Question

Why is the function not a _fastcall calling convention?

a.

because it receives parameters through ECX and EDX

b.

because it uses a standard EBP stack frame

c.

because it doesn't take any registers from the caller

d.

because it initializes every register that is accessed within the function itself

Posted under Reverse Engineering

Answer: (c).because it doesn't take any registers from the caller Explanation:Every register that is accessed within the function itself is initialized, indicating that it's not a _fastcall calling convention, which receives parameters through ECX and EDX.

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