strcpy_s for char** and char[][]

Question

I used strcpy_s as below:

char names[2][20];
strcpy_s(names[0],"Michael");
strcpy_s(names[1],"Danny");

and it worked all right.

But when I changed to char **,

int size1=2;
int size2=20;

char **names=new char*[size1];
for(int i=0;i<size1;i++)
  names[i]=new char[size2];
strcpy_s(names[0],"Michael");
strcpy_s(names[1],"Danny");

It gives me this error message:

error C2660: 'strcpy_s' : function does not take 2 arguments

Why is this happening? I need to dynamically create char arrays, so what should I do?

`strcpy` != `strcpy_s`. Which one did you mean to use? – CB Bailey Jun 03 '12 at 09:24 — CB Bailey, Jun 03 '12 at 09:24
As this is C++, consider using a `std::vector`. – hmjd Jun 03 '12 at 09:30 — hmjd, Jun 03 '12 at 09:30
@hmjd Don't consider using it. Use it. – Jun 03 '12 at 09:34 — , Jun 03 '12 at 09:34
@user1203803 high and clear! – Anxon Pués Nov 08 '19 at 12:12 — Anxon Pués, Nov 08 '19 at 12:12

score 9 · Accepted Answer · answered Jun 03 '12 at 09:28

9

There are two forms of strcpy_s (at least on Windows): one for pointers and one for arrays.

errno_t strcpy_s(
   char *strDestination,
   size_t numberOfElements,
   const char *strSource 
);
template <size_t size>
errno_t strcpy_s(
   char (&strDestination)[size],
   const char *strSource 
); // C++ only

When using pointers you have to specify the number of elements of the destination buffer:

strcpy_s(names[0], size2, "Michael");
strcpy_s(names[1], size2, "Danny");

answered Jun 03 '12 at 09:28

Pubby

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+1 was unaware of the template version (even though it was on page I linked in my deleted answer!) – hmjd Jun 03 '12 at 09:28
@Michael No problem. Don't forget to click the green check mark to accept the answer as correct. – Pubby Jun 03 '12 at 12:56

strcpy_s for char** and char[][]

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