print 0.29 * 50;;
14.5
And: printf "%.20f\n", 0.29 * 50;;
14.49999999999999800000
is what gets printed out; not what is calculated and stored internally.
All floating point calculations are done internally using IEEE754 semantics; if you want to see the full internal results use printf to display the results.
As for your perceived difference when gratuitously using eval: printf "%.20f\n", eval eval( 0.29 ) * eval( 50 );;
14.50000000000000000000
The first thing to note is that the first two evals make no difference whatsoever: printf "%.20f\n", eval 0.29 * 50;;
14.50000000000000000000
And the reason for that difference is because (in this form) eval expects a string. So the result of the multiplication (eval( 0.29 ) * eval( 50 )) is converted to a string -- using the same rules as print -- before being passed to the final eval.
You get identically different results if you do: $n = 0.29 * 50; printf "%.20f\n", "$n";;
14.50000000000000000000
Because the internal (IEEE754) representation has been lost by the conversion to a string.
Bottom line: Stop worrying about it. Do your floating point math in the usual way (no eval's; no stringifications), secure in the full IEEE754 semantics will be used internally; and then use printf/sprintf (*only*) when you need to display them.
With the rise and rise of 'Social' network sites: 'Computers are making people easier to use everyday'
Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.
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