函数返回引用怎么理解,sysop函数返回的引用是谁的引用？looper的还是sysopyer的？

Question

// strtref.cpp -- using structure references

#include <iostream>

using namespace std;

struct sysop

{

    char name[26];

    char quote[64];

    int used;

};

const sysop & use(sysop & sysopref);  // function with a reference return type

int main()

{

// NOTE: some implementations require using the keyword static

// in the two structure declarations to enable initialization

    sysop looper =

    {

        "Rick \"Fortran\" Looper",

        "I'm a goto kind of guy.",

        0

    };

    use(looper) ;            // looper is type sysop

    cout << "Looper: " << looper.used << " use(s)\n";

    sysop copycat;

copycat = use(looper);

    cout << "Looper: " << looper.used << " use(s)\n";

    cout << "Copycat: " << copycat.used << " use(s)\n";

    cout << "use(looper): " << use(looper).used << " use(s)\n";

    return 0;

}

// use() returns the reference passed to it

const sysop & use(sysop & sysopref)

{

    cout << sysopref.name << " says:\n";

    cout << sysopref.quote << endl;

    sysopref.used++;

    return sysopref;

}

Answer 1

不好意思，不会C++