what is the type of block expression that contains a diverging expression

[xmh0511]

Consider this example:

fn show()->std::io::Result<()>{
    let r = panic!();
    let i = 0;
}

According [items.fn.body], the function's body is conceptually rewritten as

return {
       let r = panic!();
       let i = 0;
};

However, https://doc.rust-lang.org/stable/reference/expressions/block-expr.html#block-expressions says:

The type of a block is the type of the final operand, or () if the final operand is omitted.

In this case, the final operand is (). However, the code could be compiled only if the block type is a never type !`, which could be coerced to any output type.

we may need to add the extra case to define the type of the block

If the final operand is evaluated, the type of a block is the type of the final operand, or () if the final operand is omitted. Otherwise, if the block contains any diverging expression, the block type is a never type.

[ehuss]

Yea, if any statement diverges, and the block is missing a tail expression, then I believe tail expression is also treated as !. I'm going to close as a duplicate of #1033 which has some more details.