How to convert fixed size array to pointer on pointer array

» cpp » How to convert fixed size array to pointer on pointer array

By : Dana Stancu

Date : November 23 2020, 04:01 AM

Any of those help Array with two dimensions and a pointer to another pointer are not the same things. Why do you need to use mat in different ways for two different functions such as double mat[N][N] in the function getCofactor() and double **mat in the determinant() function ? Since you know what is the value of N variable, you can use it as double mat[N][N] in the determinant() function as well.

code :

double determinant(double mat[N][N], int n)
    {
        double D = 0; // Initialize result

        //  Base case : if matrix contains single element
        if (n == 1)
            return mat[0][0];

        double temp[N][N]; // To store cofactors

        int sign = 1;  // To store sign multiplier

        // Iterate for each element of first row
        for (int f = 0; f < n; f++)
        {
            // Getting Cofactor of mat[0][f]
            getCofactor(mat, temp, 0, f, n);    //ERORRRRRRRRRRR
            D += sign * mat[0][f] * determinant(temp, n - 1);

            // terms are to be added with alternate sign
            sign = -sign;
        }

        return D;
    }

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Destructor for pointer to fixed size array

By : user3593877

Date : March 29 2020, 07:55 AM

like below fixes the issue I see no static array. I see a fixed-size array. Also memory for data is allocated as part of the object.
You must not explicitely delete a member of the class: the delete operator will take care of that IFF the instance was dynamically allocated.

code :

 {
      MyClass x; // auto variable
 } // x destructor run, no delete operator

 {
      MyClass* x = new MyClass(); // heap allocation variable
      delete x; // x destructor run, ::delete de-allocates from heap
 }

Allocating 2D array with pointer to fixed-size array

By : rbl

Date : March 29 2020, 07:55 AM

Any of those help The only real issue is if you request more memory than the malloc call can satisfy (you may have enough memory available, but not in a single, contiguous block).
To save your sanity, do something like this instead:

code :

T (*arr)[cols] = malloc( sizeof *arr * rows );

Is it necessary to free sub-pointer before calling realloc on pointer-to-pointer array to shrink its size?

By : user3867795

Date : March 29 2020, 07:55 AM

wish help you to fix your issue Yes, you have to free all pointers if the reallocated array will have size that is less than the original size. For example

code :

for (i = 50; i < len; i++)
{
    free(arr[i]);
}

char** temp = realloc(arr, 50 * sizeof(char*));

Why can't I pass the address of a pointer to a fixed size array into a function expecting a pointer to a pointer in C?

By : jeffwiedenfeld

Date : March 29 2020, 07:55 AM

it helps some times You're making the mistake of believing a pointer and an array are the same things. They are not. They can be used in the same way (which is completely different from saying "they are the same thing") in some contexts but not others. Your examples are in contexts in which pointers and arrays are different things, and cannot be used as if they are the same thing.
In your first case, passing foo to singleLeftPadZero() works by first performing an "array to pointer conversion" - foo is an array of ten char, and is converted to a pointer of type char * with value equal to &foo[0]. A char * is not a char **, hence the warning about incompatible type and the note.

How do I convert a char * string into a pointer to pointer array and assign pointer values to each index?

By : Daham Ilac

Date : March 29 2020, 07:55 AM

I wish did fix the issue. I have a char * that is a long string and I want to create a pointer to a pointer(or pointer array). The char ** is set with the correct memory allocated and I'm trying to parse each word from the from the original string into a char * and place it in the char **. , Assuming that you know the number of words, it is trivial:

code :

char **newtext = malloc(3 * sizeof(char *));   // allocation for 3 char *
// Don't: char * pointing to non modifiable string litterals
// char * t1 = "fus", t2 = "roh", t3 = "dah";
char t1[] = "fus", t2[] = "roh", t3[] = "dah"; // create non const arrays

/* Alternatively
char text[] = "fus roh dah";    // ok non const char array
char *t1, *t2, *t3;
t1 = text;
text[3] = '\0';
t2 = text + 4;
texts[7] = '\0';
t3 = text[8];
*/
newtext[0] = t1;
newtext[1] = t2;
newtext[2] = t2;