Re: sizeof

Andreas Schwab wrote:
 >On Mai 11 2022, Russell Shaw wrote:
 >
 >> "int()" is not an expression, but a type-id for a function pointer "int(*)()"
 >
 >Where does it say that?

*****************
7.2 Properties of expressions
7.2.1 Value category
2) Every expression belongs to exactly one of the fundamental classifications in 
this taxonomy: lvalue, xvalue, or prvalue. This property of an expression is 
called its value category.
*****************

Does "int()" look like a lvalue, xvalue, or prvalue ?


*****************
9.3.3 Ambiguity resolution [dcl.ambig.res]
1)
...
Just as for the ambiguities mentioned in 8.9, the resolution is to consider any 
construct that matches the syntax of a declaration to be a declaration.
*****************

Note that a type-id is a declaration syntax.


The parameter "int()" of "void f(int())" is a type-id because:

template-parameter :
   type-parameter

type-parameter :
   type-parameter-key identifier(opt) = type-id


Therefore, decaying "int()" to "int(*)()" is not performed on an expression, so 
none of the prohibitions about standard conversions (an expression concept) applies.



*****************
9.3.3 Ambiguity resolution [dcl.ambig.res]
2)
...
void foo(signed char a)
{
   sizeof(int());                // type-id (ill-formed)
   ...
}
*****************

The standard has an error because "int()" is a valid type-id

void f(int());