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Thanks for the explanation. I also missed that the parameter in question was an output parameter which explains why its prototype was different.
On 6/14/18, 19:03, "C400-L on behalf of Barbara Morris" <c400-l-bounces@xxxxxxxxxxxx on behalf of bmorris@xxxxxxxxxx> wrote:
On 2018-06-14 10:38 AM, Neale Ferguson wrote:
> I exercised the API and int * is correct.
Both "int" and "int *" behave the same for a linkage(OS) call.
https://www.ibm.com/support/knowledgecenter/ssw_ibm_i_73/rzarh/cpprog278.htm
It says that all arguments except pointers and aggregates are copied to
a temporary and the address of the temporary is passed.
So if the prototype has just "int", it will pass a pointer to the
integer parameter. If the parameter is "int *", it's already a pointer,
so it will just pass the parameter directly.
-- Barbara
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