CODEWITHSUNDEEP
clinuxredhatcompiler-warningscc
Morag Hughson
Morag Hughson

Reputation: 7579

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.

Upvotes: 5

Views: 81

Answers (1)

dbush
dbush

Reputation: 224842

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.

Upvotes: 10

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