| Re: confusion around a pointer to an array of structures [message #45144 is a reply to message #45001] |
Tue, 09 August 2005 14:42   |
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Paul Van Delst[1]
Messages: 1157
Registered: April 2002
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Senior Member |
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JD Smith wrote:
> On Tue, 09 Aug 2005 16:07:14 -0400, Paul Van Delst wrote:
>
>
>> Edd wrote:
>>
>>> Henry <henrygroe@yahoo.com> wrote:
>>>
>>>
>>>> But, I can see no other way around this, aside from redesigning how I'm
>>>> storing the information. Does anyone see how to do the equivalent of
>>>> "((*a)[0]).x = 99d" in the above example? (without the awkward three
>>>> line dumb hack I've shown)
>>>> I'm sure I'm just not seeing something simple....
>>>
>>>
>>> Is there anything incorrect about saying (*a)[0].x=99d ?
>>>
>>
>> I tried the OP's method and got his error,
>>
>> IDL> ((*a)[0]).x = 99d
>> % Attempt to store into an expression: Structure reference.
>> % Execution halted at: $MAIN$
>>
>> and thought that the "correct" syntax would be something like ((*a).x)[0]
>>
>> This is what I get,
>>
>> IDL> a = ptr_new( replicate( {x:0d}, 5 ) )
>> IDL> (*a).x = dindgen(5)
>> IDL> help, ((*a).x)[0]
>> <Expression> DOUBLE = 0.0000000
>> IDL> ((*a).x)[0] = 99d
>> % Internal error: The Interpreter stack is not empty on exit.
>> IDL> help, ((*a).x)[0]
>> <Expression> DOUBLE = 99.000000
>>
>> So that works too..... sort of.
>>
>> Veird.
>
>
> Actually, it's pretty much as expected. What you wanted is:
>
> IDL> (*a)[0].x=99D
>
> Doing it your way first creates a temporary array ((*a).x) and then
> indexes and assigns a member, which is slow, and leads to the reported
> error. The original error was even more severe: you can't (usually)
> include temporary expressions on the LHS of the assignment.
>
> Since `[]' and `.' are at the same level of precedence, all you need
> to remember is to use parentheses to group `*' with all the pointers
> in the expression. No other parentheses are needed, and in fact using
> any other parentheses will likely create temporary variables, which is
> not usually what you want (and can lead to errors of the type
> mentioned).
>
> Imagine a bizzarely deeply nested data structure like this:
>
> IDL> a=ptr_new(replicate({b:ptr_new([ptr_new(replicate(1,5)),$
> ptr_new(replicate({c:ptr_new(14.)},3))])},$
> 4))
>
>
> Suppose we want the "c" field of the last of that final array of
> structures. How to approach this? One way is first to pretend there
> is no such thing as operator precedence, and just write it out without
> thinking:
>
> val=****a[1].b[1][2].c
>
> This of course, won't work. Now go through and group, using (), all
> pointers together with their leading dereference operator `*',
> starting with the innermost. There are a total of 4 pointers we need
> to consider. Here is the result:
>
> val=*(*(*(*a)[1].b)[1])[2].c
>
> See how `a' is a pointer, so `(*a)' is the group, `(*a)[1].b' is a
> pointer so `(*(*a)[1].b)' is the group, etc.? You can leave off any
> "outer" parentheses that are not followed by anything (and probably
> should if you are assigning to the element, just to get in the habit).
Oh, man! Can't you use HISTOGRAM to sort all this stuff out? :o)
paulv
--
Paul van Delst
CIMSS @ NOAA/NCEP/EMC
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