reversing float

[HeLLBoY1]

Take floating point number from user and reverse it
For example user enter
123.45

Your program should reverse it
Output should be 54.321
with basic if else while....

[48T7M4Gy]

Excellent challenge. We look forward to seeing your program, perhaps even pseudocode. My guess is your solution will incorporate <strings>'s

[Marcus Aseth]

Basic if else while?
Can't you just do:

float ReverseFloat(string s)
{
	reverse(s.begin(), s.end());
	return stof(s);
}

int main()
{
	string strNum{};
	cin >> strNum;
	
	float number = ReverseFloat(strNum);

	return 0;
}

Well, maybe is an homework, so have fun :P

[HeLLBoY1]

#include <iostream>
using namespace std;

int main()
{
float reversedNumber = 0.0, remainder;
float n;

cout << "Enter an integer: ";
cin >> n;

while(n != 0.0)
{
remainder = n%10.0;
reversedNumber = reversedNumber*10 + remainder;
n /= 10;
}

cout << "Reversed Number = " << reversedNumber;

return 0;
}
why this one doesnt work

[integralfx]

Please use code tags. http://www.cplusplus.com/articles/jEywvCM9/
You can edit your post, highlight your code and click the <> button on the right.

remainder = n%10.0;
operator% isn't defined for floating point numbers. You'll need to use fmod instead.

while(n != 0.0)
{
remainder = n%10.0;
reversedNumber = reversedNumber*10 + remainder;
n /= 10;
}

This will become an infinite geometric progression, but will eventually end due to the finite precision of float/double. For example, take n = 12.34, just dividing n by 10.

12.34
1.234
0.1234
0.01234
0.001234
0.0001234
... etc

As suggested above, convert the float to a string(stringstream or to_string) and reverse that string(reverse or rbegin/rend).