Re: sort error when NaN and NA are both present

[Mark Messer]

| > Every comparison with NaN is false. NA is just another name for NaN in

| > Octave. So NaN are unsorted. This is part of the IEEE 754 standard.

| > NA is a particular NaN value, so it would be possible to sort NA

| > vs. other NaN values. But I'm not sure it is worth the effort.

 

 

It looks like this is not a bug. Even though NaN and NA have different bit patterns, both are NaN values. Octave functions and operators treat NA just like NaN.

octave:772> num2hex(NaN)

ans = 7ff8000000000000

octave:773> num2hex(NA)

ans = 7ff840f440000000

octave:774> isnan(NA)

ans =  1

octave:775> isequalwithequalnans(NaN,NA)

ans =  1

octave:776> NaN==NaN

ans = 0

octave:777> NaN!=NaN

ans =  1

octave:778> NaN==NA

ans = 0

octave:779> NaN!=NA

ans =  1

octave:780> NA==NA
ans = 0
octave:781> NA!=NA
ans =  1

 

2012/1/17 John W. Eaton <address@hidden>

 

On 17-Jan-2012, Jordi Gutiérrez Hermoso wrote:

 

| On 17 January 2012 12:22, Mark Messer <address@hidden> wrote:

| > I am not sure if this is worth a bug report or not. These ought to have the

| > same answer.

| >

| > I am using 32 bit Windows Vista, and Octave 3.4.3.

| >

| >

| > octave:713> sort([NaN,NA,inf,1,-inf])

| > ans =

| >

| >   -Inf     1   Inf   NaN    NA

| >

| > octave:714> sort([NA,NaN,inf,1,-inf])

| >

| > ans =

| >

| >   -Inf     1   Inf    NA   NaN

|

| Every comparison with NaN is false. NA is just another name for NaN in

| Octave. So NaN are unsorted. This is part of the IEEE 754 standard.

 

 

NA is a particular NaN value, so it would be possible to sort NA

vs. other NaN values.  But I'm not sure it is worth the effort.

 

jwe