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c++ - 在 std::cout 中递增变量时指针不显示更新值

转载 作者:塔克拉玛干 更新时间:2023-11-03 01:34:30 26 4
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<分区>

#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 c 在这两种情况下都没有更新为新值吗?

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