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How to erase memory fields allocated by calloc?

Hi,
I am using calloc in a function to allocate memory to my array. This function writes to the array different number of elements each time the function is called. Here is my problem. Say for the first time function writes to all the 3 locations and next time it writes to only first 2 locations. After the second call, the 3rd memory locations still contains the value written by the first function call. I understand that is the way it works.But I want to know if there is a way to erase all the locations before calling the function again? Is there any built-in function available? Also when I print the values of array initially it doesn't print zeroes. I have read that calloc initializes the memory fields to 0's. Following is the function code.

function write_to_array(int value)
{
 int xdata *ascii_value,i;
 ascii_value = calloc(3, sizeof (int));
 for(i=0;value!=0;i++)
  {
    mod = value%10;
    c = mod+'0';
    ascii_value[i] = toascii(c);
    value/=10;
  }
 }

Parents
  • How about memset()?

    memset (&myStruct, 0, sizeof(myStruct));

    If you want the structure initialized to something other than zeroes, you might want to memcpy() from a constant.

    Very small structures might be more efficiently initialized with a series of assignments.

    calloc() is essentially malloc() followed by a call to memset() to set all bytes to zero.

Reply
  • How about memset()?

    memset (&myStruct, 0, sizeof(myStruct));

    If you want the structure initialized to something other than zeroes, you might want to memcpy() from a constant.

    Very small structures might be more efficiently initialized with a series of assignments.

    calloc() is essentially malloc() followed by a call to memset() to set all bytes to zero.

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