C programming access violation segmentation fault question

**Isaiah** · 01-20-2012

Hi everyone,
I'm new to this sight and have a smallish amount of programming knowledge, so a simpler answer would be great.

For some reason, whenever I try to run this code with a H 13 or higher, it says that the program stopped working. I am using the Dev compiler and have tried debugging, and it says it thinks the problem is in line "divisorList[arrayCounter]=divisorTest;". However, I can't seem to figure out what the problem is...
Anything would be helpful.

Code:

#include <stdio.h>


main()
{
       int H, numOfHiddenDots, arraySlots, counterH, arrayCounter, arrayResetCounter, divisorTest, divisorTestTest, divisorRunthrough;
       numOfHiddenDots=0;
       arraySlots=1;
       int divisorList[arraySlots];
       
       H=(whatever you want)
       arraySlots=H+1;
       
       for (counterH=1; counterH<=H; counterH++)
       {
           arrayCounter=1;
           for (arrayResetCounter=0; arrayResetCounter<=counterH+1; arrayResetCounter++)
           {
               divisorList[arrayResetCounter]=0;
           }
           if (counterH%100==0) printf ("%d milestone reached in calculations.\n", counterH);
           
           for (divisorTest=1; divisorTest<counterH; divisorTest++)
           {
               if (counterH%divisorTest==0)
               {
                  divisorList[arrayCounter]=divisorTest;
                  arrayCounter++;
                  
                  for (divisorTestTest=2; divisorTestTest<arrayCounter-1; divisorTestTest++)
                  {
                      if (divisorList[divisorTestTest]!=0&&divisorTest%divisorList[divisorTestTest]==0) divisorList[divisorTestTest]=0;
                  }
               }
           }
           
           for (divisorRunthrough=1; divisorRunthrough<arrayCounter; divisorRunthrough++)
           {
               if (divisorList[divisorRunthrough]<=1) numOfHiddenDots=numOfHiddenDots+divisorList[divisorRunthrough]*6;
               else numOfHiddenDots=numOfHiddenDots+(divisorList[divisorRunthrough]-1)*6;
           }
       }
       
       printf ("The number of dots hidden from the center is: %d\n", numOfHiddenDots);
       getchar();
}

Thanks

**memcpy** · 01-20-2012

The array divisorList is always one place long. You can't declare a fixed array and then hope the boundaries will change when the variable denoting the size does.

Your problem is most likely an array indexing issue, by trying to read/write over the bounds of stack memory that the array is in.

**grumpy** · 01-20-2012

You have defined divisorList as an array of one element.

Doing

Code:

    int arraySlots = 1;
    int divisorList[arraySlots];
    arraySlots = some_value_greater_than_1;

does not magically resize the array divisorList. Any access of divisorList[i] where i is 1 or more will always fall off the end of the array.

You just got lucky (or unlucky, depending on how you look of it) in having the code work for H < 13 with your compiler. The code is equally invalid for all values. With another compiler, it might fail for all values of H.

**quzah** · 01-20-2012

memcpy is correct. You need to know the size of your VLA before you declare it. You could do something like this:

Code:

int size;
printf( "How big is your array? " );
fflush( stdout );
scanf( "%d", &size );
... error check ...


type yourarray[ size ];
... do stuff ...

You cannot however change the size of the array once you've made it. It has to have a set size when you declare it, and it stays that size until it goes out of scope. So you could do something like this:

Code:

void foo( int size )
{
    type array[ size ];
    ...
}

foo( 3 );
foo( 100 );

Not that this is necessarily what you want to do in this case, it's more an illustration of how VLAs work.

Quzah.

**Isaiah** · 01-20-2012

So, basically I should declare the array to be bigger than H right from the outset?
Thanks anyways

**memcpy** · 01-20-2012

Only if you know the initial size of H, and also know that it won't change.

Otherwise, get the value first.

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