Why is my variable getting initialized in Debug mode, but not in Release mode?

Question

I have a windows form application built using VC++ dot net 3. There is a integer variable say int x.

When I run the application in debug mode the value for x =0; But when I run the application without debugging the value is huge 113901996.

The variable is initialzed as int x = 0

Why is this so? Some setting I need to check?

Answer 1

EDIT: As per your new info, that you're forgetting to initialize the variable, the behavior is expected. You're reading garbage memory.

Now, even if you did initialize the variable, you could still replicate this behavior by not using it in a manner that affects the observable behavior of the program. This is what my original answer treated - which I will leave here for future reference.

Are you actually using x? If you're just initializing it, the optimizer will most likely exclude the whole thing.

If it's not influencing the observable behavior of the program, the optimizer is free to cut anything from the binary.

The watch might also be lying to you, I've seen that happening in Release mode.

Can you post some minimal code that reproduces the problem?

If you test it with some output, i.e. :

cout << x;

I'm sure you'll see that x is 0.

EDIT:

Just to clear things out:

int main()
{
   int x = 0;
   x = 3;
   return 0;
}

will translate to (in release mode, with full optimization):

00401000  xor         eax,eax 
00401002  ret  

The variable not only isn't assigned, it doesn't even exist; as opposed to the debug version:

00411370  push        ebp  
00411371  mov         ebp,esp 
00411373  sub         esp,0CCh 
00411379  push        ebx  
0041137A  push        esi  
0041137B  push        edi  
0041137C  lea         edi,[ebp-0CCh] 
00411382  mov         ecx,33h 
00411387  mov         eax,0CCCCCCCCh 
0041138C  rep stos    dword ptr es:[edi] 
0041138E  mov         dword ptr [x],0 
00411395  mov         dword ptr [x],3 
0041139C  xor         eax,eax 

where the variable x is created and assigned twice (see mov dword ptr [x],0 and mov dword ptr [x],3 ).

If observable output is modified by the variable, i.e.:

int main()
{
   int x = 0;
   x = 3;
   cout << x;
   return 0;
}

the generated binary will look like this:

00401000  mov         ecx,dword ptr [__imp_std::cout (40203Ch)] 
00401006  push        3    
00401008  call        dword ptr [__imp_std::basic_ostream<char,std::char_traits<char> 
0040100E  xor         eax,eax 

see that even with this, x isn't actually generated, but rather the 3 value is put on the stack and printed directly.

But the observed behavior in this case is the same, since 3 is printed. To the user, whether the value of x, which should be 3, is printed, or 3 is printed directly, is irrelevant.

The bottom line is: if the behavior of the program is not changed by some statements, they are excluded. In your case, the variable x is not even created.