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What does floating point error -1.#J mean?

2011-11-02 20:15 323 查看

What does floating point error -1.#J mean?

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Recently, sometimes (rarely) when we export data from our application, the export log contains float values that look like "-1.#J". I haven't been able to reproduce it so I don't know what the float looks like in binary, or how Visual Studio displays it.

I tried looking at the source code for printf, but didn't find anything (not 100% sure I looked at the right version though...).

I've tried googling but google throws away any #, it seems. And I can't find any lists of float errors.

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May 8 '09 at 14:32




Srekel

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See also:
stackoverflow.com/questions/5541975/what-does-1-mean/… – Loki Astari
Apr 4 at 18:54
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It can be either negative infinity or NaN (not a number). Due to the formatting on the field printf does not differentiate between them.

I tried the following code in Visual Studio 2008:

double a = 0.0;
printf("%.3g\n", 1.0 / a);  // +inf
printf("%.3g\n", -1.0 / a); // -inf
printf("%.3g\n", a / a);    //  NaN

which results in the following output:

1.#J
-1.#J
-1.#J

removing the .3 formatting specifier gives:

1.#INF
-1.#INF
-1.#IND

so it's clear 0/0 gives NaN and -1/0 gives negative infinity (NaN, -inf and +inf are the only "erroneous" floating point numbers, if I recall correctly)

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