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Please explain the Output ..

Discussion in 'C' started by raj_ksrt, Feb 26, 2009.

  1. raj_ksrt

    raj_ksrt New Member

    Hi all,

    I am executing the following code.
    Code:
    int main()
    {
        const int i=10;
        int *j= const_cast<int *>(&i);
    
        *((int *)&i)=20;
        printf("%d %d",*(&i),*j);
        printf("\n%x %x",&i,j);
        return 0;
    }
    
    I observe the output as 10 20.The addresses of i and j are same but their values are different. How can this be possible?

    Please provide some explanation.
    Sarma.
     
  2. xpi0t0s

    xpi0t0s Mentor

    It's undefined behaviour because you're modifying a constant.

    It's up to the compiler of course, but one way to implement consts is as a typed #define and wherever it sees "i" referenced it just replaces it with a typecheck and the value 10 (and it could also exist in memory in case anyone wants to take its address). And you're not going to fool a compiler with *(&i) - it knows these are inverse operations of each other so this is directly equivalent to i, which is 10 (because i is a const).

    Looking at the generated assembly answers all. This is from Visual Studio 2005:
    Code:
        printf("%d %d\n",*(&i),*j);
    00411762  mov         esi,esp 
    00411764  mov         eax,dword ptr [j] 
    00411767  mov         ecx,dword ptr [eax] 
    00411769  push        ecx  
    0041176A  push        0Ah  
    0041176C  push        offset string "%d %d\n" (41CB8Ch) 
    00411771  call        dword ptr [__imp__printf (4204A4h)] 
    
    and you can see at 0041176A that the literal value 10 is pushed onto the stack, NOT what is in memory at &i.
     
  3. asadullah.ansari

    asadullah.ansari TechCake

    Last edited: Mar 2, 2009

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