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?

Answer 1

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;
}

Answer 2

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;
}

Answer 3

#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;
}