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Initializer Lists in Operator Expressions: A Dilemma
In C , initializer lists, enclosed in curly braces, provide a convenient way to initialize variables. However, an intriguing question arises when attempting to use initializer lists in operator expressions. Consider the following code:
class foo { }; struct bar { template<typename... T> bar(T const&...) { } }; foo& operator<<(foo& f, bar const&) { return f; } int main() { foo baz; baz << {1, -2, "foo", 4, 5}; // This line gives an error return 0; }
Why does this code yield an error, specifically with regards to the line containing the expression baz << {1, -2, "foo", 4, 5};? The reason lies in the C Standard's definition of initializer lists.
Initializer Lists: An Exception to Expressions
Under §5 of the Standard, expressions are precisely defined. Surprisingly, initializer lists are not considered expressions themselves. Binary operators, including <<, expect expressions as their arguments. Hence, since initializer lists do not meet this criterion, they cannot be used directly as operands of binary operators.
Limited Exceptions
The Standard provides several exceptions for initializer-list usage, including as function arguments and in assignment statements. However, the exception does not extend to binary operators.
Reasons for the Restriction
The rationale for this restriction is manifold. According to the draft/discussion paper N2215, allowing initializer lists as both left-hand and right-hand operands of operators would introduce grammatical complexities that would challenge parsers. The authors of the paper ultimately decided that allowing them on the right-hand side but not the left-hand side would be too cumbersome.
Conclusion
While initializer lists offer convenience in certain contexts, their restricted use on the right-hand side of operators stems from syntactic challenges and a desire for consistency in the language's grammar. As an alternative to initializer lists as operands, one can employ constructor invocations or pass bar instances directly as arguments to operator<<(), as seen in the modified code:
baz << bar{1, -2, "foo", 4, 5};
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