return value

Scanf's return value is the number of specifiers that you give it.

how can i make my program accept 2 integers and quit if it is a non - integer value. using scanf

I've tried

Code:

 while ( scanf("%d %d", &num, &num_max) == 2)


----body of code

---return (0);

}

However that still made the program terminate wrong.


Here is my code:

Code:

/* numberSQUARED.c shows a integer, its squared and cubbed sequence in a given length*/

#include <stdio.h>
#include <stdlib.h>  /* system needed */


int main(void)

{


    int num;
    int ndx;
    int num_max;


    printf("Enter a min number and a max.");
    scanf("%d %d", &num, &num_max);



    /* format the screen */

    printf("num SQUARED CUBBED\n\n");

    /* output the results */

    for (ndx = num; ndx <= num_max; ndx++)
       {
           printf("%d %6d %6d", ndx, ndx * ndx, ndx * ndx * ndx);
    printf("\n");


    }



    printf("\n");
    system("pause");
    return (0);





}

I don't think that it can be done like that. scanf will ( I think ) automatically convery doubles/floats/etc to ints if the user inputs one (like rounding).

however, I'm not sure. don't pay much attention to me

read this page for all you need to know about scanf()
http://www.mcsr.olemiss.edu/cgi-bin/man-cgi?scanf+3

Thanks but not very helpful man.

You seemed to have all the bits, except your code didn't contain your while loop.

> Scanf's return value is the number of specifiers that you give it.
No it isn't - it's the number of conversion-assignments.

If you have
while ( scanf("%d %d", &num, &num_max) == 2)

and you type in 123 456
Then the return result will be 2

if you type in 123 quit
then the return result will be 1

Originally posted by kinghajj
I don't think that it can be done like that. scanf will ( I think ) automatically convery doubles/floats/etc to ints if the user inputs one (like rounding).

however, I'm not sure. don't pay much attention to me

Then why bother posting?

Why don't you use if>else statements?

On another note, if your variables are type int you can automatically convert whatever they put to int with somehting like this


number=(int)(number2);

Or better yet use atoi functions..


Hope that helps..

You seemed to have all the bits, except your code didn't contain your while loop.

I swear to god, last night i included this statement and it didn't work. I must have been tired or something. Thanks for your help. Here is the desired code.

Code:

/* numberSQUARED.c shows a integer, its squared and cubbed sequence in a given length*/

#include <stdio.h>
#include <stdlib.h>  /* system needed */


int main(void)

{


    int num, num_max, ndx; /* first num, maximum number and ndx counter */


    printf("Enter a min number and a max.");

    while (scanf("%d %d", &num, &num_max)== 2){

           /* format the screen */

            printf("num SQUARED CUBBED\n\n");

            /* output the results */

            for (ndx = num; ndx <= num_max; ndx++)
                {
                     printf("%d %6d %6d", ndx, ndx * ndx, ndx * ndx * ndx);
                     printf("\n");
                }



           printf("\n");
           system("pause");
           return (0);


     }  /* end of while loop */


return (0);


}

> Scanf's return value is the number of specifiers that you give it.
No it isn't - it's the number of conversion-assignments.

Isn't the correct term ' conversion specifier' ? That is what I meant in my original post.

>>>Scanf's return value is the number of specifiers that you give it.
>>No it isn't - it's the number of conversion-assignments.
>Isn't the correct term ' conversion specifier' ? That is what I meant in my original post.

The scanf function returns the value of the macro EOF if an input failure occurs before any conversion. Otherwise, the scanf function returns the number of input items assigned, which can be fewer than provided for, or even zero, in the event of an early matching failure.