内置类型的模板参数默认值

Question

Consider this function:

template <typename T>
T foo(const T& var = T()) {
  return var;
}

This call is obviously safe:

foo<std::string>()

Are those?

foo<int>()
foo<bool>()
...

Answer 1

Yes, they are perfectly fine. An expression of the form T() creates an object of type T and value-initializes it. Value-initializing an int or bool is the same as zero-initializing them. That is, the parameter var will have value 0.

The expression T() , where T is a simple-type-specifier or typename-specifier for a non-array complete object type [...] creates a prvalue of the specified type, whose value is that produced by value-initializing (8.5) an object of type T

Value-initialization is defined as:

To value-initialize an object of type T means:

• if T is a (possibly cv-qualified) class type [...]

• if T is a (possibly cv-qualified) non-union class type [...]

• if T is an array type, [...]

• otherwise, the object is zero-initialized.

Zero-initialization is defined as:

To zero-initialize an object or reference of type T means:

• if T is a scalar type (3.9), the object is set to the value 0 (zero), taken as an integral constant expression, converted to T ;

• [...]

The temporary object created by T() is then bound to the const reference, which extends its lifetime.

There are two contexts in which temporaries are destroyed at a different point than the end of the full-expression. [...] The second context is when a reference is bound to a temporary.