A "void" function is only checked at compile time, and may still throw

[MircoBabin]

Affected page

https://www.php.net/manual/en/language.types.void.php

Current issue

• There is no mention that a "void" function is only checked by the compiler. And that the check is that "return expression;" is forbidden for a void function.
• There is no mention that no type inference is performed.
• There is no mention that a "void" function will never produce a TypeError at runtime.
• There is no mention that a "void" function can still throw an Exception.
• There is no mention that "void" is not a type, but only a return-only type.

Suggested improvement

Mention these 5 points, because they are really important.

Additional context (optional)

No response

[DanielEScherzer]

There is no mention that a "void" function is only checked by the compiler. And that the check is that "return expression;" is forbidden for a void function.

Yeah, like for #5390, we should add that, something like "Attempting to include a return statement with an expression in a void function is a compile-time error, even if the return would not be reached during execution."

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There is no mention that no type inference is performed.

What type inference is omitted that would normally be performed?

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There is no mention that a "void" function will never produce a TypeError at runtime.

We don't normally document that errors will not be thrown, what makes this special?

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There is no mention that a "void" function can still throw an Exception.

The default is that exceptions can be thrown - we also don't mention that an "int" function can still throw an Exception. What makes void different?

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There is no mention that "void" is not a type, but only a return-only type.

From the first sentence, bolding added

void is a return-only type declaration indicating the function does not return a value, but the function may still terminate.