c++ - 取消引用时如何将空指针转换为指针？

Question

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

我有一个指向 int 指针的 void 指针。当我想取消引用时正确转换 void 指针的语法是什么？上面的代码退出:

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

谢谢!

Answer 1

对于空指针，你真的不需要知道有多少间接寻址。

#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();
}

输出:

x = 5
x = 5

See this here .