On Mar 8, 10:34 pm, "R.G. Stockwell" <n...@email.please> wrote:
> "yp" <Yaswant.Prad...@gmail.com> wrote in message
>
> news:1173383911.537437.293170@h3g2000cwc.googlegroups.com...
>
>
>
>>>> > sza = 45.0D ;C (static)
>>>> > int_LUT, wave, nwave, 0.0D, 0.0D, 0.0D, chl[i], foq0 ;Case1
>>>> > int_LUT, wave, nwave, sza[i], vza[i], dphi[i], chl[i], foq
> ...
>> Not really.
>> In Case 1 the arguments are passed by value (which is shown in the
>> code)
>> in Case 2 (comment the line left to Case1 and uncomment the line left
>> to Case2) the arguments are passed by regerence.
>
> I still do not follow what you mean. If you look at the three lines
> above, you have sza = 45.0D.
>
> Then you execute the following 2 procedure calls:
> IDL> int_LUT, wave, nwave, 0.0D, 0.0D, 0.0D, chl[i], foq0
> the next call you have shown is:
> IDL> int_LUT, wave, nwave, sza[i], vza[i], dphi[i], chl[i], foq
>
> In the first case, the 3rd parameter is 0.0D.
> In the second case, the 3rd parameter is sza[0] = 45.0D.
>
> Anyways, I assume what you actually are doing is something
> different than what is shown in the messages you posted.
>
> Why not reduce this problem to a very short example, and
> post the code here so we can take a look at it.
>
> Cheers,
> bob
Hi Bob,
Sorry about the confusion. Here I go again...
--------------
null=0.0D
int_LUT, 412.5, 1, 0.0D, 0.0D, 0.0D, 0.03D, result1
int_LUT, 412.5, 1, null, null, null, 0.03D, result2
print,'RESULT 1: ', (result1[*])
print,'RESULT 2: ', (result2[*])
---------------
RESULT 1: 0.090099994
RESULT 2: 0.090071241
I hope this is clearer now.
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