integer pointer to constant integer

Question

I'd like to know what is happening internally and its relation to values displayed. The code is:

# include <iostream>

int main(){
    using namespace std;
    const int a = 10;

    int* p = &a;  //When compiling it generates warning "initialization from int* to 
              //const int* discard const -- but no error is generated

    cout << &a <<"\t" << p <<endl; //output: 0x246ff08 0x246ff08 (same values)
    cout << a << "\t" << *p << endl; //output: 10 10

    //Now..
    *p = 11;
    cout << &a <<"\t" << p <<endl; //output: 0x246ff08 0x246ff08 (essentially, 
                               //same values and same as above, but..)
    cout << a << "\t" << *p << endl; //output: 10 11

    return 0;
}

QUESTION: If p = address-of-a, how come a=10, but *p = (goto address of a and read value in the memory location) = 11?

Answer 1

 cout << a << "\t" << *p << endl; //output: 10 11

You lied to the compiler and it got its revenge.

With:

const int a = 10;

you promised you'll never modify a object.

Answer 2

You promised not to modify a, and the compiler believed you. So it decided to optimise away the reading of a, because it trusted you. You broke your promise.

Having said that, a real C++ compiler won't compile your code because int* p = &a is illegal. So perhaps you are lying to us as well. Or perhaps you need to get a real C++ compiler.