how to reverse digits of a real no? - c++

Question

i know to reverse the digits of a number(integer) in c but my questions is how can we reverse the digits of a real no
for e.g 1200.23 gets reversed as 32.0021
code for integers
#include <stdio.h>
/* Iterative function to reverse digits of num(integer)*/
int reversDigits(int num)
{
int rev_num = 0;
while(num > 0)
{
rev_num = rev_num*10 + num%10;
num = num/10;
}
return rev_num;
}
/*Driver program to test reversDigits*/
int main()
{
int num = 4562;
printf("Reverse of no. is %d", reversDigits(num));
getchar();
return 0;
}

Answer 1

Obviously your paramater needs to be floating point type.
You could convert double to string then reverse the string then convert back to double:
#include <string>
#include <sstream>
template<typename T>
std::string toString(T t)
{
std::stringstream ss;
ss << t;
rteurn ss.str();
}
double reversDigits(double num)
{
std::string s = std::to_string(num);
std::reverse(s.begin(), s.end());
double d;
std::stringstream ss(s);
ss >> d;
return d;
}

Answer 2

You could use the following code, where the iostream library is being used:
#include <iostream>
#include <math.h>
using namespace std;
int main() {
double pass = 0;
double n;
double result=0;
cout << "Please enter a number to reverse" << endl;
cin >> n;
while (trunc(n / pow(10.0, pass)) > 0) {
result = fmod((n / pow(10.0, pass)), 10.0);
cout << trunc(result);
pass = pass + 1;
}
cout << endl;
system("pause");
return 0;
}
This uses the methods (12345/10), which gives you 1234.5 which then truncates into 1234. The above code also use the modulo method ('%' is the operator for modulo). For example, if you do 12345 % 10, the program displays the remainder of 12345/10, which in this case, would be 5.
I used this method in a loop to get the desired output.
PLEASE NOTE THAT I MADE THIS PROGRAM FOR A CLASS, SO IF THIS IS NOT EXACTLY WHAT YOU ARE LOOKING FOR, KINDLY DISREGARD THIS POST.

Answer 3

Try this code-
#include <stdio.h>
#include<stdlib.h>
#include<string.h>
float reversDigits(float num)
{
char buffer[20],ch,*ptr;
int i,j,len;
snprintf(buffer,20,"%.3lf",num);
len = strlen(buffer);
for(i=0,j=len-1;i<j;i++,j--)
{
ch=buffer[i];
buffer[i]=buffer[j];
buffer[j]=ch;
}
buffer[len] = '\0';
return atof(buffer);
}
int main()
{
float num = 45000.062;
printf("Reverse of no. is %.5lf\n", reversDigits(num));
return 0;
}
Am passing a number with 3 digits after ., so i am using %.3lf in snprintf, if you simply use %lf means, while converting it to string it will add zeros and some numbers. To avoid that am using, else you need to do a little bit of more work to remove it!
Note: while converting string to float else float to string, in decimal position it will add some extra number or zero. The programmer need to take care of it!