Why is `printme({\'a\', \'b\', \'c\'});` Illegal in C Template Type Deduction?

Type Deduction in Functions with Initializer Lists

Consider the function template:

template<typename T>
void printme(T&& t) {
  for (auto i : t)
    std::cout << i;
}

This function expects one parameter of a type with a begin() and end() function.

Question:

Why is the following call illegal?

printme({'a', 'b', 'c'});

Answer:

The call printme({'a', 'b', 'c'}) is illegal because the template argument T cannot be inferred. Without specifying the template argument explicitly, the compiler cannot determine the type of the parameter, as it could be any type with a begin() and end() function. This issue does not arise in the following cases:

• Explicit template argument specification: printme>({'a', 'b', 'c'})
• Parameter types with type deduction: printme(std::vector({'a', 'b', 'c'}))
• Using auto to deduce the type: const auto il = {'a', 'b', 'c'}; printme(il);
• Explicitly specifying the template argument: printme>({'a', 'b', 'c'})

The exception to this is the deduction of auto as std::initializer_list, which differs from the template argument deduction. This behavior is explicitly specified in the C 11 standard (§ 14.8.2.5/5), which states that the template argument for a function parameter that is an initializer list but not of type std::initializer_list is not deduced. However, § 7.1.6.4/6 of the standard provides explicit support for auto to deduce std::initializer_list<>.

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