Ambiguity when non constant pointer is pointing to constant variable

Vinoth Rajendran4

Hi All,

In the code above, a non constant pointer is pointing to constant variable.
I was trying to change the variable value using *ptr = 100.
When I check the value of *ptr and a, it was different. (Line number 20 and 21)
But both ptr and &a is pointing to same memory location. (Line number 23)

I am not able to understand how, two different values are assigned to same memory address ? does compiler creates temporary address to hold both different values here (100 and 10) ?

Any help is appreciated. Thanks!

JonB

@Vinoth-Rajendran4
I believe your code does write into a via *ptr. I wonder if the compiler has kept const int a in a register, does not expect it to be changed since it's const, and reports the original value for cout << a?

I wonder if you can fool it with cout << *&a?

J.Hilk

@Vinoth-Rajendran4 thats not what const canst was defined for, and is undefined behaviour, switch the compiler, optimisation level or to release/debug build, you may get a different outcome.

Vinoth Rajendran4

@JonB : for cout << *&a , I am getting value as 100.

KroMignon

@Vinoth-Rajendran4 said in Ambiguity when non constant pointer is pointing to constant variable:

for cout << *&a , I am getting value as 100

As @J-Hilk already said, your code is almost unsafe. Depending on C++ compiler type (GCC, MSVC, etc.), code optimization level and build mode (release/debug) you will probably got different results.

Declaring a variable as const will allow compiler to do some optimization. Violate this const state will introduce unpredictable behavior.

It is not because you can write a code that it implies that this code is valid, for example following code is also valid but will crash:

char * ptr = nullptr;

cout << ptr;

Vinoth Rajendran4

@J-Hilk : I thought const_cast<> is used to remove or add constant to the expression. If its okay, can you please explain me the actual use of const_cast<> ?

KroMignon

@Vinoth-Rajendran4 said in Ambiguity when non constant pointer is pointing to constant variable:

I thought const_cast<> is used to remove or add constant to the expression. If its okay, can you please explain me the actual use of const_cast<> ?

The easiest is to look at C++ documentation (cf. https://en.cppreference.com/w/cpp/language/const_cast):

const_cast makes it possible to form a reference or pointer to non-const type that is actually referring to a const object or a reference or pointer to non-volatile type that is actually referring to a volatile object. Modifying a const object through a non-const access path and referring to a volatile object through a non-volatile glvalue results in undefined behavior.

JonB

@Vinoth-Rajendran4 said in Ambiguity when non constant pointer is pointing to constant variable:

@JonB : for cout << *&a , I am getting value as 100.

Which was my point/guess! At some level cout << a is taking advantage of the const int a = 10; it saw earlier, it's not really going to fetch the current contents of a which you actually altered via *ptr = 100. Some optimization (on your compiler) decided you would not alter a.

My guess was that *&a would force it to generate code to actually look in the a memory address for the value, by-passing the "optimization". And so it has proved to be --- again, on your compiler.

As the others have said, this whole thing is compiler-specific/undefined/"very naughty", for precisely the above reasons (e.g. see @KroMignon's quote) and the way you are seeing a != *&a. Which is why you shouldn't be doing it! :)

jsulm

@Vinoth-Rajendran4 See here: https://en.cppreference.com/w/cpp/language/const_cast
What you are doing here is exactly what is mentioned there: "Modifying a const object through a non-const access path and referring to a volatile object through a non-volatile glvalue results in undefined behavior."

fcarney

@KroMignon said in Ambiguity when non constant pointer is pointing to constant variable:

cout << ptr;

This won't crash. It will just print 0. But this: cout<< *ptr; will.

Kent-Dorfman

This post is deleted!

KroMignon

@fcarney said in Ambiguity when non constant pointer is pointing to constant variable:

This won't crash. It will just print 0. But this: cout<< *ptr; will.

You are right :)