Question

假设我在C ++中有以下示例程序：

#include <iostream>
using namespace std;

int main (int argc , char** argv) {

int a = 1;
int b = 2;
int& aRef = a;
int& bRef = b;
aRef = bRef; // This just sets aRef to point to b?
aRef = 3; // Now aRef points to a new int 3 not stored in a other variable?
// a = 3 b = 2
bRef = 4;
// a = 3 b = 4
aRef = long(&bRef); // Why do we need long casting here?
bRef = 5;
// a: varying b = 5 // Why is a varying?
aRef = bRef;
bRef = 6;
// a = 5 b = 6 // Why a no more varying?

}

有人可以逐行解释并可能揭示错误吗？我已经添加了一些注释，这些注释对我来说特别不清楚。

Answer 1

#include <iostream>
using namespace std;

int main (int argc , char** argv) {
    int a = 1;
    int b = 2;
    int& aRef = a;         // aRef is now another name for a
    int& bRef = b;         // bRef is now another name for b
    aRef = bRef;           // same as a = b
    aRef = 3;              // same as a = 3
    bRef = 4;              // same as b = 4
    aRef = long(&bRef);    // &bRef is the same as &b - i.e. take address of b - stores address in a
    bRef = 5;              // same as b = 5
    aRef = bRef;           // same as a = b
    bRef = 6;              // same as b = 6

}

参考类型示例问题

1 个答案: