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Skip array by reference using C

Yes, I read this question and answers: Passing an array by reference in C? . I have a similar problem and implemented the same idea from this question.

However, I still get the error message from the following code:

#include <iostream>

void FillArray(int** myArray)
{
   free( *myArray ) ;
   * myArray = (int*) malloc(sizeof(int) * 2);

   *myArray[0] = 1;
   *myArray[1] = 2;
}

int main()
{
   int* myArray = NULL;
   FillArray(& myArray);    
   return 0;
}

I got the following runtime error right after the completion of the FillArray function:

Unhandled exception at 0x773115de in Program.exe 0xC00000005: location of access violation record 0xcccccccccc.

Visual Studio, Visual ++ Empty Project. main.cpp. , ++ C? , , C? main.cpp main.c, . ( , "pass by reference" , C ++.)

.

,

.

+3
3

, . :

(*myArray)[0] = 1;
(*myArray)[1] = 2;

(: iostream C. ++.)

+10

- .

[] , *,

*myArray[1] 


*(myArray[1]).

, , . :

(*myArray)[1] = 2;

C ++, . , ++, new delete over malloc free, .

+2

Good:

int main()
{
   int* myArray = NULL;
   FillArray(& myArray);   
   ...

Good:

void FillArray(int** myArray)
{
   * myArray = (int*) malloc(sizeof(int) * 2);

Catastrophically bad:

void FillArray(int** myArray)
{
   free( *myArray ) ;
   * myArray = (int*) malloc(sizeof(int) * 2);
   ...

it's better:

void FillArray (int** myArray)
{
   if (myArray)
     free (*myArray);
   *myArray = (int*) malloc(sizeof(int) * 2);
   ...

and

   *(myArray[0]) = 1;
   *(myArray[1]) = 2;
+1
source

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