Is enum { a } e = 1; valid?

Question

A simple question: is enum { a } e = 1; valid?

In other words: does assigning a value, which isn't present in the set of values of enumeration constants, lead to well-defined behavior?

Demo:

$ gcc t0.c -std=c11 -pedantic -Wall -Wextra -c
<nothing>

$ clang t0.c -std=c11 -pedantic -Wall -Wextra -c
<nothing>

$ icc t0.c -std=c11 -pedantic -Wall -Wextra -c
t0.c(1): warning #188: enumerated type mixed with another type
# note: the same warning for enum { a } e = 0;

$ cl t0.c /std:c11 /Za /c
<nothing>

Answer 1

From the C18 standard in 6.7.2.2:

Each enumerated type shall be compatible with char, a signed integer type, or an unsigned integer type. The choice of type is implementation-defined, but shall be capable of representing the values of all the members of the enumeration.

So yes enum { a } e = 1; is valid. e is a 'integer' type so it can take the value 1. The fact that 1 is not present as an enumeration value is no issue. The enumeration members only give handy identifiers for some of possible values.

Answer 2

Valid in C

This is valid in C as per e.g. the 202x working draft:

6.7.2.2/4 Each enumerated type shall be compatible with char, a signed integer type, or an unsigned integer type. The choice of type is implementation-defined, but shall be capable of representing the values of all the members of the enumeration

particularly due to the "compatible type" requirement:

6.2.7/1 Two types have compatible type if their types are the same. Additional rules [...]

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... implementation-defined and possible unspecified/undefined behavior in C++14 and earlier/C++17 and later

Whilst out of scope for this Q&A (C tag, not C++), it may be interesting to point out that the same does not hold for C++, where "C style" unscoped enums with no fixed underlying type have subtle differences from C. For details about the C++ case, see the following Q&A:

• Is enum E { a } e = E(2); valid?