返回值问题

[mengde007]

关于返回值的存活期问题，我最近困了，请同志们帮忙，两个程序如下：
(一）：#include<iostream>
using namespace std;
struct sysop
{
char name[20];
char quote[64];
int used;
};
const sysop & use (sysop & sysopref);//function with a reference return type
int main()
{
sysop looper={"rick \"fortran\"looper",
"i'm a goto kind of guy.",
0
};
use (looper);//looper is a type sysop
cout <<"looper :"<<looper.used<<"use(s)\n";
return 0;
}
//use ()returns the reference passed to it
const sysop & use(sysop & sysopref)
{
sysop newguy;
newguy=sysopref;
return newguy;
}
但是上述程序的use却不可用，原因是函数返回一个临时变量（newguy)的引用，函数运行完毕后，它将不复存在;请看（二）：
#include<iostream>
using namespace std;
int value(int,int);
int main()
{
int a=3,b=4,c;
c=value(a,b);
cout<<c;
return 0;
}
int value(int a,int b)
{
c=a+b;
return c;
}
但是次程序却得到了结果，我想这两个程序都是在子函数中定义了一个数据，它们都是在调用后被释放了，为什一个可以得到返回值，而另一个却没有得到返回值呢？+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++

[ttkk_2007]

不一样啊，你第一个返回的是引用，第二个是按值返回
const sysop & use(sysop & sysopref)
{
sysop newguy;
newguy=sysopref;
return newguy;
}
=============================
newguy返回的时候销毁了，有返回引用，那么引用的就是一块无效内存

int value(int a,int b)
{
c=a+b;
return c;
}
=======================
这个是按值返回，返回的是c的拷贝， c=value(a,b); 然后再将临时变量拷贝给c

[ryfdizuo]

函数返回引用时，不能将临时变量返回，
可以改为全局，或静态变量；

[zywfriend]

上面返回是引用.下面返回的是值,解决办法,上面返回指针

[Kratos]

引用

const sysop & use(sysop & sysopref)
{
sysop newguy;
newguy=sysopref;
return newguy;
}

sysopref是不是要改成const型的？期待高人解答。

[Kratos]

晕，才看了几分钟竟然这么多人回答。

[ttkk_2007]

[Quote=引用 4 楼 cinanynames 的回复:]
引用 const sysop & use(sysop & sysopref)
{
sysop newguy;
newguy=sysopref;
return newguy;
}

sysopref是不是要改成const型的？期待高人解答。
[/Quote]
要不要改为const，看你要不要再函数里修改它，如果不修改，最好是const