[SOLVED] Avoiding strcpy overflow destination warning

Avoiding strcpy overflow destination warning

With a structure such as the following

typedef struct _mySTRING
{
  unsigned short StringLength;
  char           String[1];
} mySTRING;

and use of this structure along these lines

mySTRING * CreateString(char * Input)
{
  int        Len     = (int)strlen(Input);
  int        Needed  = sizeof(mySTRING) + Len;
  mySTRING * pString = (mySTRING *)malloc(Needed);
  :
  strcpy(pString->String, Input); 
}

results, on Red Hat Linux cc compiler, in the following warning, which is fair enough.

strings.c:59:3: warning: âstrcpyâ writing 14 bytes into a region of size 1 overflows the destination [-Wstringop-overflow=]
   strcpy(pString->String, Input);

I know that, in this instance of code, this warning is something I don't need to correct. How can I tell the compiler this without turning off these warnings which might usefully find something, somewhere else, in the future.

What changes can I make to the code to show the compiler this one is OK.

Solution

You're using what's commonly referred to as the "struct hack" in order to have an array at the end of the struct have a variable size. That method has always been dubious as far as the C standard is concerned.

The proper way to do that as of the C99 standard is with a flexible array member, which leaves the size of the array unspecified:

typedef struct _mySTRING
{
  unsigned short StringLength;
  char           String[];
} mySTRING;

You'll also need to add 1 to the allocated size, as sizeof(mySTRING) doesn't include the String member and you need space for the null terminating byte.

int Needed  = sizeof(mySTRING) + Len + 1;

This makes the rest of your code compliant to the C standard.