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Consider the following C++ code:
void* a = &a; Why doesn't the compiler complain for using an undeclared identifier?
Also, what does the compiler consider the variable a to be? Is it a pointer to a void object or is it a pointer to a void* pointer?
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void (C++)When used as a function return type, the void keyword specifies that the function doesn't return a value. When used for a function's parameter list, void specifies that the function takes no parameters.
void* is a pointer type that doesn't specify what it points to.
yes it is safe.
Void functions, also called nonvalue-returning functions, are used just like value-returning functions except void return types do not return a value when the function is executed. The void function accomplishes its task and then returns control to the caller. The void function call is a stand-alone statement.
The scope of declaration of variables in C++ can be pretty surprising:
void* a = &a; ^~~~~~~~~~~~~~~~~ a declared as `void*` from here on Therefore, &a is void** but since any pointer type is implicitly convertible to void*...
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It is equivalent to
void* a; a = &a; Therefore, a has been declared. So a gets the address of a written in a. So it is a pointer to a void pointer. (You did not define any objects yet.)
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