#Sequence的平均值

Question

I'm having trouble figuring out the loop in the average function. User enters sequence and if the number of numbers in the sequence is greater than 0, output avg of the numbers and repeat for another sequence. If the sequence has no numbers, exit program.

I understand that I'm telling the average function to return 0 when sum = 0 and that's why it exits after the 1st sequence (I think).

Any suggestions as to how to avoid that? Pseudocode if possible!

#include<stdio.h>
#include<stdlib.h>

double average();

int main () 
{
    while( average() ) 
    {
    }
}//end main

double average() 
{
    double n, sum = 0;
    int count = 0;

    printf ( "Enter sequence: " );
    while( scanf( "%lf", &n ) )
    {
        sum += n;
        count++;
    }

    if( sum > 0 )
    {
        printf( "average is %.2f\n", sum/(double)count );
        return 1;
    }
    else
        return 0;
    }
}

Here is my output:

Enter sequence: 3 4 5 x
    average: 4.00
Enter sequence: Press any key to continue . . .

Answer 1

#include <stdio.h>
#include <stdlib.h>

int average(void);

int main(void) 
{
    while (average()) 
        ;
    return 0;
}

int average(void) 
{
    double n, sum = 0;
    int count = 0;

    printf("Enter sequence: ");
    while (scanf( "%lf", &n ) == 1)
    {
        sum += n;
        count++;
    }
    int c;
    while ((c = getchar()) != EOF && c != '\n')
        ;

    if (sum > 0)
    {
        printf("average is %.2f\n", sum / count);
        return 1;
    }
    else
        return 0;
}

This reads anything on the line up to the newline after a conversion fails, thus setting you up for reading the next sequence. The loop test for scanf() is improved; it will exit on EOF, too. The cast in the division was unnecessary; the compiler has to convert count to double because sum is a double , even without you telling it to do so explicitly. The return type of average() is not a double; it is a boolean, which is classically spelled int . In C99 or later (which the code assumes you have; if you don't, you're stuck on Windows and need to move int c; to the top of the function), then you could #include <stdbool.h> and use a return type of bool and replace return 1; by return true; and replace return 0; by return false; .

Answer 2

I think you can initially get input for the variable named count asking for the user to enter the total no of numbers in sequence,and then get those values in sequence. And if the count is 0,then exit the program.. else continue finding average.

This is because in your case, you have to enter a non numeric char each time to end the sequence.

Answer 3

Well I ended up with this and its working perfectly:

#include<stdio.h>
#include<stdlib.h>
#include<ctype.h>

double average();

int main () {
    while( average() ) {}
}//end main

double average() {
    double n, sum = 0;
    int count = 0;

    printf ( "Enter sequence: " );
    getchar();
    while( scanf( "%lf", &n ) ){
        sum += n;
        count++;
    }//end while

    if( n > 0 ){
        printf( "    average: %.2f\n", sum/(double)count );
        return 1;
    }//end if
    else
        return 0;
}//end function