How to use a loop function with user input in C?

Question

Here is the average program I created.

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

 int main()
 {
    float f1,f2,f3;

    /* Program to calculate averages. */

    /*Asks for the numbers.*/

    printf(" Please enter three numbers.\n");
    printf ("\t" "First number please.\n");
    scanf("%f", &f1);
    printf ("\t" "Second number please.\n");
    scanf ("%f", &f2);
    printf("\t" "Third number please.\n");
    scanf("%f", &f3);

    /* Now it averages it.*/
    printf(" Thank you, wait one.\n");
    printf(" Excellent, your sum is.\n");
    printf("%f""\n", f1+f2+f3);


    printf("Your average of the sum is now!!!!\n");
    printf("%f", (f1+f2+f3)/3);

    return 0;
}

Now would I turn this into a do-while? Or an if else?

int i = 0; do{printf... scanf... i = i + 1}while(i < 3); have a 'sum' variable which will aggregate the sum of your numbers and divide it by 3 at the end (instead of three variables which is kind of bad idea). — sashkello
– sashkello, Commented Apr 27, 2013 at 5:56
In addition to @sashkello's point, you can automatically keep a sum of the entered values, as well as a count of how many have been entered. Then you can do the appropriate math at the end (careful of divide by 0, and if it were a real world project then be careful of value overflows). for (i = 0; i < COUNT; ++i) { /* ... */ } would be better if you actually know the count up front. — pickypg
– pickypg, Commented Apr 27, 2013 at 5:58
Learning C by trial and error is a bad idea, because C has a concept known as "undefined behaviour", which allows mistakes to go relatively unnoticed because the code appears to work. Coincidentally, upgrades to a compiler or porting the code to other systems reveals that undefined behaviour breaks in devastating ways, and you want to avoid it! Reading a book is a great way to learn how to write well defined code. Which book are you reading? — autistic
– autistic, Commented Apr 27, 2013 at 5:59
So like int sum sum = my numbers divided by three? As I said, extremely novice and apologize for the amount of questions I might ask. Why would you do i=i+1? And would that be (i=i+1) At the end of the syntax before I star the while (1<3);? Jeesh if that even makes sense what I'm asking. — Markus Andrew White
– Markus Andrew White, Commented Apr 27, 2013 at 6:00

Jonathan Leffler · Accepted Answer · 2013-04-27 06:21:55Z

6

If you want to repeat the whole entry and averaging process, you can wrap a loop around the code:

#include <stdio.h>

int main(void)
{
    float f1,f2,f3;

    while (1)
    {
        printf("Please enter three numbers.\n");
        printf("\tFirst number please.\n");
        if (scanf("%f", &f1) != 1)
            break;
        printf("\tSecond number please.\n");
        if (scanf("%f", &f2) != 1)
            break;
        printf("\tThird number please.\n");
        if (scanf("%f", &f3) != 1)
            break;

        printf("Your sum     is %f\n", f1+f2+f3);
        printf("Your average is %f\n", (f1+f2+f3)/3);
    }

    return 0;
}

Note that this code checks the return value from scanf() each time it is used, breaking the loop if there's a problem. There's no need for string concatenation, and a single printf() can certainly print a string and a value.

That's a simple first stage; there are more elaborate techniques that could be used. For example, you could create a function to prompt for and read the number:

#include <stdio.h>

static int prompt_and_read(const char *prompt, float *value)
{
    printf("%s", prompt);
    if (scanf("%f", value) != 1)
        return -1;
    return 0;
}

int main(void)
{
    float f1,f2,f3;

    while (printf("Please enter three numbers.\n") > 0 &&
           prompt_and_read("\tFirst number please.\n", &f1) == 0 &&
           prompt_and_read("\tSecond number please.\n", &f2) == 0 &&
           prompt_and_read("\tThird number please.\n", &f3) == 0)
    {
        printf("Your sum     is %f\n", f1+f2+f3);
        printf("Your average is %f\n", (f1+f2+f3)/3);
    }

    return 0;
}

If you want to get away from a fixed set of three values, then you can iterate until you encounter EOF or an error:

#include <stdio.h>

static int prompt_and_read(const char *prompt, float *value)
{
    printf("%s", prompt);
    if (scanf("%f", value) != 1)
        return -1;
    return 0;
}

int main(void)
{
    float value;
    float sum = 0.0;
    int   num = 0;

    printf("Please enter numbers.\n");

    while (prompt_and_read("\tNext number please.\n", &value) == 0)
    {
        sum += value;
        num++;
    }

    if (num > 0)
    {
        printf("You entered %d numbers\n", num);
        printf("Your sum     is %f\n", sum);
        printf("Your average is %f\n", sum / num);
    }

    return 0;
}

You might also decide to replace the newline at the ends of the prompt strings with a space so that the value is typed on the same line as the prompt.

If you want to check whether to repeat the calculation, you can use a minor variant on the first or second versions of the code:

#include <stdio.h>

static int prompt_and_read(const char *prompt, float *value)
{
    printf("%s", prompt);
    if (scanf("%f", value) != 1)
        return -1;
    return 0;
}

static int prompt_continue(const char *prompt)
{
    printf("%s", prompt);
    char answer[2];
    if (scanf("%1s", answer) != 1)
        return 0;
    if (answer[0] == 'y' || answer[0] == 'Y')
    {
        int c;
        while ((c = getchar()) != EOF && c != '\n')      // Gobble to newline
            ;
        return 1;
    }
    return 0;
}

int main(void)
{
    float f1,f2,f3;

    while (printf("Please enter three numbers.\n") > 0 &&
           prompt_and_read("\tFirst number please.\n", &f1) == 0 &&
           prompt_and_read("\tSecond number please.\n", &f2) == 0 &&
           prompt_and_read("\tThird number please.\n", &f3) == 0)
    {
        printf("Your sum     is %f\n", f1+f2+f3);
        printf("Your average is %f\n", (f1+f2+f3)/3);
        if (prompt_continue("Do you want to try again?") == 0)
            break;
    }

    return 0;
}

edited Apr 27, 2013 at 6:21

answered Apr 27, 2013 at 6:02

Jonathan Leffler

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5 Comments

Markus Andrew White Over a year ago

:/ I think my question wasn't asked properly, my apologies, and I appreciate this immensely. What i want is, after the average is made, is for my program to say something like printf("Would you like to use this program again? y/n?" Then loop the program after a user input.

Markus Andrew White Over a year ago

Oh man, Jonathan Leffler you definitely just answered my question right on the dot! I appreciate it so much, and while it IS complicatedly over my head, I will study this, and look up certain keywords, aka void, static int, etc. And I will learn this inside out, thank you so much!

Rich Scriven Over a year ago

I'm wondering if the numbers can be store in an array, to be sorted further down the code? I don't regularly use C, but I need to implement your examples along with an insertion sort, if possible. I have the sort already but am having trouble with the inputs from stdin. I may post a question in a minute.

Jonathan Leffler Over a year ago

@RichardScriven: yes, you could. You'd probably use a loop, and something like: for (int i = 0; i < 10; i++) { if (prompt_and_read("Next number please: ", &a[i]) != 0) { …handle error… } }.

Rich Scriven Over a year ago

Mine is a bit like that. I've posted a question

Jonathan Leffler · Accepted Answer · 2013-04-27 06:29:16Z

1

You can do this:

int main()
{
    float number, sum=0.0f;
    int index=0;
    do
    {
        printf ("\t" "Enter number please.\n");  //Asking for a number from user
        scanf("%f", &number); //Getting a number from a user
        sum+=number; //Add number entered to the sum
        i++;
    } while (i < 3);
    printf("Excellent, your average is %f\n", sum/3);
    return 0;
}

edited Apr 27, 2013 at 6:29

Jonathan Leffler

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answered Apr 27, 2013 at 6:01

Houssam Badri

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3 Comments

Markus Andrew White Over a year ago

I appreciate all of your responses, and I am glad that you guys have helped me in this, I will take these examples and build programs around them. Thanks everyone!

Houssam Badri Over a year ago

@MarkusAndrewWhite : there are many ways to improve you code, don't be afraid take your time to get improved

Markus Andrew White Over a year ago

@Houssem Bdr Thanks so much, I actually was scared at first because I've seen people get torn apart on here on how "stupid" they are, but I work hard every night at this and I actually really really enjoy all of this.

Fawzan · Accepted Answer · 2017-01-17 23:35:08Z

0

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

int main()
{
float f1,f2,f3;
char c='Y';

/* Program to calculate averages. */

/*Asks for the numbers.*/
do
{

printf(" Please enter three numbers.\n");
printf ("\t" "First number please.\n");
scanf("%f", &f1);
printf ("\t" "Second number please.\n");
scanf ("%f", &f2);
printf("\t" "Third number please.\n");
scanf("%f", &f3);

/* Now it averages it.*/
printf(" Thank you, wait one.\n");
printf(" Excellent, your sum is.\n");
printf("%f""\n", f1+f2+f3);

printf("Your average of the sum is now!!!!\n");
printf("%f", (f1+f2+f3)/3);

printf ("Do you wana continue [Y/N]...\n");
scanf("%c", &c);

}while(c!='N'&&c!='n');


return 0;
}

edited Jan 17, 2017 at 23:35

Fawzan

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answered Apr 27, 2013 at 6:05

Civa

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1 Comment

Jonathan Leffler Over a year ago

One concern with this solution is that the scanf("%c", &c) is going to pick up the newline left in the input by the previous scanf("%f", &f3), and the newline is not going to be either N or n so the loop will continue (indeed, it won't wait for the user to type before continuing). That's why my analogous code uses %1s for the format instead; the %s conversion specifier skips leading white space (such as newlines) which %c does not. You could also put a blank in the format before the %c: scanf(" %c", &c). You should really check the value returned by scanf() too.

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