Converting integers into a char* c++

Question

Here is the function I have, "Sprite" is an object in the program, and "GetSpriteAtPosition" just returns a pointer to the correct sprite at the coordinates.

My problem is that I store a letter in each sprite, in the form of an integer. 0 is a, and 25 is z, with everything in between respectively. I need my function to return a char* that gives me the letters of a row of sprites, so if in the program the sprites spell out "abcdefgh", then that's what I need this function to print out. There's an 8x8 grid of sprites, and I'm getting the coordinates correctly, but I get an error that I can't convert an int to a char* in the marked line. What can I do to get this to work?

Thanks in advance!

char* RowLetters(int row)
{
    char* pointer;
    for( int i = 0; i < 8; i++)
    {
        Sprite* selectedSprite = SpriteAtPosition(row*50, i * 50);
        if(selectedSprite != NULL)
        {
            char* temp = (char)(selectedSprite->Frame() + 97); //error here
            pointer = strcat(pointer, temp);
        }
        else
        {
            pointer = strcat(pointer, "test");
        }
    }
    return pointer;
}

maybe you want to convert to a char* , **(char)**(selectedSprite? — joy, Nov 27 '12 at 21:34
Why are you even using C strings in a C++ program ? Use `std::string` and save yourself a lot of grief. — Paul R, Nov 27 '12 at 21:57
A couple of suggestions, not directly related to the problem. First, store a `char` instead of an `int`. Second, store the character that you want, `'a'`, `'b'`, etc. That way you don't have to go through the non-portable conversion of adding 97. — Pete Becker, Nov 27 '12 at 22:00

hall.stephenk · Accepted Answer · 2012-11-28T01:25:39.660

3

Try this:

        char temp = (char)(selectedSprite->Frame() + 97);
        pointer = strcat(pointer, &temp);

I've changed the variable into a standard char rather than a pointer and then passed a reference to strcat() with the & operator.

EDIT:

As pointed out in the comments, this doesn't work because &temp isn't NULL terminated. I used to get around this when I programmed more C by doing the following.

        char temp[2];
        temp[0] = (char)(selectedSprite->Frame() + 97);
        temp[1] = '\0';
        pointer = strcat(pointer, temp);

Of course, the temp array could be declared outside the for() loop for a little better performance (in theory).

None of this addresses the other problems with the code like pointer never being declared. I think a broader understanding of the calling function would be in order to determine whether pointer should be allocated within this function or passed in by the caller.

edited Nov 28 '12 at 01:25

answered Nov 27 '12 at 21:34

hall.stephenk

1,175
7
10

@Chad: Yep, that seems like it would be a problem once you run the code. But, the above modification should at least allow it to compile, which solves the original problem in the post. – hall.stephenk Nov 27 '12 at 21:38
Since `&temp` will likely not be null-terminated, this will also be a problem. While this fixes the compiler error, it leads to other kinds of bad behavior. – Chad Nov 27 '12 at 21:45
A compile time error is infintely better than a runtime error. See this related question: http://stackoverflow.com/questions/4834811/strcat-concat-a-char-onto-a-string – Chad Nov 27 '12 at 21:51

score 2 · Answer 2 · answered Nov 27 '12 at 21:41

Your code as written, will have undefined behavior because pointer is not initialized, and does not point to any valid memory that you have allocated (to hold the appropriate length of letters in the row.

If this truly is C++, as you state, then you don't want to return a char* from this function, as that implies that you have a static string already allocated within that function (yuck), or you will be dynamically allocating the string in that function and the caller must free it (yuck).

Neither of these options is ideal.

I'd suggest a very simple change to return a std::string, like this:

std::string RowLetters(int row)
{
    std::string pointer;
    for( int i = 0; i < 8; i++)
    {
        Sprite* selectedSprite = SpriteAtPosition(row*50, i * 50);
        if(selectedSprite != NULL)
        {
            pointer.push_back((char)(selectedSprite->Frame() + 97));
        }
        else
        {
            // ???
            // pointer = strcat(pointer, "test");
        }
    }

    return pointer;
}

Converting integers into a char* c++

2 Answers2