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if I multiply 0.29 with 50, I get 14.5: Only with this line: my \$num4 = sprintf ("%.64f", \$num7); I get the exact result in "double" (14.999...).

The only difference between:

```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.

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In reply to Re: IEEE-754 calculation - best way? by BrowserUk
in thread IEEE-754 calculation - best way? by Biker1971

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