C++ rounding to nearest multiple of 5

figured it out

logic for me went

first round up any decimal points

then take the number, then using remainders and an if statement to find the closest multiple of 5.

i wont post code, since (no offense to the maquarie guys here) but I dont wanna be burnt by plagarising.

 

well bugger it, heres my solution to the issue

first turn it into a whole number
numberToRound = ceil(x-0.5)

ensure it needs further rounding
if (numberToRound%5 != 0)

then the statement to round to the closest multiple of 5

int rem, x;
rem = numberToRound%5;
if (rem > 2)
{
x=5-rem;
numberToRound = numberToRound+x;
}
else
{
numberToRound = numberToRound-rem;
}


and it works too (thats the important part)

the problem is that the lecturer would prefer a solution that DOESNT use if statements... and thats something thats got me completely baffled. any suggestions on that (mind you im sticking with this solution as im pretty happy with it )

 

How does it handle negative numbers?
If you don't have to handle negative numbers then its a piece of piss if you do then i don't think it can be done

Camh's solution handles positive numbers without an if statement.

 

no negative numbers need be calculated.

im gonna stew on this a bit more today.

 

You can cheat and use the ternary operator ?: - its an if but not an if

int n = static_cast<int>(rounf(f))
return ((n+2*(n ? n/abs(n) : 1))/5)*5)

Assuming sane division semantics with negative numerator. Otherwise you'll need to use div(3).

Omit the first line for integer input.

It's essentially the same as my first post, except for negative numbers you need to subtract 2 instead of adding 2. The expression n/abs(n) yields -1 or 1, which can be used to reverse the sign of +2 for negative inputs. Unfortunately n==0 needs to be handled without a divide by zero needing the ternary operator ?:

 

yea, see they say, sure use any solution BUT we PREFER (will award marks) if you code a solution using ONLY what you've learnt in lectures thus far...

no typecasting, no if statements to name a few.

heck, it works, and Im happy.

 

Hmm no if statements and no typecasting, i like a challenge

Try this - i haven't simplified it and watch the brackets

( ( ( (x+2)/5 ) - ( x/5 ) ) + ( x / 5 ) ) * 5

Only works with unsigned integers though

 

The -x/5 and the +x/5 will cancel each other out leaving you with camh's code on the first page

Just like to add awake for over 42 hours and it stands out like a sore thumb.

 

Heh i was up waaay to late last night and had a few beers, i logged on now because i just relised what i had posted was really stupid and needed changing - anyway back to bed for me

 

the final solution ended up being something like


floor((number/5)+0.5)*5

yea... I cant believe its that simple lol (but im awful at maths)