Question

Trying to create a pointer to an array (string of characters) that the user enters.

Here i define my pointer and array in my class:

private:

char userWord[200];     // input word
char * userWordPtr; // Pointer to the beginning of the input word

Here is the function that reads in the information:

void WordFont::inputWord()
{

    cout << "What word would you like to draw?\n"
         << "The useable characters are: A E I O U Y B C D L M N R S T\n" ;
    cout << "Please enter your word: ";
    cin.get(userWord, 200);
    cin.clear();
    cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');

    userWordPtr = &userWord;

}

The user enters a word and my code does something (long if-else ladder) based on the letters of the word. But in passing a * Ptr to the if-else ladder, I can’t get the if-else ladder to recognize the null (/0) character when I’m incrementing the position of the pointer. I also get an error using the ‘&’ to reference the address of the beginning of the char array. I can’t figure out why I’m getting that error, and why the if-else ladder below doesn’t exit til ‘i’ gets to whatever random location in memory has a ‘/0’ character. If I do something as simple as entering ‘AAA’, it will well past the 4th position where the ‘/0’ character is.

Here’s the loop:

void WordFont::buildOutputArray(char * Ptr)
{

    length = 0;
    bool switchLooper = true;
    int i = 0;

    do
    {
        if(*Ptr + i == 'a' || *Ptr + i == 'A')
            constructA();
        else if (*Ptr + i == 'e' || *Ptr + i == 'E')
            constructE();
        else if (*Ptr + i == 'i' || *Ptr + i == 'I')
            constructI();
        else if (*Ptr + i == 'o' || *Ptr + i == 'O')
            constructO();
        else if(*Ptr + i == 'u' || *Ptr + i == 'U')
            constructU();
        else if(*Ptr + i == 'b' || *Ptr + i == 'B')
            constructB();
        else if(*Ptr + i == 'c' || *Ptr + i == 'C')
            constructC();
        else if(*Ptr + i == 'd' || *Ptr + i == 'D')
            constructD();
        else if(*Ptr + i == 'l' || *Ptr + i == 'L')
            constructL();
        else if(*Ptr + i == 'm' || *Ptr + i == 'M')
            constructM();
        else if(*Ptr + i == 'n' || *Ptr + i == 'N')
            constructN();
        else if(*Ptr + i == 'r' || *Ptr + i == 'R')
            constructR();
        else if(*Ptr + i == 's' || *Ptr + i == 'S')
            constructS();
        else if(*Ptr + i == 't' || *Ptr + i == 'T')
            constructT();
        else if(*Ptr + i == '\0')
            switchLooper = false;
        i++;
    }while(switchLooper);


}

Its probably a easy newbie mistake, and I’m sure you’ll spend notice lots of other problems, but any help I greatly appreciate.

Thanks,
Tom

Answer 1

The user enters a word and my code does something (long if-else ladder) based on the letters of the word. But in passing a * Ptr to the if-else ladder, I can’t get the if-else ladder to recognize the null (/0) character when I’m incrementing the position of the pointer.

Actually, what you are doing by using *Ptr + i is:

• dereferencing your Ptr pointer (you are accessing the value at the address pointed by Ptr)
• then you are adding i to the value contained at the address Ptr

Serge gave you the answer, you have to use *(Ptr + i) to get the character at that address.

Here, you are shifting the Ptr pointer of i “blocks”.

I also get an error using the ‘&’ to reference the address of the beginning of the char array. I can’t figure out why I’m getting that error,

An array of char (or whatever) is “like” a pointer (there are a bunch of differences but I let you the pleasure of discovering them ), so when you are using str[i] it’s quite the same thing than *(str + i).

And why the if-else ladder below doesn’t exit til ‘i’ gets to whatever random location in memory has a ‘/0’ character. If I do something as simple as entering ‘AAA’, it will well past the 4th position where the ‘/0’ character is.

You shouldn’t have this problem anymore.

What if the user enters 200 characters?