C#•2y ago

Why are out parameters inside an if statement condition not local to the if block? [Answered]

if (dictionary.TryGetValue(key, out var val))
{

}

// why can I access val here?

if (dictionary.TryGetValue(key, out var val))
{

}

// why can I access val here?

Is this because of how it ends up compiling val above the if statement?

7 Replies

333fred•2y ago

You can access val there because the if (!TryGetValue) return; pattern is super common in .NET, and we wanted that pattern to work with out variable declarations

nathanAjacobs•2y ago

I see, so in that case when it fails, you wouldn't have to call it again to access the out parameters. That makes sense. Thanks!

333fred•2y ago

Or declare them ahead of time, like

int i;
if (!TryGetValue(out i)) return;
// Do something with i

int i;
if (!TryGetValue(out i)) return;
// Do something with i

nathanAjacobs•2y ago

I feel like that's more explicit, but I can see how that could get annoying.

333fred•2y ago

Well, that was the old way But C# 6 (iirc) added inline out variable declaration, and we wanted it to work for all scenarios The leaky scoping was controversial, but imo for the best

nathanAjacobs•2y ago

Ahh didn't know it used to be like that. If it was to make it work for all scenarios, then I totally agree.

Accord•2y ago

✅ This post has been marked as answered!

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Why are out parameters inside an if statement condition not local to the if block? [Answered]