Thread: help on assignment

Code:

scanf("%1.1i", &integer);

I think that reads in one char with scanf(), because the maximum chars read is 1, and so is the minimum.

[edit]
Ignore the above, I misread your post.

12345 / 100 % 10 = 2

Does that give you any ideas?
[/edit]

Well, you could check out some of the tutorials on the CProgramming home page to be better prepared for your next assignment. I'll help you out here for this one.

100 % 10 is a use of the modulus operator which I'm surprised you don't know about. The modulus operator divides two numbers, but the answer returned is the remainder from that division. I'll get to that in a second.

You know how to scan the number in, I assume. If not, it's simple.

Code:

int entry;

scanf("%i", &entry);

But, then you have to single out each digit. It's not that hard, once you have an algorithm. That's what dwks gave. I'll break it down.

Code:

(12345 / 100) % 10 //Dividing by 100gets rid of the last three digits.
(12) % 10 //Modulus by 10, removes the first digit.
2 //You are left with the second digit, 2

To get the first digit, you would change the 100 and the 10. See if you can figure it out.

Come on, it's not that hard. If you can't figure this out, seriously, consider a different hobby / career. Don't you see any pattern in the above code? You don't see anything that you could repeat? Any steps for which you could do something over and over, while you don't have your answer?


Quzah.

And in that code of yours, which also works, by the way, you do not do anything with those values. They're just sitting there and since you don't assign them to a variable, they are lost.

Code:

int digits[5];
digit[0] = ...
digit[1] = ...
digit[2] = ...
digit[3] = ...
digit[4] = ...

If you want to see how that works, work it out on a piece of paper. It's 3rd grade math.

Don't be surprised when you fail this assignment. That's a horrible way to do it. On that note:

Code:

printf("%d+",number % 10000 % 1000 % 100 % 10); /* prints the 1st digit + */

What's the "first digit"? The left, or the right?

Left:

Code:

printf("%d+", number / 10000 );

Right:

Code:

printf("%d+", number % 10 );

In either case, there's a much better way to do what you're doing.

You can avoid that whole mess by reading my last reply.


Quzah.

Beauty personified.

Code:

I got it : 
 
#include <stdio.h>  
#include<conio.h>  
int main(){  
 
int number,total=0;  
 
 
printf("enter a five digit integer: "); 
 
scanf("%d",&number);  
 
printf("%d+",number % 10000 % 1000 % 100 % 10); /* prints the 1st digit + */ 
 
printf("%d+",number % 10000 % 1000 % 100 / 10 ); /* 10ths + */ 
 
printf("%d+",number % 10000 % 1000 / 100 ); /* 100th + */ 
 
printf("%d+",number % 10000 / 1000 ); /* 1000th + */ 
 
printf("%d=",number /10000 %10); /* 10000th = */ 
 
total=("%d+",number % 10000 % 1000 % 100 % 10)+ /* adds the 1st digit + */ 
 
(number % 10000 % 1000 % 100 / 10 )+ /* 10ths + */ 
 
(number % 10000 % 1000 / 100 )+ /* 100th + */ 
 
(number % 10000 / 1000 )+ /* 1000th + */ 
 
(number /10000 %10); /* 10000th = */ 
 
printf ("%d\n",total); 
 
 
 
getch();  
}

I'd pass you. lol

Code:

#include <stdio.h>
#include <conio.h>  
int main()
{
   
    int treenef;
    printf("Enter ANY digit number:");
    scanf("%d",&treenef);
    int tree_n_ef_ff=treenef;int tre_e_n_ef_ff=0;
    while (tree_n_ef_ff>0)
    {tree_n_ef_ff=tree_n_ef_ff/10;tre_e_n_ef_ff++;}
    int treee=0;int treee_e_e=1;
    for (int i=0; i<tre_e_n_ef_ff; i++)
    {int tree_n_ef=treenef/treee_e_e;
    int j=tree_n_ef%10;printf("%d + ",j);
    treee=treee+j;treee_e_e=treee_e_e*10;}   
    printf(" = "); printf("%d ",treee);
    getch();
}