ref和指针之间的不同行为 [英] Different behavior between ref and pointer

问题描述

  int main（）$ b这是一个奇怪的事，我遇到了， $ b {
 int i {5}; 
 
 void * v =& i; 
 
 int * t = reinterpret_cast< int *>（v）; 
 int u = reinterpret_cast< int&>（v）; 
 
 int i2 = * t; 
 
} 
  

t正确为5，u为垃圾。

发生了什么事？

解决方案

code>存储 i 的地址，这就是你所说的垃圾。 （它不是真的垃圾，但是价值本身是无意义的，除了在程序的这个特定运行中的地址 i ）。 u 存储与 v 相同的值（按位），但重新解释为整数，因此它是垃圾。

t 不是5。或者如果是，这是一个极不可能的巧合。 t 是 i 的地址， i 是5 ，因此 * t （即 t 解除引用）为5。

This is just an oddity I ran into and can't quite understand what's happening.

int main()
{
    int i{ 5 };

    void* v = &i;

    int* t = reinterpret_cast<int*>(v);
    int u = reinterpret_cast<int&>(v);

    int i2 = *t;

}

t is correctly 5 and u is garbage.

What's going on?

解决方案

v stores the address of i, which is, as you put it, garbage. (it's not really garbage, but the value itself is meaningless, except as it is the address of i in this particular run of the program). u stores the same value (bitwise) as v, but reinterpreted as an integer, so it is "garbage".

t is not 5, as you claim. Or if it is, it's an extremely unlikely coincidence. t is the address of i, and i is 5, so *t (that is, t dereferenced) is 5.

这篇关于ref和指针之间的不同行为的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！