C++ Does Const_Cast induce new variable?

Question

#include <iostream>

using namespace std;

int main() {
        const int a = 2;
        const int *p = &a;
        int *p2 = const_cast<int*>(p);
        *p2=5;
        char *p3 = (char *)&a;
        cout << "p2 is" << *p2 << endl;
        cout << "p2 address " << p2 << endl;
        cout << "a is " << a << endl;
        cout << "a address " << &a << endl;

        return 0;
}

Hi all!

According to the output, *p2 and a has different values, *p2 is 5 and a is 2.

However, p2 and &a are the same. I'm confused...

Could you please help me understand where this is the case?

Thank you very much!

It's undefined behaviour to change a constant. – chris Oct 19 '12 at 19:29 — chris, Oct 19 '12 at 19:29

Luchian Grigore · Accepted Answer · 2012-10-19T19:40:37.893

Undefined behavior means that anything can happen. Including this.

5.2.11 Const cast [expr.const.cast]

7) [ Note: Depending on the type of the object, a write operation through the pointer, lvalue or pointer to data member resulting from a const_cast that casts away a const-qualifier⁷³ may produce undefined behavior (7.1.6.1). —end note ]

The underlying reason might be that the compiler, seeing how a is const, optimizes cout << "a is " << a << endl; to a simple cout << "a is " << 2 << endl;.

For example, even in a debug build, I get:

        cout << "a is " << a << endl;
00CE1581  mov         esi,esp  
00CE1583  mov         eax,dword ptr [__imp_std::endl (0CFD30Ch)]  
00CE1588  push        eax  
00CE1589  mov         edi,esp  
//...
00CE158B  push        2 
//...
00CE158D  push        offset string "a is " (0CE7840h)  
00CE1592  mov         ecx,dword ptr [__imp_std::cout (0CFD308h)]  
00CE1598  push        ecx  
00CE1599  call        std::operator<<<std::char_traits<char> > (0CE1159h)  
00CE159E  add         esp,8  
00CE15A1  mov         ecx,eax  
00CE15A3  call        dword ptr [__imp_std::basic_ostream<char,std::char_traits<char> >::operator<< (0CFD304h)]  
00CE15A9  cmp         edi,esp  
00CE15AB  call        @ILT+415(__RTC_CheckEsp) (0CE11A4h)  
00CE15B0  mov         ecx,eax  
00CE15B2  call        dword ptr [__imp_std::basic_ostream<char,std::char_traits<char> >::operator<< (0CFD300h)]  
00CE15B8  cmp         esi,esp  
00CE15BA  call        @ILT+415(__RTC_CheckEsp) (0CE11A4h)

I highlighted the essential part - 2 is pushed directly on the argument stack of operator<<, instead of the value of a being read.

I got it. Thank you for your help!! BTW, rather than give me fish, could you explain a bit how to fish? How to debug using assembly like this? Could you give some hint? Thanks! — user1582802, Oct 19 '12 at 19:52
@user1582802 depends on the IDE/tools you have. I use MS visual studio, and you can switch to assembly during debugging - right clock -> show disassembly. — Luchian Grigore, Oct 19 '12 at 20:03

score 1 · Answer 2 · answered Oct 19 '12 at 19:37

This is where you've gone astray:

int main() {
        const int a = 2;
        const int *p = &a;
        int *p2 = const_cast<int*>(p);
        *p2=5;

On the last line here, you've assigned a variable thu a pointer that points to something that is actually const. That is, a is const. p2 points to a. You can't change the value of a, even through p2, without invoking Undefined Behavior.

C++ Does Const_Cast induce new variable?

2 Answers2

5.2.11 Const cast [expr.const.cast]