In C programming why is my printf not printing after a loop?

Question

I need to print the sum and average of user input array. So if user inputs 2,4,6,9,10 it should print 6. However, after the loop ended my printf is not printing anything.

Even if I put the printf inside the array it only prints out 0.

#include <stdio.h>
#include <math.h>

int main()
{
    int i;
    double num[6],average, sum=0, closest;
    printf("Enter 6 doubles\n");

    for (i=0; i<6; i++)
    {
    scanf("%lf",&num[i]);
    sum += num[i];

    }
    average = sum/i;
    printf("Average %d", average); 
}

Answer 1

There are a few things you need to do in the code. You should be making sure they enter 6 numbers (in your opening post you only list 5, this will create problems). I changed the printing to use this and stripped out some variables that you don't use.

#include <stdio.h>
#include <math.h>

int main()
{
    int i;
    double sum = 0;

    printf("Enter 6 doubles\n");
    for (i = 0; i < 6; i++)
    {
        double value;
        scanf("%lf", &value);
        sum += value;
    }

    printf("Average = %f", sum / i);
}

Enter 6 doubles:

2 4 6 9 10 10

Average = 6.833333

Answer 2

This question is not a duplicate but I found the answer on StackOverflow here

The stdout stream is buffered, so will only display what's in the buffer after it reaches a newline (or when it's told to). You have a few options to print immediately:

Print to stderr instead using fprintf:

fprintf(stderr, "I will be printed immediately");

Flush stdout whenever you need it to using fflush:

printf("Buffered, will be flushed"); fflush(stdout); // Will now print everything in the stdout buffer

You can also disable buffering on stdout by using setbuf:

setbuf(stdout, NULL);

Then regarding your code, here are a few remarks:

• As described in man 3 printf the conversion specifier f already convert to double floating point values, so no need for a length modifer flag.
• The average value is also a double so if you print it as an integer %d you will lose the real valued part, consider using %f as well.

Answer 3

the following proposed code:

1. uses a proper signature for main()
2. corrects the format used in the call to printf()
3. appends a '\\n' to the format string in 'printf()' so the data is immediately output rather than after the program exits
4. gives 'magic' numbers (IE 6) meaningful names
5. properly checks for I/O errors and handles any such error
6. eliminates unneeded variables
7. does not include header files those contents are not used
8. documents why each header file is included
9. properly limits the scope of local variable 'i'
10. cleanly compiles
11. performs the desired functionality

and now, the proposed code:

#include <stdio.h>   // printf(), scanf(), perror()
//#include <math.h>
#include <stdlib.h>  // exit(), EXIT_FAILURE


#define MAX_ENTRIES 6

int main( void )
{
    //int i;
    // double num[6];
    double num;
    double average;
    double sum=0.0;
    // double closest;

    printf("Enter %d doubles\n", MAX_ENTRIES );

    for (int i=0; i< MAX_ENTRIES; i++)
    {
        if( scanf( "%lf", &num ) != 1 )
        {
            fprintf( stderr, "scanf for number failed\n" );
            exit( EXIT_FAILURE );
        }

        sum += num;
    }

    average = sum / MAX_ENTRIES;
    printf("Average %f\n", average); 
}

a typical run of the code results in:

Enter 6 doubles
1.0
2.0
3.0
4.0
5.0
6.0
Average 3.500000