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Spartacus

__fastcall???

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Can someone tell me what happens when a function is defined as __fastcall? What are the differences between a __fastcall defined function and a function that is not defined as __fastcall? Thanks, -René

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It simply tells the compiler to pass the arguments of the fonction by the register instead of the stack if possible.

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However note that Visual C++ takes no notice of it :-)

In other compilers you must remember that you can only __fastcall a function if there are fewer than three parameters.

Also remember that any function in a class has one parameter more than it looks like (''this'' is passed a parameter to member functions).

Alan

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