Does C++ do value initialization of a POD typedef?
Does C++ do value initialization on simple POD typedefs?
Assuming
typedef T* Ptr;
does
Ptr()
do value-initialization and guarantee to equal (T*)0 ?
eg
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.
#include <iostream>
struct Foo {
char bar;
char baz;
char foobar;
// the struct is a POD
//virtual void a() { bar='b'; }
};
int main() {
Foo o1;
Foo o2 = Foo();
std::cout << "O1: " << (int)o1.bar <<" "<< (int)o1.baz <<" "<< (int)o1.foobar << std::endl;
std::cout << "O2: " << (int)o2.bar <<" "<< (int)o2.baz <<" "<< (int)o2.foobar << std::endl;
return 0;
}
Output:
O1: -27 -98 0
O2: 0 0 0
Adding () propagates initializer calls to all POD members. Uncomenting the virtual method changes output to:
O1: -44 -27 -98
O2: -71 -120 4
However adding destructor ~Foo() does not suppress the initialization, although it creates non-POD object (output similar to first one).
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