InterlockedIncrement usage
It depends on your compiler settings. However, by default, anything eight bytes and under will be aligned on a natural boundary. Thus an "int" we be aligned on a 32-bit boundary.
Also, the "#pragma pack" directive can be used to change alignment inside a compile unit.
I would like to add that the answer assumes Microsoft C/C++ compiler. Packing rules might differ from compiler to compiler. But in general, I would assume that most C/C++ compilers for Windows use the same packing defaults just to make working with Microsoft SDK headers a bit easier.
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Kiran Kumar
Software developer from Bangalore, my area of interest is mainly C++. 33rd to C++ Gold Badge
Updated on April 15, 2022Comments
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Kiran Kumar about 2 years
While reading about the function InterlockedIncrement I saw the remark that the variable passed must be aligned on a 32-bit boundary. Normally I have seen the code which uses the InterlockedIncrement like this:
class A { public: A(); void f(); private: volatile long m_count; }; A::A() : m_count(0) { } void A::f() { ::InterlockedIncrement(&m_count); }
Does the above code work properly in multi-processor systems or should I take some more care for this?