Why does "case::LABEL:" compile in g++?

Question

I was surprised to learn that g++ (4.9) was compiling this (while gcc would not):

#include <stdio.h>

enum
{
    ONE   = 1,
    TWO   = 2,
    THREE = 3
};

int main(int argc, char* argv[])
{
    int sw = 2;

    switch (sw)
    {
    case::ONE:
    {
        printf("1\n");
        break;
    }

    case::TWO:
    {
        printf("2\n");
        break;
    }

    case::THREE:
    {
        printf("3\n");
        break;
    }

    default:
    {
        printf("default\n");
    }
    }
}

How is the g++ preprocessor able to separate the "case" from the "::ONE:"?

score 4 · Accepted Answer · answered Jan 06 '16 at 14:18

4

How is the g++ preprocessor able to separate the "case" from the "::ONE:"?

It's not the preprocessor, but simply the compiler, that interprets :: to refer to the global namespace.

The case labels will be parsed as

case :: THREE :

and that's perfectly OK, as your enum values appear in the global namespace (::).

answered Jan 06 '16 at 14:18

πάντα ῥεῖ

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Once Cecilio fixes the 'root' namespace, your answer will become a duplicate ;) – SergeyA Jan 06 '16 at 14:26
I meant to say "parser", not "preprocessor". – John Jan 06 '16 at 14:40
@John You have to be precise about that, the preprocessor is a different thing. – πάντα ῥεῖ Jan 06 '16 at 14:41

score 3 · Answer 2 · answered Jan 06 '16 at 14:17

3

It is not the preprocessor. The C compiler (its tokenizer) sees that as:

case :: ONE

which is ok in c++. The :: operator means the root namespace. C has no such thing.

answered Jan 06 '16 at 14:17

Cecilio Pardo

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There is no root namespace, it is a global namespace. – SergeyA Jan 06 '16 at 14:26
"C has no such thing": thus it's the C++ compiler, not the C compiler :) – mikedu95 Jan 06 '16 at 14:30

score 0 · Answer 3 · answered Jan 06 '16 at 14:21

You have a simple enum, not a scoped enum (see: enum class). Plain enums are in the root scope, hence you can write ::ONE.

With scoped enums, you can do the following:

enum class Scoped { FIRST, LAST = 9};
//...
case Scoped::FIRST : //...

The tokenizer in g++ (as this is a C++ feature) can resolve even if you don't leave whitespaces between the case keyword and the enumerator.

score 0 · Answer 4 · edited May 23 '17 at 12:31

During the lexical analysis phase of the compilation process, the stream of characters in the program file is converted to tokens. One of the popular algorithms used here is maximal munch in which the longest matching legal token is considered as the intended token. In this example, it breaks the case::ONE: into case, ::, ONE and :.

There are also some disadvantages: x = y/*z is an error since /* is the longest matching legal token, while the user probably wanted to say x = y / *z. You can find many such curious examples, couple of them are here and here.

Why does "case::LABEL:" compile in g++?

4 Answers4