How can you switch a float to a time format? - c++

Question

Hello I am a begineer so please be nice :)..
In C++, if I have a float like 12.5 or 13.25, how can I convert this to show as 12:30 and 13:15?
Thank you so much.

Answer 1

float time = ...;
Simly casting the float to an int will chop off the fractional part, leaving you with the hours.
int hour = static_cast<int>(time);
If you subtract the hours, then all that's left is the minutes as fractions of an hour. If you multiply this by 60, you get the remainder in minutes. Again, cast to an int to chop off the seconds, leaving only minutes.
int minute = static_cast<int>((time-hour)*60);
Then display the hours, and a colon. To display the minutes properly, tell it to always use a width of 2 characters, and fill in the blanks with a '0'. This makes it say 02 instead of just 2.
std::cout << hour << ':' << std::setw(2) << std::fill('0') << minute;

Answer 2

As with any problem, you need to break it down into its component parts and attack each one separately.
What you have:
A floating-point number of hours.
What you need:
To print the integer number of hours.
To print a ":"
To print the fractional number of hours as a proportion of 60 minutes.
So, now, we can address each part of the problem.
Let's say the input is:
const float time = 13.25;
The first part is quite easy — truncating a floating-point variable can be done using the mathematical floor function, but all you really need to do is cast to int to get the same effect:
std::cout << (unsigned int)time;
The second part is also really easy:
std::cout << ':';
The third part takes a little more work. We need to discard everything but the fractional part. We can do that by subtracting the integer part:
time - (unsigned int)time
Then we must transform the value so that instead of being a proportion of the range [0.00,1.00), it's a proportion of the range [0,60), simply by multiplying by 60:
60 * (time - (unsigned int)time)
We're left with:
const float time = 13.25;
std::cout << (unsigned int)time;
std::cout << ':';
std::cout << 60 * (time - (unsigned int)time);
// result: 13:15
(live demo)
For a general solution, we also want to show a leading zero if there's only one digit:
const float time = 12.10;
std::cout << (unsigned int)time;
std::cout << ':';
std::cout << std::setw(2) << std::setfill('0');
std::cout << 60 * ((time - (unsigned int)time) / 100);
// result: 12:06
In reality, to avoid rounding errors and possible overflows, you'd drop the float altogether and simply store integer minutes:
const unsigned int time_mins = (12*60) + 6;
std::cout << (time_mins / 60);
std::cout << ':';
std::cout << std::setw(2) << std::setfill('0');
std::cout << (time_mins % 60);
// result: 12:06
(live demo)
Or, y'know, use an actual time/date type.