Change “const int” via an “int *” pointer. Surprising and interesting [duplicate]

Answer 1

This probably due to optimizations. Since you declare const int n=9, you're basically making a promise that you won't modify n. So the compiler is free to optimize cout << n << endl; to a simple cout << 9 << endl;. I don't think the old value is stored anywhere else.

Nevertheless, this is undefined behavior.

I can confirm that the cause of this behavior is optimization, even on a debug build:

    cout< >::operator<< (11582ACh)]  
0115143E  cmp         edi,esp  
01151440  call        @ILT+325(__RTC_CheckEsp) (115114Ah)  
01151445  mov         ecx,eax  
01151447  call        dword ptr [__imp_std::basic_ostream >::operator<< (11582A8h)]  
0115144D  cmp         esi,esp  
0115144F  call        @ILT+325(__RTC_CheckEsp) (115114Ah)  

n doesn't even come into discussion.

If n wasn't const, the optimization would be illegal:

00D51426  mov         esi,esp  
00D51428  mov         eax,dword ptr [__imp_std::endl (0D582B4h)]  
00D5142D  push        eax  
00D5142E  mov         edi,esp  

//HERE
//
00D51430  mov         ecx,dword ptr [n]  
00D51433  push        ecx  
//
//

00D51434  mov         ecx,dword ptr [__imp_std::cout (0D582B0h)]  
00D5143A  call        dword ptr [__imp_std::basic_ostream >::operator<< (0D582ACh)]  
00D51440  cmp         edi,esp  
00D51442  call        @ILT+325(__RTC_CheckEsp) (0D5114Ah)  
00D51447  mov         ecx,eax  
00D51449  call        dword ptr [__imp_std::basic_ostream >::operator<< (0D582A8h)]  
00D5144F  cmp         esi,esp  
00D51451  call        @ILT+325(__RTC_CheckEsp) (0D5114Ah)  

Here, the value of n is pushed on the argument stack because it's not const.