Variable not initialized as C ++ condition

Question

I want to do if you want:

if(x has not been initialized) {... } 

Is it possible? Thanks

Answer 1

"There is no way to check the contents of an undefined variable or not. The best you can do is assign a signal / sentinel value (for example, in the constructor) to indicate that further initialization will be performed."

Alexander Gessler

from here

Answer 2

You can use pointers to implement this behavior, the default is 0.

For instance:

 int *number = 0; // ... if (!number) { // do something } 

You can use this trick for any type, and not just for integers:

 Cat *kitty = 0; // ... if (!kitty) { // do something } 

Answer 3

a) For primitive data types, such as int, float, it cannot be recognized if it is initialized or not.

b) For pointers you can check if its nullptr or not

 if(ptr != nullptr) { } 

c) For a custom class, you need to introduce a bool element that can be set to true in the constructor so that we can use it to check if the object is initialized or not.

 if(obj.isInitialized()) { } 

Answer 4

• If you have a pointer, you can use nullptr to mean “not initialized”.
• If you don't have a pointer, you can use boost::optional<T> (or std::optional<T> when it becomes available).
• Otherwise, the naive flag <<23> must be executed. optional<T> is just an encapsulation of this flag.

Answer 5

With C ++ - 11, you can consider storing a variable using smart pointers. Declare yor x (assuming x will be int ) as

 std::shared_ptr<int> x; 

When assigning the variable x use

 x = std::make_shared<int>(newValueOfX); 

and then you can determine if x ever been assigned by checking

 if (this->x) { ... } 

See Checking variable initialization for a more detailed example.

Answer 6

Well, something like this will work for pointers. Say:

 int* x = NULL; //initialize if(x == NULL) { //dostuff } 

or simply

 if(!x) { //dostuff } 

Not sure if there is a way for a regular int though

EDIT: Now that I think about it, Lutyan is right. This will be undefined behavior. As the others said, you must initialize a known value.