从字符串转换为void *并返回

Question

is possible to re-map STL class object from void* ?

#include <string>

void func(void *d)
{
    std::string &s = reinterpret_cast<std::string&>(d);
}

int main()
{
    std::string s = "Hi";
    func(reinterpret_cast<void*>(&s));
}

Answer 1

Use static_cast to convert void pointers back to other pointers, just be sure to convert back to the exact same type used originally. No cast is necessary to convert to a void pointer.

This works for any pointer type, including pointers to types from the stdlib. (Technically any pointer to object type, but this is what is meant by "pointers"; other types of pointers, such as pointers to data members, require qualification.)

void func(void *d) {
  std::string &s = *static_cast<std::string*>(d);
  // It is more common to make s a pointer too, but I kept the reference
  // that you have.
}

int main() {
  std::string s = "Hi";
  func(&s);
  return 0;
}

Answer 2

I re-wrote as following,

#include<string>

void func(void *d)
{
    std::string *x = static_cast<std::string*>(d);
/* since, d is pointer to address, it should be casted back to pointer
   Note: no reinterpretation is required when casting from void* */
}

int main()
{
    std::string s = "Hi";
    func(&s); //implicit converssion, no cast required
}

Answer 3

You code shouldn't compile. Change

std::string &s = reinterpret_cast<std::string&>(d);

to

std::string *s = static_cast<std::string*>(d);

EDIT: Updated code. Use static_cast instead of reinterpret_cast

Answer 4

Yes, it is possible, but you are trying to cast from a pointer to void * , then to a reference . The reinterpret_cast operator only allows casting back to exactly the same type that you started with. Try this instead:

void func(void *d)
{
    std::string &s = *reinterpret_cast<std::string*>(d);
}