Does C++ do value initialization on simple POD typedefs?
Assuming
typedef T* Ptr;
does
Ptr()
do value-initialization and guarantee to equal (T*)0
?
e.g.
Ptr p = Ptr();
return Ptr();
It does. For a type T
, T()
value-initializes an "object" of type T
and yields an rvalue expression.
int a = int();
assert(a == 0);
Same for pod-classes:
struct A { int a; };
assert(A().a == 0);
Also true for some non-POD classes that have no user declared constructor:
struct A { ~A() { } int a; };
assert(A().a == 0);
Since you cannot do A a()
(creates a function declaration instead), boost has a class value_initialized
, allowing to work around that, and C++1x will have the following, alternative, syntax
int a{};
In the dry words of the Standard, this sounds like
The expression T(), where T is a simple-type-specifier (7.1.5.2) for a non-array complete object type or the (possibly cv-qualified) void type, creates an rvalue of the specified type, which is value-initialized
Since a typedef-name is a type-name, which is a simple-type-specifier itself, this works just fine.
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