Copy a string to a malloc'd array of strings - c

Question

I thought I understood the answer to this question but I don't. I understand the first result but I still don't know how to do the copy correctly. I tried the following code:
// TstStrArr.cpp : Defines the entry point for the console application.
//
#include "stdafx.h"
#include <string.h>
#include <malloc.h>
int main()
{
char ** StrPtrArr;
char InpBuf0[] = "TstFld0";
char InpBuf1[] = "TstFld1";
StrPtrArr = (char **)malloc(2 * sizeof(char *));
StrPtrArr[0] = (char *)malloc(10 + 1);
printf("inpbuf=%s sizeof=%2d ", InpBuf0, sizeof(StrPtrArr[0]));
strncpy_s(StrPtrArr[0], sizeof(StrPtrArr[0]), InpBuf0, _TRUNCATE);
printf("strptrarr=%s\n", StrPtrArr[0]);
StrPtrArr[1] = (char *)malloc(10 + 1);
printf("inpbuf=%s sizeof=%2d ", InpBuf1, sizeof(*StrPtrArr[1]));
strncpy_s(*StrPtrArr[1], sizeof(*StrPtrArr[1]), InpBuf1, _TRUNCATE); // error here
printf("*strptrarr=%s\n", StrPtrArr[1]);
free(StrPtrArr[0]);
free(StrPtrArr[1]);
free(StrPtrArr);
return 0;
}
The result I got was:
inpbuf=TstFld0 sizeof= 4 strptrarr=Tst
inpbuf=TstFld1 sizeof= 1
and the following error:
Exception thrown: write access violation.
destination_it was 0xFFFFFFCD.
The result I thought I'd get was either of the following:
inpbuf=TstFld1 sizeof=11 *strptrarr=TstFld1
inpbuf=TstFld1 sizeof= 1 *strptrarr=T
I understand the first copy copied the input buffer to the 4 byte pointer which was incorrect. I thought the second copy would copy the input buffer to the value of the dereferenced pointer of a size of 11 but it didn't. I'm guessing the copy was to the first character of the string in the array. I don't understand memory enough to know the significance of the address 0xFFFFFFCD but I guess it's in read-only memory thus causing the error.
What is the correct way to do the copy?
(I don't think it matters, but I'm using VS 2015 Community Edition Update 3.)

Answer 1

Why
strncpy_s(*StrPtrArr[1], sizeof(*StrPtrArr[1]), InpBuf1, _TRUNCATE);
?
*StrPtrArr[1] should be StrPtrArr[1] because StrPtrArr is of type char** and you need char* here.
and sizeof(*StrPtrArr[1]) - is quite strange....
actually sizeof(StrPtrArr[1]) also cannot provide correct value.
You should remember size of allocated memory and then use it like:
size_t arrSize = 10 + 1;
StrPtrArr[1] = (char *)malloc(arrSize);
. . .
strncpy_s(StrPtrArr[1], arrSize, InpBuf1, _TRUNCATE);

Answer 2

The problem is that you are using sizeof when deciding how many characters to copy. However, you allocated a fixed number of characters which is not known to sizeof operator: sizeof StrPtrArr[0] is equal to the size of char pointer on your system (four bytes, judging from the output), not 10 + 1. Hence, you need to specify that same number again in the call to secure string copy.

Answer 3

It isn't as complicated as people seem to think.
char* array = calloc( n, sizeof(array[0]) ); // allocate array of pointers
// assign a dynamically allocated pointer:
size_t size = strlen(str) + 1;
array[i] = malloc(size);
memcpy(array[i], str, size);
I intentionally used calloc during allocation, since that sets all pointers to NULL. This gives the advantage that you can harmlessly call free() on the pointer, even before it is assigned to point at a string.
This in turn means that you can easily (re)assign a new string to an index at any time, in the following way:
void str_assign (char** dst, const char* src)
{
size_t size = strlen(src) + 1;
free(*dst);
*dst = malloc(size);
if(*dst != NULL)
{
memcpy(*dst, src, size);
}
}
...
str_assign(&array[i], "something");