Explicitly saying ref or out when invoking a function

[Michiel Helvensteijn]

Ary Borenszweig wrote:

> In C# when you define a function that takes an out or ref parameter,
> when invoking that function you must also specify ref or out. For example:
> 
> void fun(ref uint x, double y);
> 
> uint a = 1;
> double b = 2;
> fun(ref a, b);
> 
> What do you think?

I see what you mean, however:

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swap(ref a, ref b);

I think that's overly verbose for a call with very descriptive function name
to begin with.

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Perhaps an IDE can use formatting to show you the ref parameters.

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The designer of the function can always request pointers, so the caller has
to explicitly pass addresses.

-- 
Michiel Helvensteijn