Strange value of the double type variable: -nan(0x8000000000000)

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I am confused by the value of "currdisk->currangle" after adding operation. Initially the value of "currdisk->currangle" is 0.77500000000000013, but after adding operation, it's changed to "-nan(0x8000000000000)", Can anyone explain ? Thanks! The following is the occasion of gdb debugging.

C++

3338          currdisk->currangle += (simtime - seg->time) / currdisk->rotatetime;
(gdb) p currdisk->currangle
$28 = 0.77500000000000013
(gdb) p (simtime - seg->time) / currdisk->rotatetime
$29 = 0.00833333333333325
(gdb) p (simtime - seg->time) 
$30 = 0.092592592592591672
(gdb) p currdisk->rotatetime
$31 = 11.111111111111111
(gdb) n

(gdb) p currdisk->currangle 
$32 = -nan(0x8000000000000)
(gdb) p/x (char[8])currdisk->currangle 
$52 = {0x0, 0x0, 0x0, 0x0, 0x0, 0x0, 0xf8, 0xff}
(gdb)

Then I change

C++

currdisk->currangle += (simtime - seg->time) / currdisk->rotatetime ;

to

C++

double tmp1 = (simtime - seg->time) / currdisk->rotatetime; 
currdisk->currangle += tmp1;

The value of currdisk->currangle is normal. Can anyone explain the confusing phenomenon ?

Posted 29-May-14 18:19pm

915086731

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Comments

[no name] 30-May-14 0:43am

Where's your code? What types are the other variables etc.

915086731 30-May-14 3:54am

All is double.

915086731 30-May-14 3:54am

My code is upload to git server: github.com/915086731/test.git . After making, please run "./disksim parv.hpc3323 outv ascii traceb 0 "

915086731 30-May-14 3:57am

Then I change
currdisk->currangle += (simtime - seg->time) / currdisk->rotatetime ;
to
double tmp1 = (simtime - seg->time) / currdisk->rotatetime;
currdisk->currangle += tmp1;
The value of currdisk->currangle is normal

[no name] 30-May-14 3:54am

Seriously people are not going to go and download your code to solve your problems. Get real and submit a complete question with enough information for others to see what's going on. You need to provide just enough code to reproduce the problem. If you want an answer that's how to go about it.

915086731 30-May-14 3:55am

Sorry, the project is big. It is a open source code.

915086731 30-May-14 4:50am

Seems it is a BUG of linux OS.

[no name] 30-May-14 8:39am

OK then post a solution with a full explanation so that the question is answered and closed.

915086731 30-May-14 8:57am

I still want to find the solution.

915086731 30-May-14 8:56am

assembly code
tmp1 = (simtime - seg->time) / currdisk->rotatetime;
0x0808fdf4 <disk_buffer_sector_done+559>: fldl 0x80b2208
0x0808fdfa <disk_buffer_sector_done+565>: mov -0x38(%ebp),%eax
0x0808fdfd <disk_buffer_sector_done+568>: fldl (%eax)
0x0808fdff <disk_buffer_sector_done+570>: fsubrp %st,%st(1)
0x0808fe01 <disk_buffer_sector_done+572>: mov 0x8(%ebp),%eax
0x0808fe04 <disk_buffer_sector_done+575>: fldl 0xc4(%eax)
0x0808fe0a <disk_buffer_sector_done+581>: fdivrp %st,%st(1)
0x0808fe0c <disk_buffer_sector_done+583>: fstpl -0x28(%ebp)
currdisk->currangle += tmp1;
0x0808fe0f <disk_buffer_sector_done+586>: mov 0x8(%ebp),%eax
0x0808fe12 <disk_buffer_sector_done+589>: fldl 0x284(%eax)
0x0808fe18 <disk_buffer_sector_done+595>: faddl -0x28(%ebp)
0x0808fe1b <disk_buffer_sector_done+598>: mov 0x8(%ebp),%eax
0x0808fe1e <disk_buffer_sector_done+601>: fstpl 0x284(%eax)

2 solutions

Solution 1

Nothing strange at all. This is NaN — Not A Number value. Floating-point data types obey the IEEE 754 standard which defines the meaning and binary representations of this value. Please see:
http://en.wikipedia.org/wiki/IEEE_754[^],
http://en.wikipedia.org/wiki/IEEE_754[^],
http://babbage.cs.qc.cuny.edu/IEEE-754.old/References.xhtml[^].

See also: http://en.wikipedia.org/wiki/Floating-point[^].

In newer cultured platforms, things like floating-point division by zero are allowed and give infinite values. You can use infinity operations further for other calculation. For example if you calculate 1/+Inf, the result will be zero. However, floating-point division of zero by zero and some other operations create uncertainty which is denoted by the NaN result.

This is implemented in the instruction set of the FPU subsystem of Intel-compatible CPU instruction sets. You can set a flag for a process which controls the behavior of the operations like those mentioned above: either throw hardware exception or return +Inf, -Inf or NaN. To me, the only cultured decision is using those infinities and NaN values.

—SA

Posted 29-May-14 19:01pm

Sergey Alexandrovich Kryukov

Updated 29-May-14 19:04pm

v2

Comments

915086731 30-May-14 1:22am

Thanks! But I change the sentence to
double tmp1 = (simtime - seg->time) / currdisk->rotatetime;
currdisk->currangle += tmp1;
the value of currdisk->currangle is not NoN. It does the same work as previous.

915086731 30-May-14 1:23am

And it's not division by zero.

[no name] 30-May-14 1:25am

All true but how does it relate to the specifics of the question?

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915086731 · Accepted Answer · 2014-05-30T18:37:00

tmp1 = (simtime - seg->time) / currdisk->rotatetime;
currdisk->currangle += tmp1;

The above code can also cause NaN in the later calling.

So I step into to the assembly.

tmp1 = (simtime - seg->time) / currdisk->rotatetime;
0x0808fdf4  <disk_buffer_sector_done+559>:  fldl   0x80b2208
0x0808fdfa  <disk_buffer_sector_done+565>:  mov    -0x38(%ebp),%eax
0x0808fdfd  <disk_buffer_sector_done+568>:  fldl   (%eax)

...Here, the content of %eax is 0x80bdfeb, which is the address of seg->time.
After the above "fldl (%eax) " executed, the content of register st0 is 0x8000000000000000, which represents NaN.
So the NaN is propagated to the following instructions. This is the key issue.

C++

0x0808fdff  <disk_buffer_sector_done+570>:  fsubrp %st,%st(1)
0x0808fe01  <disk_buffer_sector_done+572>:  mov    0x8(%ebp),%eax
0x0808fe04  <disk_buffer_sector_done+575>:  fldl   0xc4(%eax)
0x0808fe0a  <disk_buffer_sector_done+581>:  fdivrp %st,%st(1)
0x0808fe0c  <disk_buffer_sector_done+583>:  fstpl  -0x28(%ebp)