Debug failure error

I'm trying to create my own string class, which I named CStr.

First, I defined it as such (I left out a lot of irrelevant code at the moment):

Code:

---

class CStr
{
        char* pData;
        int nLength;
        int stlen(char* c){for(int i = 0; *c != '\0'; c++, i++); return i;}
        char* stcpy(char*, const char*);
        char* stcat(char*, const char*);
public:
        CStr(char*);
        void set(char* c){stcpy(pData, c);}
};

---

Then the member function definitions:

Code:

---

char* CStr::stcpy(char* dest, const char* src)
{
        char* p;
        p = dest;
        if(nLength > 0)
        {
                delete [] pData;
                pData = new char[stlen(src)];
                p = dest = pData;
        }
        while(*(p++)=*(src++));
        return dest;
}

CStr::CStr(char* c)
{
        pData = new char[stlen(c)];
        stcpy(pData, c);
        nLength = stlen(c);
}

---

Now here's the code from the main() function.

Code:

---

        char* p = "First input";
        char q[256];
        CStr c(p);
        while(gets(q))
                c.set(q);

---

My problem is that I get a "Debug Error" message (in VC++ 2003)) which states as follows:
DAMAGE: After Normal Block(#42) at XXXXX.

This occurs at the line "delete [] pData" in the definition of stcpy() when it is called by "c.set(q)".

Please tell me, what am I doing wrong here?

you don't allow room for the null-terminator

Code:

---

pData = new char[stlen(src)+1];

---

why does strcpy() method have two parameters? It only needs one -- the source string. Destination string should always be pData, which does not have to be passed to the function.

Thanks, your correction worked perfectly.

As for my stcpy function, I was curious how the actual strcpy function in string.h worked so I thought I might as well try my hand at mimicing it. I could have written one specifically for this class, but just wanted to make something generic.

Hi,

The defining aspect of a C-style string is that it ends in a '\0' character. However, that '\0' is not included in the strlen() count. That will constantly trip you up when dealing with C-style strings. The '\0' is required at the end of C-style strings because it indicates to all the <cstring> functions where the characters in the array end. There's no bounds checking in C++, so a while loop will go right past the end of an array and into uncharted memory if you let it. cout<< also uses that '\0' to determine when to stop outputing characters to the screen. So, don't forget about the '\0'!

Your strcat() function will provide a little twist on that concept because you don't want to end up with:

"string1\0string2\0"

The <cstring> functions and cout<< use a while loop to advance along the array one character at a time, and the loop will terminate when a '\0' character is encountered. So those functions won't be able to see anything past the first '\0'.