How do you reverse a 6 digit number in C plus plus?

Answer 1

The number of digits is immaterial, the principal is exactly the same. The key to reversing any number is in the following user-defined function:

uint reverse(uint num)

{

uint rev=0;

while(num)

{

rev*=10;

rev+=num%10;

num/=10;

}

return(rev);

}

The following code demonstrates how to reverse a 6 digit number entered from the console.

#include<iostream>

#include<string>

#include<sstream>

typedef unsigned int uint;

uint reverse(uint num)

{

uint rev=0;

while(num)

{

rev*=10;

rev+=num%10;

num/=10;

}

return(rev);

}

int getrange(uint digits,uint& min,uint&max)

{

if(!digits)

return(-1);

min=1;

while(--digits)

min*=10;

max=min*10-1;

return(0);

}

uint inputnumber(const std::string prompt,const uint min,const uint max)

{

using namespace std;

uint result=0;

while(1)

{

cout<<prompt;

string s;

getline(cin,s);

if(s.size())

result=stoul(s,0,10);

if(result<min result>max)

cout<<"The number must be in the range "

<<min<<".."<<max<<"\n"

<<"Please enter another number.\n"<<endl;

else

break;

}

return(result);

}

uint inputnumber(const uint digitcount)

{

using namespace std;

uint min=0, max=0;

if(getrange(digitcount,min,max)==-1)

return(0);

std::stringstream prompt;

prompt<<"Enter a "<<digitcount<<" digit number: ";

return(inputnumber(prompt.str(),min,max));

}

int main()

{

using namespace std;

uint num=inputnumber(6); // 6 digits.

if(num)

{

uint rev=reverse(num);

cout<<"Reverse: "<<rev<<"\n"<<endl;

}

return(0);

}

Output:

Enter a 6 digit number: 123456

Reverse: 654321