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Problem of INT() is that it is not necessarily predictable in the sense e.g. INT(5.D0/1.D0) is not necessarily 5 but cal also happen to be 4. NINT() does not really help as NINT(4.8) = 5. In other words, I would need exact arithmetics. So I tried to write a simple function:

FUNCTION f_int(x) RESULT(i_x) REAL(KIND=8) x, DEPS INTEGER i_x PARAMETER ( DEPS = 5.D-15 ) IF ( ABS( NINT(x) - x ) .GT. DEPS) THEN i_x = INT(x) ELSE i_x = INT(NINT(x)) ENF IF END FUNCTION f_int

My question is, does this make sense for the purpose I need and what can I be missing (while I realise that if I wanted to do things like e.g. 5.99999999978/1.999999999999999 on purpose, it would probably fail but I can live with this)?

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>>INT(5.D0/1.D0) is not necessarily 5 but cal also happen to be 4.

Incorrect, the above will produce 5.

However note, INT(SomeEpressionWithErrorThatMathematicallyShouldResultIn5.0D0 / 1.0D0) may not necessarily be 5.

Sketch of questionable code

ValueWithError = 0.1D0 ! infinitely repeating fraction (IOW contains error in LSB)

...

Result = DoubleFunctionThatProduces~5.0D0using(ValueWithError)

I_X = INT(Result / 1.0D0) ! not necessarily 5.

Consider using:

nBitsPrecision = ... ! you specify here

if(Result .NE. 0.0D0) Result = Result + (NEAREST(Result,Result) * (2.0D0**nBitsPrecision))

I_X = INT(Result)

You can also use EPSILON to similar effect.

Jim Dempsey

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My question is, does this make sense for the purpose...

You have not yet declared what your purpose is.

In other words, I would need exact arithmetics.

Not available with machine floating point.

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@Leos P.,

It would appear you're looking for the ANINT standard intrinsic function in Fortran: https://software.intel.com/en-us/node/679004

If not, you would need to furnish more details.

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