Question

#include <iostream>
#include <string>

using namespace std;

int main()
{
  int a = 5;
  int& b = a;
  int* c = &a;

  cout << "CASE 1" << endl;
  cout << "a is " << a << endl << "b is " << b << endl << "c is " << *c << endl;
  b = 10;
  cout << endl << "a is " << a << endl << "b is " << b << endl << "c is " << *c << endl << endl;

  cout << "CASE 2";
  a = 5;
  cout << endl << "a is " << a << endl << "b is " << b << endl << "c is " << *c << endl;
  b = 10;
  cout << endl << "a is " << a << endl << "b is " << ++b << endl << "c is " << *c << endl << endl;

  cout << "CASE 3";
  a = 5;
  cout << endl << "a is " << a << endl << "b is " << b << endl << "c is " << *c << endl;
  b = 10;
  cout << endl << "a is " << a << endl << "b is " << b++ << endl << "c is " << *c << endl;
}

输出：

案例1：

a is 5. b is 5. c is 5.
a is 10. b is 10. c is 10.

案例2：

a is 5. b is 5. c is 5.

a is 11. b is 11. c is 10.

案例3：

a is 5. b is 5. c is 5.
a is 11. b is 10. c is 10.

我理解案例1.但我很难理解案例2和案例3.有人可以解释为什么在这两种情况下都不会c更新新值？

Answer 1

操作数的评估顺序未指定，您正在修改对象并在不对这些操作进行排序的情况下读取它。

因此，您的程序未定义为cout << a << a++;，任何事情都可能发生。

Answer 2

我认为你的问题是由于所谓的序列点。您可以在this answer中详细阅读该文章，但简而言之，它基本上说明了表达式元素的顺序或评估。

在你的情况下

更新这个订单是未定义的，尽管有些编译器似乎是从右到左。

在std :: cout中递增变量时，指针未显示更新的值

2 个答案: