Possible Memory Leak: new char[strlen()] - c++

Question

This is a fairly basic question and I am pretty sure I know the answer, but seeing as the consequence for being wrong is a segfault I figure I should ask. I have been using strlen() and the new char[] operator in the following way for quite some time now and just noticed something that threw up a red flag:
void genericCopy(char *somestring, char *someOtherString) {
someOtherString = new char[strlen(somestring)];
strcpy(someOtherString,somestring);
}
My question is, seeing as a string should be null terminated, should I be doing this as such:
void genericCopy(char *somestring, char *someOtherString) {
someOtherString = new char[strlen(somestring)+1];
strcpy(someOtherString,somestring);
someOtherString[strlen(someOtherString)] = '\0';
}
So far I have never had a problem with the first method, but that doesn't mean I'm doing it right. Since the length being return by strlen()is the number of characters in the string without the null terminator so new isn't reserving space for '/0'... At least I don't think it is.

Answer 1

First of all, you should know that this function of yours is pointless to write, just use strdup (if available on your system).
But yes, you need an additional byte to store the \0, so always do something like new char[strlen(somestring)+1];. However, there is no need to manually add the \0; strcpy already does this.
You should use something like Valgrind to discover this and similar bugs in your code.
There is however an additional problem in your code; your code will always leak someOtherString; it will not be returned to where you called it from. You either need to change your method to something like:
char *genericCopy(char *something) {
char *copy = new char[strlen(somestring)+1];
strcpy(copy,somestring);
return copy;
}
and then get the copy as follows:
copy = genericCopy(something);
Or you need to change your method to something like:
void genericCopy(char *something, char **copy) {
*copy = new char[strlen(somestring)+1];
strcpy(*copy,somestring);
}
and call it as:
genericCopy(something, &copy);
If you'll be using C++ you could also just change the method prototype to:
void genericCopy(char* somestring, char*& someOtherString)
and call it as:
genericCopy(something, copy);
Then someOtherString will be passed as a reference, and the new value you allocate to it will propagate outside of your method.

Answer 2

Yes, your suspicion is correct. You should be allocating an additional character, and making sure the copied string is null-terminated. (strcpy() itself will do this, but when someone advises to you that you switch to strncpy(), as they no doubt will (it's safer!) you'll need to be extra careful, because it is NOT guaranteed to copy the '/0'.)
If you're already using C++, though, you may be well-advised to switch to using std::string. It's often an easier, less error-prone method of manipulating character arrays.
However, here's the further problem that you need to address. You are assigning your new character array to a COPY of someOtherString. You need to make some changes:
void genericCopy(char *somestring, char **someOtherString) {
*someOtherString = new char[strlen(somestring)+1];
strcpy(*someOtherString,somestring);
(*someOtherString)[strlen(somestring)] = '\0';
}
This way you will get back the new character buffer outside your function call.