C - turn 3.00 into 3 [closed] - c

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I've got multiple float numbers to print:
{3.00, 3.20, 2.23, 5.00, 3.40 }
They all have 2 decimal places, however if there are two zero's after a decimal, it needs to be displayed as a whole number.
How can I convert 3.00 into 3 and 5.00 into 5?
it is conversion not printing

Answer 1

There is no way to directly (i.e. using only the format specifier) print either two decimals or zero for a number ending in e.g. 3.20.
You need to check if the number ends with .00 beforehand; for example:
if((int)round(num * 100) % 100 == 0) // if decimals are `.00`
printf("%.0f", num);
else // not `.00`
printf("%.2f", num);
If 3.2 is okay for 3.20, then the %g specifier might also be of some use.

Answer 2

Just assign them to an int array or just type cast them in int for your purpose.
int i = 3.00; //lhs is int and rhs is double i will store 3 only
(int)3.00 //cast
printing them with %.f will also work

Answer 3

I think you cannot do this with a general format specifier or similar. You might need to treat each number separately and adjust the printf accordingly.
First you need to detect if you have digits that are 0.
You could do it like this
if (fabs(number - round(number)) < 0.01)
printf("%.2f", number);
else
printf("%d", (int) number);
This is just meant as a hint in the right direction. You might improve it a bit on your own.
Edit:
I assume that you want two digits for 3.20 and not 3.2 only.

Answer 4

Casting to integer type will truncate the positive integers down to integer part:
float f = 2.00;
printf("f: [%f], [%d]", f, (int) f);
f: [2.000000], [2]
If you want to leave 2.50 as 2.5 then checking for .00 is necessary:
if( (int) round(num * 100) % 100 == 0) // if decimals are .00
printf("%.0f", num); // truncate all
else // not .00
printf("%.2f", num); // truncate to 2 decimal places