访问由 void 指针引用的 memory

Question

I am getting my head around void pointers in C++ and as an exercise have written the following code:

void* routine(void* number){

    int n = (int)number;
    int* np = (int*)number;

    cout<<"void pointer: "<<number<<endl; 
    cout<<"casted int pointer: "<<np<<endl; 
    cout<<"int pointer content: "<<*np<<endl; 


    return (void*)NULL;
}

int  main(){
    int num = 1;
    routine((void*)num);
}

This breaks in runtime as the pointer np , casted from void* to int* is dereferenced.

I attempt to pass the argument number as a void pointer and dereference its value inside the function. Arithmetic on void pointers are illegal, therefore what I try to do is cast the pointer to an int* and attempt to dereference this casted copy.

I somewhat expect this not to work, but if that were the case I would also expect some sort of compiler error as is the case when this attempt is made without the int* cast. The reason I sort of expect it to work is that the casted pointer retains the original void pointer address, meaning that after the cast it appears to be just a regular memory address, which I see no reason to be inaccessible by regular dereference. The n variable is a copy of the referenced value, but is just that - I cannot change the actual value in the memory address.

Is this possible with a void pointer? What am I missing?

Answer 1

The usual &num will do.

routine(&num);

Any object pointer implicitly converts to void* . On the callback function you'll have to convert it back to the correct pointer type (use static_cast not C-style cast, that will prevent you from accidentally mixing up pointers containing addresses with integer values that are not memory addresses).

void* routine( void* user_ptr )
{
    int* np = static_cast<int*>(user_ptr);

---

Other pointers which aren't object pointers won't implicitly convert to void* , including function pointers and pointer-to-members. Adding a cast will force the compiler to do the wrong thing and be quiet about it, but (apart from the return value of dlsym() or its Windows equivalent GetProcAddress() ) it isn't safe to mix void* with these other pointer varieties anyway.

Answer 2

void* routine(void* number){

    int n = (int)number;
    int* np = (int*)number;

    cout<<"void pointer: "<<number<<endl; 
    cout<<"casted int pointer: "<<np<<endl; 
    cout<<"int pointer content: "<<*np<<endl; 


    return (void*)NULL;
}


int  main(){
    int num = 1;
    int* num_p = &num;
    routine((void*)num_p);
}

Pointing to num s address before passing does the trick. The casting of an integer to a void pointer makes no sense.