Hi,
Because the association of a const variable is with the name of the variable and not the address of variable, at compile time.
So the no matter you try to modify it by other methods, it will remain same.
Regards.
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const in C++
I am trying to understand the exact semantics of const in C++. So I wrote the following program:
#include <iostream>
using namespace std;
int main() {
const int x = 2;
int *p = (int *)&x;
*p = 1;
cout << x << endl;
}
My expectation was that the program should print 1 but it prints 2. Is this correct behavior according to the C++ standard? Surely when the programmer has explicitly overridden the constness of a variable, the compiler must obey. But this does not seem to be the case. Are there any explanations? I am using "g++ -g test.cpp" to compile the program.
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Answers
That's because you've used a C style cast, which doesn't care about const-ness. Try it with a static cast.
int main() {
const int x = 2;
int *p = static_cast<int *>(&x);
*p = 1;
cout << x << endl;
return 1;
}
The compiler will now say that it can't cast from const int * to int *. Of course, you can override that with a const_cast, but that's beside the point. You can cast to const int *, but then the compiler will disallow *p = 1
Moral of the story : Never use C style casts unless they're absolutely necessary.
Yes "s_senthil_kumar" is right, try it out what he said. and u will find why it was like this.
even in the symbol tables i guess the X will have the value 2 and though as per the assembly shown above it's just reading the symbol table and try to print the associated value of x in symbole table !
|+|ayank
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by: AlexFMPosted on 2004-09-27 at 04:54:34ID: 12159210
Why does this happen? This is Assembly code produced by C++ compiler:
13: const int x = 2;
00401798 mov dword ptr [ebp-4],2
14: int *p = (int *)&x;
0040179F lea eax,[ebp-4]
004017A2 mov dword ptr [ebp-8],eax
15: *p = 1;
004017A5 mov ecx,dword ptr [ebp-8]
004017A8 mov dword ptr [ecx],1
16: cout << x << endl;
004017AE push offset @ILT+195(std::endl) (004010c8)
004017B3 push 2
...
push 2 - compiler passes hard-coded 2 instead of c to cout. I tested with VC++ 6.0 and 7.1 with the same result. Since VC++ 7.1 is almost 100% standard compliant, I guess this is correct behaviour.
Compiler doesn't show error message or warning because it cannot recognize constness violation done using pointer.