解引用时如何将void指针转换为指针？

Question

int x = 5;
int *xPtr = &x;
void **xPtrPtr = &xPtr;
printf("%d\n", *(int*)*xPtrPtr);

I have a void pointer pointing to an int pointer. What is the syntax for properly casting the void pointer when I want to dereference? The above code quits with:

error: invalid conversion from ‘int**’ to ‘void**’

Thank you!

Answer 1

For void pointers, you don't really need to know how many indirections there are.

#include <iostream>
int main(void){
    int x = 5;
    int* pX = &x;
    void* pV = &pX;
    std::cout << "x = " << **(int**)pV << std::endl;
    // better use C++-Style casts from the beginning
    // or you'll be stuck with the lazyness of writing the C-versions:
    std::cout << "x = " << **reinterpret_cast<int**>(pV) << std::endl;
    std::cin.get();
}

Output:

x = 5
x = 5

See this here .

Answer 2

指向int的指针是*((int **)xPtrPtr)

Answer 3

void **xPtrPtr = (void**)&xPtr;

Answer 4

    int x = 5;
    int *xPtr = &x;
    void **xPtrPtr = reinterpret_cast<void**>(&xPtr);
    int y = **reinterpret_cast<int**>(xPtrPtr);
    cout << y;

Compile: No Error. No Warning!

Output:

5

Code at ideone : http://www.ideone.com/1nxWW

Answer 5

This compiles correctly:

#include <stdio.h>

int main(int argc, char *argv[]) {
   int x = 57; 
   int *xPtr = &x; 
   void **xPtrPtr = (void**)&xPtr; 
   printf("%d\n", *((int*)*xPtrPtr));
}