什么是“（void）指针;”在c ++中是什么意思？

Question

There are some c++ code

struct data {
    /* some fields */
};

typedef struct data * pData;

int function(pData pointer) {
    if(pointer == NULL)
        return ERROR;
    (void)pointer;
    /* other work */
}

What does "(void)pointer" mean?

Just for your information, there are some int/char*/etc, some functions pointers which are used as callback functions in the structure.

Answer 1

It is used to circumvent an unused-variable warning.

If you do use the variable, it is a no-op.

Mostly unused variables are parameters, that are required to fulfil a signature for a callback function, but not needed in your actual implementation.

Cf.

• unused parameter warnings in C code
• GCC manual : -Wunused-variable (enabled by -Wall , too).

Update:

Just because it was not mentioned otherwise: The type of the variable might be anything. It is not constricted to pointer types.

Answer 2

It doesn't mean a whole lot.

It evaluates the expression pointer , then explicitly ignores it by casting it to void .

Sometimes you see this construct when trying to convince a compiler not to warn for an un-used argument, but in your code the argument is already used since it's being NULL -checked.

Answer 3

It casts the pointer value to a "no type" value, or that it is "absent of a type".

void foo(); // absent of a return value 
int bar(void); // absent of parameters
void *var; // Type of the variable pointed to is absent
(void) var; // The type of the variable is absent

Answer 4

It is a typical way to suppress unused variable compiler warning.

However, since the pointer is actually used as if(pointer == NULL) , I see no reason do that.

If I have to guess, I would guess that the NULL check and return is added later than the warning suppression.