Why doesn't this enum convert to int?

Question

Why does the following code not compile under g++ (C++14), MSVC (C++14), or ARM (C++03)?

The named Error instance calls the integer constructor, but the anonymous Er

Answer 1

The problem is that code

Error(value_1);

is a declaration of a variable value_1 of type Error.

This is a legacy from C language that uses expressions as part of type declaration.

For example int *i is a pointer to int because it says that expression *i should evaluate to type int. More examples of this:

• int (*func)() is a pointer to function returning int because expression (*func)() evaluates to type int.
• int *p[8] is an array of pointers to int because expression *p[x] evaluates to type int.
• int (*p)[8] is a pointer to array of 8 int's (int[8]) because expression (*p)[x] evaluates to type int.
• int (*(*p[8])())() is an array of 8 pointers to functions returning pointers to a function returning int because expression (*(*p[x])())() evaluates to type int.

Similarly int (i) is a plain variable of type int as expression (i) evaluates to type int.

And so because C++ inherits this from C, it uses parenthesis as part of type declaration, but also adds more syntax on top, leading to some unexpected results.

The rule applied by C++ here says to treat everything that can be a declaration as a declaration.

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Similar confusion if often caused by code like this:

Error ec();

which is a forward declaration of a function ec that returns Error.