Split number into digits

[yasirassassin]

hey sir i am having problem in splitting a number into its digits if i use remainder and division operation then i can split a number and print it in reverse order but dont know how to split that number and print that in correct order
here is my program

#include<iostre am>
using namespace std;
int main()
{
int x=12345,r,c=1;
while(c<=5){
r=x%10;
x=x/10;
cout<<r;
c++;


}

cin.get();
return 0;
}

[alexis4]

Follow this algorithm:

(12345/10000)%10 = 1
(12345/1000)%10 = 2
(12345/100)%10 = 3
(12345/10)%10 = 4
(12345/1)%10 = 5


Implement this algorithm in code and off you go!

EDIT: Obviously the "%10" in line 1 and the "/1" in line 5 is not needed. But I gave it this way to make the algorithm more obvious, in case you implement a loop.

[yasirassassin]

well mmm i did this before i read your reply i think i succeeded anyway thanks for reply i'm really grateful to you

#include<iostre am>
using namespace std;
int main()
{
int x=12345,c=1,n=1 0000,fin;
while(c<=5){
fin=x/n;
n=n/10;

c++;
cout<<fin<<endl ;
}
cout<<endl;
cout<<"at the end here you go "<<fin<<end l;

cin.get();
return 0;
}

[alexis4]

while(c<=5){
fin=x/n;
n=n/10;

c++;
cout<<fin<<endl ;
}

How can that work? You are printing fin right? So:

Loop 1: fin=12345/10000 = 1
Loop 2: fin = 12345/1000 = 12

So how could you separate the digit without the remainder?

[weaknessforcats]

This is C++. You might try:

stringstream s;
 s << 12345;
 string str;
 s >> str;

[donbock]

Does your function need to handle negative inputs?

Look at your code in the original post. Looping 5 times only works if the input is a 5-digit number. Instead, loop until x/10 is zero.

Your original algorithm collects digits from right to left. Make the change I suggested above and your original algorithm works for all positive inputs regardless of how many digits they have. On the other hand, the revised algorithms that collect digits from left to right are tuned to a specific number of digits. I suggest you use the original algorithm to fill an array with digits and then print the array contents in reverse order. That leaves you with the most flexible function.

[yasirassassin]

ok how about this one check this out
#include<iostre am>
using namespace std;
int main()
{
int x=12345,c=1,n=1 0000,fin;
while(c<=5){
fin=x/n;
n=n/10;
fin%=10;
c++;
cout<<fin<<endl ;
}
cout<<endl;
cin.get();
return 0;
}

[alexis4]

It seems correct. Does it work OK?

[nehatrivedi]

#include <iostream>
#include <cstdlib>
#include <math.h>
using namespace std;
int main()
{
int c=0,fin,r;
int x,xx,n;
cout<<"Enter the number ; ";
cin>>x;
xx=x;
while(x!=0)
{
r=x%10;
c++;
x=x/10;
}
c--;
n=pow(10,c);
while(n!=0)
{
fin=xx/n;
fin=fin%10;
cout<<fin<<endl ;
n=n/10;
}
cout<<endl;
system("PAUSE") ;
return EXIT_SUCCESS;
}

[shilpa george]

while(c<=5){
fin=x/n; //************** fin=(int)x/n;
n=n/10;

c++;
cout<<fin<<endl ;
}