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Why does a modified constant variable retain its original value at compile time, even though a non-const pointer allows for modification of its actual value?

Mary-Kate Olsen
Mary-Kate OlsenOriginal
2024-11-01 19:33:02469browse

Why does a modified constant variable retain its original value at compile time, even though a non-const pointer allows for modification of its actual value?

Modifying a Constant through a Non-Const Pointer

In the given code, a constant variable e is initialized to 2. A non-const pointer w is then cast to point to the address of e, implicitly removing the const-ness of the variable. By dereferencing w and assigning a new value to it, the value at the address of e is modified from 2 to 5.

However, when the values of *w and e are printed, they appear different:

<code class="cpp">cout << *w << endl; // Outputs 5
cout << e << endl; // Outputs 2</code>

This behavior can be explained by considering the different ways in which *w and e are evaluated:

  • *w is being evaluated at runtime: The compiler recognizes that *w is a non-const pointer and dynamically retrieves the value at the address pointed to by w. Since the value at that address has been modified to 5, *w correctly outputs 5.
  • e is being treated as a compile-time constant: The compiler recognizes that e is a constant variable and evaluates its value at compile time. Since the value of e has not been modified at compile time, the compiler continues to treat it as 2.

Therefore, the difference in values arises from the distinct evaluation times of *w and e. Despite the modification to the value at the address of e, the constant variable itself remains unmodified at compile time. However, the non-const pointer w allows for modification of the value at the address it points to, resulting in the output of 5 for *w.

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