Lecture 2 Arrays

Question

I tried using the code David show to average a set of scores in an array.

#include <stdio.h>
#include <cs50.h>

float average (void);
int main(void)

{
    int n = get_int("Total scores: ");
    int scores[n];
    for(int i = 0; i < n; i++)
    {
        scores[i] = get_int("Score %i\n", i+1);
    }
    printf("Average %f", average);
}

float average(int n, int scores[])
{
    int sum = 0;
    for(int i = 0; i < n; i++)
    {
        sum = sum + scores[i];
    }
    return sum / (float) n;
}

This is the error I get.

arrayPractice2.c:14:26: error: format specifies type 'double' but the argument has type 'float (*)(void)' [-Werror,-Wformat]
    printf("Average %f", average);
                    ~~   ^~~~~~~
arrayPractice2.c:17:7: error: conflicting types for 'average'
float average(int n, int scores[])
      ^
arrayPractice2.c:4:7: note: previous declaration is here
float average (void);
      ^

I never define average as a double so I don't understand why it says there is a conflict. Any help you could provide would be greatly appreciated.

Answer 1

The error you are getting is not saying that you are passing a double instead of a float, but that a double is expected. You are passing average, which is not a variable but a function. Your prototype declaration for the function (float average (void)) is also throwing an error because it does not match the function, which need has a definition of float average (int n, int scores[]). So, you need to apply two fixes to get this to work:

First, correct your prototype declaration of average to be float average (int n, int scores[]);. Remember that your prototype declaration has to be identical to the function's declaration that comes later in your code or the compiler will think that you have two different functions.

Then, change the line printf("Average %f", average); to printf("Average %f", average(n, scores);. This will pass the necessary parameters that the function average requires, allowing it to return the average that you want to print. Remember, to call a function that has declared parameters, you need to provide the parameters.

Hope this helps and makes sense. If it does, please feel free to click the check mark next to my answer. If not, let me know and I will try to help further.

Answer 2

The code shown is passing the address of the address function to print. I suspect this is a typo, the intent is to call the average function, with the return value being used by print. Note this line has an extra set of parenthesis:

printf("Average %f", average());