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Erik Carstensen
Erik Carstensen

Reputation: 888

Why can enum values have large (>int) values?

After I have defined an enum like this:

typedef enum { x = 1ULL<<40 } e;

it seems that the expression 'x' evaluates to 1UL<<40.

However, C11 6.4.4.3 says "An identifier declared as an enumeration constant has type int."

Incidentally this saves my day, since I want enum values > 1<<32. However, I need to know how this can be; I'm a bit nervous to write code that isn't standards compliant.

I tried this code:

int main() {
        printf("%zu %zu %lx\n", sizeof(e), sizeof(x), x);
        return 0;
}

Compiling with gcc -std=c11 -Wall (gcc 8.3.1, linux x86_64), it outputs: 8 8 10000000000.

If I change %lx to %llx, I get a compile warning and the same output, which indicates x is long-sized.

Based on my reading of the spec, I would expect compile warnings and output 4 4 0.

Upvotes: 3

Views: 345

Answers (1)

Carl Norum
Carl Norum

Reputation: 224964

Compiler extensions. For example, with clang:

$ CFLAGS=-Weverything make example && ./example
cc -Weverything    example.c   -o example
example.c:3:16: warning: ISO C restricts enumerator values to range of 'int'
      (1099511627776 is too large) [-Wpedantic]
typedef enum { x = 1ULL<<40 } e;
               ^   ~~~~~~~~
1 warning generated.
8 8 10000000000

And with gcc:

$ CFLAGS='-Wall -pedantic' make example && ./example
cc -Wall -pedantic    example.c   -o example
example.c:3:20: warning: ISO C restricts enumerator values to range of ‘int’ [-Wpedantic]
 typedef enum { x = 1ULL<<40 } e;
                    ^~~~
8 8 10000000000

Upvotes: 3

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