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I've a code like this:
struct
{
enum
{
entry,
} en;
} data;
void foo()
{
switch(data.en)
{
}
}
that gives me a warning:
main.cpp:13:11: warning: enumeration value 'entry' not handled in switch [-Wswitch]
switch(data.en)
which is expected. I'm curious if I can add case entry: without making my struct named one (which obviously works).
This:
struct
{
enum
{
entry,
} en;
} data;
void foo()
{
switch(data.en)
{
case entry:
break;
}
}
gives an error + warning:
main.cpp: In function 'void foo()':
main.cpp:15:14: error: 'entry' was not declared in this scope
case entry:
^~~~~
main.cpp:13:11: warning: enumeration value 'entry' not handled in switch [-Wswitch]
switch(data.en)
^

You can write:
case decltype(data.en)::entry:
however I think it would not be considered good code.

In C you can do it the following way
#include <stdio.h>
struct
{
enum
{
entry,
} en;
} data = { entry };
void foo()
{
switch ( data.en )
{
case entry:
puts( "Hello, World!" );
break;
}
}
int main( void )
{
foo();
}
In C++ you do it the following way
#include <iostream>
struct
{
enum
{
entry,
} en;
} data = { decltype( data.en )::entry };
void foo()
{
switch ( data.en )
{
case data.entry:
std::cout << "Hello, World!" << std::endl;
break;
}
}
int main()
{
foo();
}