Dealing with accuracy problems in floating-point numbers?

A very simple and effective way to round a floating point number to an integer.

A very simple and effective way to round a floating point number to an integer: int rounded = (int)(f + 0.5); Note: this only works if f is always positive. (thanks j random hacker).

Assuming f is positive. – j_random_hacker Feb 26 '09 at 15:34 Yes "columns" is always positive in this application. – AndyUK Feb 26 '09 at 16:26 @j_random_hacker - you could use absolute value, in theory.

– Moshe May 28 '10 at 15:42 2 @Moshe: Not sure abs() would buy you much, since presumably you want the final answer to have the original sign and that will mean you need to "invert" the abs() by multiplying by the original sign. Probably simpler just to replace the 0.5 by (0.5 - (f – j_random_hacker May 30 '10 at 11:06.

When you use floating point arithmetic strict equality is almost meaningless. You usually want to compare with a range of acceptable values. Note that some values can not be represented exactly as floating point vlues.

See What Every Computer Scientist Should Know About Floating-Point Arithmetic and Comparing floating point numbers.

If you haven't read it, the title of this paper is really correct. Please consider reading it, to learn more about the fundamentals of floating-point arithmetic on modern computers, some pitfalls, and explanations as to why they behave the way they do.

That's a great article. – Scottie T Feb 26 '09 at 14:57.

There's no accurracy problem. The result you got (1.9999999999999996) differed from the mathematical result (2) by a margin of 1E-16. That's quite accurate, considering your input "4.600".

You do have a rounding problem, of course. The default rounding in C++ is truncation; you want something similar to Kip's solution. Details depend on your exact domain, do you expect round(-x)== - round(x)?

1 for noticing the real problem. – David Thornley Feb 26 '09 at 19:05.

If accuracy is really important then you should consider using double precision floating point numbers rather than just floating point. Though from your question it does appear that you already are. However, you still have a problem with checking for specific values.

You need code along the lines of (assuming you're checking your value against zero): if (abs(value).

1 I think you mean "abs(computed_value - expected_value).

On computers, floating point numbers are never exact. They are always just a close approximation. (1e-16 is close.) Sometimes there are hidden bits you don't see.

Sometimes basic rules of algebra no longer apply: a*b! = b*a. Sometimes comparing a register to memory shows up these subtle differences.

Or using a math coprocessor vs a runtime floating point library. (I've been doing this waayyy tooo long.) C99 defines: (Look in math. H) double round(double x); float roundf(float x); long double roundl(long double x); .

Or you can roll your own: template inline int ROUND(const TYPE & x) { return int( (x > 0)?(x + 0.5) : (x - 0.5) ); } For floating point equivalence, try: template inline TYPE ABS(const TYPE & t) { return t>=0? T : - t; } template inline bool FLOAT_EQUIVALENT( const TYPE & x, const TYPE & y, const TYPE & epsilon ) { return ABS(x-y).

You can read this paper to find what you are looking for. You can get the absolute value of the result as seen here: x = 0.2; y = 0.3; equal = (Math. Abs(x - y).

– MSalters Feb 26 '09 at 15:25 -1, for exactly the reason MSalters gave. Decimal numbers are useful for working with money not because they have superior precision but because your imprecise calculations will be identical to everyone elses'. In all other respects decimal numbers suffer from the exact same problems.

– j_random_hacker Feb 26 '09 at 15:32 Although there are some libraries that store fractions.4.6 would be 4 + 3/5 in one of those. They only fall apart when given an operation impossible to manage as a fraction, like multiplying by pi. – Zan Lynx Feb 26 '09 at 15:40 They solve the OP's problem, isn't that the point?

– Can Berk Güder Feb 26 '09 at 16:01 1 @Can: They may solve this particular instance, but there definitely exist values of mw, p, g and h for which the exact same problem will recur. That's what makes this solution a hack -- it only works for a few cases, not for all cases. – j_random_hacker Feb 26 '097 at 5:24.

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