strdup

Hi,

An exercise from my book is to implement my own version of the strdup function, which returns a copy of a string allocated on the freestore.

This is my code:

Code:

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char *strdup(const char* ch)
{
        const char* p = ch;
        int size = 0;
        while(*p){
                ++size;
                ++p;
        }

        int index = 0;
        char* name = new char[size];
        while(*ch){
                name[index] = *ch;
                cout << *ch;
                ++ch;
                ++index;
               
        }

        return name;
               
}

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My question is that when I create name, I allocate memory on the free store. When I return this, I take it that it is still allocated on the freestore and when whatever has called it finishes using it, that is when I call delete[]? Is that correct?

BTW, the exercise states that I cannot use any standard library function, hence my own version of size.

Thanks.

Looks good to me - your analysis is correct.

Quote:

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Originally Posted by Salem

Looks good to me - your analysis is correct.

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Thanks, I'm on the right track then! :)

Quote:

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BTW, the exercise states that I cannot use any standard library function, hence my own version of size.

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I would write my own strlen and strcpy then... :)

It also appears that you might not be allocating space for and copying the null terminator. (It would be more immediately visible if you used your own versions of strlen and strcpy with the exact same behavior.)

And I guess a strdup user would assume that the buffer was allocated with malloc (they need to know the details)...