C pointers with arrays in functions with realloc

Question

Hi I am slowly getting my head around pointers and indirection but am still having a little trouble.

In my main function i am creating an array of structs (computers) using malloc

int arraySize = 0;//This may be the first issue
Computer    *ptrComputer = NULL; 
ptrComputer = (Computer*) malloc(sizeof(Computer) * arraySize);

I am then passing the pointer to a function that reads data out of a file and into the array

arraySize = readFileToArray(&ptrComputer, arraySize);

int readFileToArray(Computer **compArray, int arraySize){

Computer newComp;
int foundARecord = 0;
/*File stuff*/

   arraySize = extendArrays(compArray, arraySize, no_elements);

    /*Use fscanf to read file data into the newComputer variable*/
    printf("g %i\n", arraySize);
    *compArray[arraySize - 1] = newComp;//set the newly created part of the array to newComputer


return arraySize;
}

int extendArrays(Computer **compArray, int arraySize){
   arraySize++;
   //Resize the computer array
   *compArray = (Computer*)realloc(*compArray, (sizeof(Computer)*(arraySize + 1)));
   return arraySize;
}

Now as far as i understand, i am passing the address of ptrComputer to the readFileToArray() function. It is then passing that same address to the extendArrays() function which resizes it. I am then trying to assign newComp to the location in memory that compArray points to. This works as long as i try to write to index 0 but any others cause xcode to throw an exc_bad_access error. This is all very confusing, can anyone with a bit of experience with this see where i am going wrong? It was working before when i was passing the actual pointer to the readFile function but it only worked once (think i was reallocating a copy of the pointer) Any help would be very muchly appreciated Thanks

The first thing you should have done was check the return value of malloc(). — Jeff Mercado
– Jeff Mercado, Commented May 25, 2011 at 1:47
Also, you shouldn't immediately assign the return value of realloc to *compArray, as it may return NULL, in which case the former value of *compArray is still valid. — jwodder
– jwodder, Commented May 25, 2011 at 1:53
I do have checking implemented on both, just didnt include them to keep the size of the code down — Darc
– Darc, Commented May 25, 2011 at 1:59
Do not cast the return value of malloc. Casting the return value of malloc serves no purpose and may hide errors the compiler would catch otherwise. — pmg
– pmg, Commented May 25, 2011 at 8:24

hsmiths · Accepted Answer · 2011-05-25 02:09:01Z

2

Instead of this:

    *compArray[arraySize - 1] = newComp;//set the newly created part of the array to newComputer

Try this:

    (*compArray)[arraySize - 1] = newComp;//set the newly created part of the array to newComputer

This way you explicitly dereference compArray before subscripting.

answered May 25, 2011 at 2:09

hsmiths

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3 Comments

Darc Over a year ago

That appears to have done it, thank you so much. Could you please explain the difference between the two?

Lawrence Velázquez Over a year ago

The array subscripting operator has a higher precedence than the dereferencing operator. So *x[i] evaluates as *(x[i]), not (*x)[i].

Darc Over a year ago

So the way i was doing was trying to deference compArray[arraySize - 1] instead of just compArray

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