Thread: How do I use this function?

I'm reading this wonderful book called The C Programming Language 2nd Edition and I'm having a hard time understanding how to use this function that is on page 97. Here it is:

Code:

int getint(int *pn)
{
    int c, sign;

    while (isspace(c = getch()))
        ;
    if (!isdigit(c) && c != EOF && c != '+' && c != '-') {
        ungetch(c);
        return 0;
    }
    sign = (c == '-') ? -1 : 1;
    if (c == '+' || c == '-')
        c = getch();
    for (*pn = 0; isdigit(c); c = getch())
        *pn = 10 * *pn + (c - '0');
    *pn *= sign;
    if (c != EOF)
        ungetch(c);
    return c;
}

Can someone show me an example of this function in action and also explain what is up with

Code:

for (*pn = 0; isdigit(c); c = getch())
        *pn = 10 * *pn + (c - '0');

? I just don't get these two lines. Thanks a lot.

Let's say the user enters the characters "123". When the function finishes, pn[0] will be equal to 1, pn[1] will be equal to two, and pn[3] will be equal to 3, right? In this case, all characters are valid numbers so I don't see how the above will happen when the condition in the for loop is isdigit(c). All characters are digits so the function will put all numbers in pn[0]! Am I missing something? Thanks.

"The following loop fills an array with integers by calls to getint:

Code:

int n, array[SIZE], getint(int *);

for (n = 0; n < SIZE && getint(&array[n]) != EOF; n++)
    ;

Each call sets array[n] to the next integer found in the input and increments n."

That's what's written on page 96 and that is what is confusing me

Have you got it figured out now?

No!

Originally Posted by csisz3r

Let's say the user enters the characters "123". When the function finishes, pn[0] will be equal to 1, pn[1] will be equal to two, and pn[3] will be equal to 3, right? In this case, all characters are valid numbers so I don't see how the above will happen when the condition in the for loop is isdigit(c). All characters are digits so the function will put all numbers in pn[0]! Am I missing something? Thanks.

>pn[0] will be equal to 1, pn[1] will be equal to two, and pn[3] will be equal to 3, right?

Not quite. array[0] (pn[0]) will hold the value of 123. As hk_mp5kpdw demonstrated, this line:

Code:

*pn = 10 * *pn + (c - '0');

...does all the work (technically this one line does not do all the work - I am using a figure of speech.. ) - now *pn is a pointer to something, and that something is passed to the function getint() at the time it is called. The line of code above does not work on multiple elements of an array, but on one element, passed when getint() is called.

So what does *pn point to? Consider the following, which is (almost) taken straight out of the book you are looking at:

Code:

int getint(int *);

int main(void)
{
         for (n = 0; n < SIZE && getint(&array[n]) != EOF; n++);
...
return 0;
}

As you can see from the getint() function prototype, the lone parameter is a pointer to type int - that is to say, that this functions expects to get the address of a type int variable. So this is what we have given it. As others have demonstrated, this can be done a few ways, depending on what you need.

Now this line:

Code:

 for (n = 0; n < SIZE && getint(&array[n]) != EOF; n++);

There is the passing of the address which the function requires. The first time through the loop, getint() is called, and the address passed is array[0]. The variable n is not incremented until getint() returns, and when getint() does return (provided that the user just entered some nice numbers) element 0 of array will hold that number in an int form.

As I mentioned earlier, this line:

Code:

*pn = 10 * *pn + (c - '0');

is working with just one address, which corresponds to whatever address was passed at the time of the function call. Therefore, by dereferencing the pointer as it is, the data being worked on at the time is in one location, or in this case, in just one element of the array.

~/