Modifying a const through a non-const pointer

Answer 1

I suspect that you're tripping up the compiler. It doesn't expect you to play dirty tricks with e, so when it sees the line:

cout << e << endl;

It simply inserts the value 2 instead of looking for the actual value. You can verify (or disprove) this by looking at the disassembly of your program.

Answer 2

I guess the compiler uses the constness to optimizes out the variable and insert a fixed value into the code.

Answer 3

The only thing I can think of is the compiler has some how optimised the code in such a way that any references to e are replaced with a value of 2 even though it assigns memory for e

so in effect (affect?) the line at comment (4) is 'optimized' to be

cout << "2" << endln;

Answer 4

I'm guessing that the compiler has optimised the value output. It sees that e is const (so, it can't change -- in theory) and changes cout << e << endl; to cout << 2 << endl;. However, e still has to exist because it's used by w, so w correctly takes its address and modifies its value, but you don't see that in the cout.

Moral of the story -- only declare things const when you actually want to be const. Casting away constness is not a good idea.

Answer 5

As I said in my comment, once you modified the const value you are in undefined behaviour land, so it doesn't make much sense to talk about what is happening. But what the hell..

cout << *w << endl;            // (3) outputs 5
cout << e << endl;             // (4) outputs 2

At a guess, *w is being evaluated at runtime, but e is being treated as a compile time constant

Answer 6

This behaviour of compiler is misleading. It should stop constant folding once address of the constant is assigned to some other const or non-const pointer. Then for both variable single value will be printed. simple for compiler and programmer both.