to an only global). (Use -usereleased to inhibit warning)
   poly_add.c:25:12: Storage sum released
poly_add.c:26:5: Unreachable code: free(sum)
  This code will never be reached on any possible execution. (Use -unreachable
  to inhibit warning)

But freeing the memory before the function return does not seem correct.

ntubski · 02-26-2023, 07:12 AM

The caller of add() needs to call free(), because the caller will be using the malloc'd return value.

Furthermore, your allocation is too small, you should have

Code:

int* sum = malloc(max(n, m) * sizeof(int));

// where max something like
int max(int n, int m) { if (n > m) return n; else return m; }
// or
#define max(n, m) (((n) > (m))? (n) : (m))

pan64 · 02-26-2023, 07:53 AM

yes, the usual way is to malloc before calling this function and you need to pass this variable to the function too, so you will never need to malloc/free inside.
The other way is a bit different, you need to use and return a struct, not int *, in that case this will be solved automatically. And the struct should contain int[N]. where N is the maximal size of A and B.

amikoyan · 02-26-2023, 08:44 AM

Quote:

Originally Posted by ntubski

The caller of add() needs to call free(), because the caller will be using the malloc'd return value.

Furthermore, your allocation is too small, you should have

Code:

int* sum = malloc(max(n, m) * sizeof(int));

// where max something like
int max(int n, int m) { if (n > m) return n; else return m; }

Thank you ntubski, I corrected to this.

Code:

int max(int m, int n) { return (m > n) ? m : n; }
 
/*A[] represents coefficients of first polynomial
* B[] represents coefficients of second polynomial
* m and n are sizes of A[] and B[] respectively */

int* add(int A[], int B[], int m, int n)
{
	
   
   int* sum = malloc(max(n, m) * sizeof(int));

pan64 · 02-26-2023, 09:03 AM

Quote:

Originally Posted by amikoyan

Thank you ntubski, I corrected to this.

Code:

int max(int m, int n) { return (m > n) ? m : n; }
 
/*A[] represents coefficients of first polynomial
* B[] represents coefficients of second polynomial
* m and n are sizes of A[] and B[] respectively */

int* add(int A[], int B[], int m, int n)
{
	
   
   int* sum = malloc(max(n, m) * sizeof(int));

Again, this is the same problem. You cannot free this in the function add, therefore either you need to malloc outside, not in the function add or implement it in another way.

Code:

int *sum = malloc(max(n, m) * sizeof(int));
call add
do whatever you want with sum
free(sum);

void add(int A[], int B[], int m, int n, int *sum)
 /* Initialize the sum polynomial */
    for (int i = 0; i < m; i++)
        sum[i] = A[i];
	
    /*Take every term of first polynomial */
    for (int i = 0; i < n; i++)
        sum[i] += B[i];

}

By the way, this function is incorrect, will not work properly if m < n.

amikoyan · 02-26-2023, 09:23 AM

Quote:

Originally Posted by pan64

yes, the usual way is to malloc before calling this function and you need to pass this variable to the function too, so you will never need to malloc/free inside.

This took me a bit longer, but

Code:

int* sum1 = malloc(max(n, m) * sizeof(int));

is now in main() not int* add(). I made some poor choices with variable names and had to rename sum to sum1 because I also used sum in main(). This confused me and caused quite a hold up.

Quote:

The other way is a bit different, you need to use and return a struct, not int *, in that case this will be solved automatically. And the struct should contain int[N]. where N is the maximal size of A and B.

I will try this with a struct later. Probably another day - my brain is full!

Thanks pan64.

amikoyan · 02-26-2023, 09:32 AM

Quote:

Originally Posted by pan64

Again, this is the same problem. You cannot free this in the function add, therefore either you need to malloc outside, not in the function add or implement it in another way.

Code:

int *sum = malloc(max(n, m) * sizeof(int));
call add
do whatever you want with sum
free(sum);

I was busy posting my last reply and didn't see this. I have used malloc outside the function now.

Quote:

Code:

void add(int A[], int B[], int m, int n, int *sum)
 /* Initialize the sum polynomial */
    for (int i = 0; i < m; i++)
        sum[i] = A[i];
	
    /*Take every term of first polynomial */
    for (int i = 0; i < n; i++)
        sum[i] += B[i];

}

By the way, this function is incorrect, will not work properly if m < n.

Sigh - so my work is not finished today. I will think some more

amikoyan · 02-26-2023, 10:04 AM

Because addition does not care which way you do it (6 + 3 is the same as 3 + 6), a property called 'commutative' - so google tells me. I added this if...else

Code:

/* Initialize the product polynomial */
 if (m >= n) {
    for (int i = 0; i < m; i++)
        sum1[i] = A[i];
	
    /*Take every term of first polynomial */
    for (int i = 0; i < n; i++)
        sum1[i] += B[i]; }
        
else {
	for (int i = 0; i < m; i++)
        sum1[i] = B[i];
	
    /*Take every term of first polynomial */
    for (int i = 0; i < n; i++)
        sum1[i] += A[i]; 
        }	
    return sum1;    
}

amikoyan · 02-26-2023, 10:08 AM

Thank you pan64 and ntubski.

pan64 · 02-27-2023, 12:02 AM

glad to help you

ntubski · 02-27-2023, 07:52 AM

You could also considering swapping the polynomials if m is smaller:

Code:

    if (m < n) {
        int tlen = m;
        m = n;
        n = tlen;
        int* tPoly = B;
        B = A;
        A = tPoly;
    }

    for (int i = 0; i < m; i++)
        sum1[i] = A[i];
	
    for (int i = 0; i < n; i++)
        sum1[i] += B[i];

amikoyan · 03-24-2023, 02:19 AM

Quote:

Originally Posted by ntubski

You could also considering swapping the polynomials if m is smaller:

Code:

    if (m < n) {
        int tlen = m;
        m = n;
        n = tlen;
        int* tPoly = B;
        B = A;
        A = tPoly;
    }

    for (int i = 0; i < m; i++)
        sum1[i] = A[i];
	
    for (int i = 0; i < n; i++)
        sum1[i] += B[i];

Sorry ntubski for the long delay in responding to your suggestion, it has been a busy month.

I like your suggestion and will experiment with it in my code. Thanks for the help.

pan64 · 03-24-2023, 02:29 AM

I still think it is overcomplicated:

Code:

void add(int A[], int B[], int m, int n, int *sum)
 /* Initialize the sum polynomial */
    memset (sum, 0, max(m, n));
	
    /*Take every term of first polynomial */
    for (int i = 0; i < m; i++)
        sum[i] += A[i];
	
    /*Take every term of second polynomial */
    for (int i = 0; i < n; i++)
        sum[i] += B[i];
}

NevemTeve · 03-25-2023, 03:56 AM

Also one could check for the degree:

Code:

int d=max(m,n);
while (d>=0 && result[d]==0) --d;

Here constant zero polynom has degree -1