int x = x; What will happen?

Question

What will happen as a result of such instructions:

int x = x;

?

And what observable behavior do you expect from this instruction?
@valuev, what's stopping yourself from trying different options and seeing what happens?
@insolor not understand assembly language at the proper level
@insolor then what are the "different options" you propose to consider?

Community spirit ♦ one · Answer 1 · 2016-05-07T18:21:27

If the variable x already defined in this scope, it will not compile:

 int x = 0; int x = x; // <-- ошибка, повторная декларация переменной

http://ideone.com/3VGhe0

If there is no variable x in this scope (for example, it is not defined at all, or defined in the outer area), then this will be compiled, but the program will have undefined behavior . This means that you are not even guaranteed that there will be any garbage in the variable. The program has the right to do anything: from the point of view of the compiler, it is meaningless. Never do that: the C ++ compiler considers you quite mature and responsible, and does not check whether you are following the rules.

Yes, I do not know why in this case the compiler does not give an error. Other languages trust the programmer less and control.

the declaration occurred before the equal sign, after which meaningless assignment is made.
In the first, overriding a variable that is prohibited by the rules.
In the second, the use of an uninitialized value, which is UB by default.
the variable x exists and is available after the = sign, not after ;
Compilers issue warnings, at least on similar code inside functions.
I already collected a collection of warnings from GCC, clang and MSVC, but ultimately decided that it did not pull the answer.
I would advise to include all warnings and treat them as errors.

hitcode hitcode 347 one 20 · Answer 2 · 2016-05-07T17:16:37

It depends on where it is, this line.

There are three types of memory, dynamic, static and heap.

1] Dynamic memory: usually allocated new, malloc, alloc..itd memory is allocated and there is garbage (remnants of previous work), but in some cases the memory 0 is filled, and the new constructor can fill the memory

2] static memory (global change, static): Memory is allocated by the compiler and filled with a zero value. It can be seen in the binary and change.

3] stack (int x inside a function): memory is allocated, but there is garbage in it.

It would be worth saying what happens in each of these cases.
Well and still, to declare a variable in dynamic memory will not work.

Answer 3 · 2016-05-07T20:00:13

In practice, exactly the same thing happens as with a simple declaration.

 int x;

However, in theory, this is undefined behavior, so whatever ...

This question was mentioned by Sutter in “ Challenges ”, namely in problem 9.1, postscript 1.

There is one more similar code, but with the new operator:

 T t; new(&t) T(t);

Assigning an uninitialized variable is UB, but a declaration without initialization is not.
@Qwertiy You hurried with a comment, I already corrected the answer by finding this place from Sutter.

int x = x; What will happen?

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