Re: st: passing indefinite no of arguments

[Nick Cox <njcoxstata@gmail.com>]

di `: list sizeof 0 '

Nick

On Fri, Mar 30, 2012 at 8:52 PM, Maarten Buis <maartenlbuis@gmail.com> wrote:
> On Fri, Mar 30, 2012 at 8:43 PM, tashi lama wrote:
>> So, say I have a do file which will add the arguments and display
>>
>> *sum
>>
>> args a b c d
>>
>> di `a' +`b'+`c'+`d'
>>
>> This one is easy since I could pass 4 arguments a, b, c and d and my do file will sum them. What if I don't know the no of arguments I am passing? I could have passed 2, 3, 5 or even 1 and find their sum.
>
> Here is a quick solution. The logic is that within a program a number
> of locals are created. The local `0' contains all arguments, and the
> local `1' contains the first argument, `2' the second, etc. This is
> not very stable, e.g. when you call -tokenize- the locals `1', `2'
> etc. will be replaced, but in a small program like the one below it is
> an easy enough solution.
>
> *---------- begin example ------------
> program drop _all
> program define sumargs
>        // count the number of arguments
>        local k : word count `0'
>
>        // add them
>        tempname sum
>        scalar `sum' = 0
>        forvalues i = 1/`k' {
>                scalar `sum' = `sum' + ``i''
>        }
>
>       // display the result
>        di `sum'
> end
>
> sumargs 2 3 7
> sumargs 4
> sumargs 2 3 4 5 6 7 8
> *---------- end example -------------
>

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