Question

I asked the question: Pass array by reference using C. I realized my problem was the usage of the pointer star in C. And eventually it turned out that, this method works fine for my program:

#include <stdio.h>

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

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

int main()
{
     int* myArray = NULL;
     FillArray(& myArray);  
     printf("%d", myArray[0]);
     return 0;
}

Everyting was fine up to that point. Then, I modified the FillArray() function like the following for a better code readability:

#include <stdio.h>

void FillArray(int** myArray)
{
     int* temp = (*myArray);

     temp = (int*) malloc(sizeof(int) * 2);
     temp[0] = 1;
     temp[1] = 2;
}

int main()
{
     int* myArray = NULL;
     FillArray(& myArray);  
     printf("%d", myArray[0]);
     return 0;
}

Now, I get the following run-time error at the printf line:

Unhandled exception at 0x773115de in Trial.exe: 0xC0000005: Access violation reading location 0x00000000.

Even though I’m not an C expert, it seemed legitimate to do this modifying. However, apparently it doesn’t work. Isn’t the syntax a little bit confusing? Do I miss something here?

Thanks for your helpful answers,

Sait.

Answer 1

temp gets a copy of the address of myArray, but then you assign some malloced memory to temp, so the original assignment was pointless and had no lasting effect. You then modify the malloced memory, but that doesn’t change myArray at all. To change myArray in main, you have to assign

*myArray = temp;

at the end of FillArray.

void FillArray(int** myArray)
{
     int* temp;
     temp = (int*) malloc(sizeof(int) * 2);
     temp[0] = 1;
     temp[1] = 2;
     *myArray = temp;
}

does what you intend.