int x; int y; int *ptr; is not initialization, right?

Question

I'm reading 'C++ All-in-One for Dummies' by J. P. Mueller and J. Cogswell and stumbled onto this:

#include <iostream> using namespace std; int main() {     int ExpensiveComputer;     int CheapComputer;     int *ptrToComp; ... 

This code starts out by initializing all the goodies involved — two integers and a pointer to an integer.

Just to confirm, this is a mistake and should read '... by declaring', right? It's just strange to me that such basic mistakes still make their way to books.

Answer 1

From the point of view of the language, this is default initialization. The problem is, they are initialized to indeterminate values.

otherwise, nothing is done: the objects with automatic storage duration (and their subobjects) are initialized to indeterminate values.

Default initialization of non-class variables with automatic and dynamic storage duration produces objects with indeterminate values (static and thread-local objects get zero initialized)

Note that any attempt to read these indeterminate values leads to UB.

From the standard, [dcl.init]/7

To default-initialize an object of type T means:

• If T is a (possibly cv-qualified) class type ([class]), constructors are considered. The applicable constructors are enumerated ([over.match.ctor]), and the best one for the initializer () is chosen through overload resolution ([over.match]). The constructor thus selected is called, with an empty argument list, to initialize the object.

• If T is an array type, each element is default-initialized.

• Otherwise, no initialization is performed.