指针上的此操作如何工作? [英] How does this operation on pointers work?

问题描述

  int x = 4;
  int* q = &x;                 // Is it always equivalent to int *q = &x;  ?
  cout << "q = " << q << endl; // output: q = 0xbfdded70
  int i = *q;                  // A
  int j = *(int*)q;            // B, when is this necessary?
  cout << "i = " << i << endl; // output: i = 4
  cout << "j = " << j << endl; // output: j = 4

我的问题是A和B行是做什么的，为什么输出都为4?

My question is what does lines A and B do, and why the outputs are both 4?

推荐答案

  int x = 4;

x是4

  int* q = &x;

q是x(包含4个)的内存位置

q is the memory location of x (which holds 4)

  cout << "q = " << q << endl; // output: q = 0xbfdded70

有您的存储位置.

  int i = *q; // A

i是内存位置q上的值

i is the value at memory location q

  int j = *(int*)q; // B

j是存储器位置q的值. q被强制转换为int指针，但这已经是原来的了.

j is the value at memory location q. q is being cast to an int pointer, but that's what it already is.

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