Help with this problem in C language!

[stayit]

I need to fix the following error in this program in C language.

The directions:

The following function supposedly computes the sum and average of the numbers in the array a, which has length n.avg and sum point to variables that the function should modify. Unfortunately, the function contains several errors; find and correct them.

#include <stdio.h>
 
void avg_sum(double a[], int n, double *avg, double *sum)
    {
        int i;
         sum = 0;
        for (i = 0; i < n; i++)
         sum += a[i];
         avg = sum / n; 
    }

The part sum+= a[i]; gives me an error [Error] invalid operands `double *' and `int' to binary `operator \'
in the line [i]sum += a;

Thanks. Please advise.

PS. I am using the problem C - Free 4.1 to write the program in a Vista 64 bit system. If you have a better program, please don't hesitate to tell me.

[JosAH]

In the body of that function replace every occurence of 'sum' by '*sum' and 'avg' by '*avg'. Both are pointers to doubles; you don't want to change the value of the pointers themselves, you want to change the doubles they point to.

kind regards,

Jos

[donbock]

@stayit
It is well worth your time to learn how to interpret error messages.

I am a little skeptical of the error message you gave us. Could it perhaps have been the following:

[Error] Invalid operands 'double *' and 'int' to binary operator '/'

and perhaps it was reported against line 9 instead of line 8?

Whenever possible you should cut the error message out of the log file and paste it into your post -- even a minor typo (like changing slash to backslash) can delay a helpful reply.

The error message tells you that the operands to the '/' operator are invalid; and that one of those operands is of type 'double *' and the other is of type 'int'. This should be enough to make you focus on the expression "sum / n". That's when you discover that the error message is right, sum is a 'double *'. As Jos said, you don't want to divide the pointer value by an integer, you want to divide the value being pointed at by an integer.

Bonus question: what happens if somebody calls avg_sum and passes a value of 0 for n?

[stayit]

Thanks, guys. That was very helpful. I think that by calling a 0 instead of n, it restricts the amount of numbers that you can average. At least, that was my initial thinking. Apologize, I am not very good at programming even though I have just a little experience in Java, Python.


@donbock

[donbock]

@stayit
I was hinting that your program fails dramatically if the caller sets n to zero -- and there is no compiler message to help you find this problem. You need to look at each spot where n is used and ask yourself what happens if it is zero.

[petpenguin]

But the logic is correct? when I use it in a main function, the math somehow returns incredibly wrong.

[donbock]

@petpenguin: the code in the original post is deeply flawed in its use of pointers.