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Home/ Questions/Q 3351626
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T01:55:53+00:00 2026-05-18T01:55:53+00:00

Consider this code: #include <iostream> using namespace std; int main() { bool lock =

  • 0

Consider this code:

#include <iostream>
using namespace std;

int main()
{
   bool lock = false;
   lock = __sync_val_compare_and_swap( &lock, false, true );
   cout << lock << endl;
}

I expect the result to be displayed as 1 but the o/p is 0. Just calling __sync_val_compare_and_swap( &lock, false, true ); (so the return value is not captured) and then displaying lock results in 1 being displayed.

What am I missing here?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-18T01:55:54+00:00Added an answer on May 18, 2026 at 1:55 am

    From the GCC doco:

    bool __sync_bool_compare_and_swap (type *ptr, type oldval type newval, ...)
    type __sync_val_compare_and_swap (type *ptr, type oldval type newval, ...)

    These builtins perform an atomic compare and swap. That is, if the current value of *ptr is oldval, then write newval into *ptr.

    The “bool” version returns true if the comparison is successful and newval was written. The “val” version returns the contents of *ptr before the operation.

    Seems to me that 0 is the right value. I think you are incorrectly assigning "…the contents of *ptr before the operation" to lock.

    This should output sensible results:

    #include <iostream>
    using namespace std;
    
    int main()
    {
       bool lock = false;
       bool oldvalue = __sync_val_compare_and_swap( &lock, false, true );
       cout << lock << ", " << oldvalue << endl;
    }
    
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