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方案一：结构体变量作为参数，进行传值。
编译器需要拷贝，不影响origin value，使用成员操作符(.)直接访问
/**********************************************************************
* 版权所有 (C)2017, Wang maochun。
*
* 文件名称：travel.cpp
* 文件标识：无
* 内容摘要：主要演示结构体作为参数以及返回值
* 其它说明："传值”
* 当前版本：V1.0
* 作 者：Wang maochun
* 完成日期：2017.7.23
*
**********************************************************************/#include
structtravel_time
{inthours;intmins;
};const int Mins_per_hr = 60;
travel_time sum(travel_time t1,travel_time t2);voidshow_time(travel_time t);intmain()
{using namespacestd;
travel_time day1= {5,45}; //5 hours 45 minutes
travel_time day2 = {4,55}; //4 housr 55 minutes
travel_time trip=sum(day1,day2);
cout<< "Two-day total:";
show_time(trip);
travel_time day3= {4,32};
cout<< "Three-day total:";
show_time(sum(trip,day3));return 0;
}
travel_time sum(travel_time t1,travel_time t2)
{
travel_time total;
total.mins= (t1.mins + t2.mins) %Mins_per_hr;
total.hours= (t1.hours + t2.hours) + (t1.mins + t2.mins) /Mins_per_hr;returntotal;
}voidshow_time(travel_time t)
{using namespacestd;
cout<< t.hours << "hours,"
<< t.mins << "minutes\n";
}
运行结果：
方案二：结构体指针作为参数，传地址。
编译器不需要拷贝，和main函数采用相同地址。为了不影响origin value，使用const修饰
使用指针指向结构体操作符(->)间接访问
/**********************************************************************
* 版权所有 (C)2017, Wang maochun。
*
* 文件名称：travel.cpp
* 文件标识：无
* 内容摘要：主要演示结构体作为参数以及返回值
* 其它说明："传地址”
* 当前版本：V1.0
* 作 者：Wang maochun
* 完成日期：2017.7.23
*
**********************************************************************/#include
structtravel_time
{inthours;intmins;
};const int Mins_per_hr = 60;
travel_time sum(travel_time* t1,travel_time*t2);void show_time(travel_time*t);intmain()
{using namespacestd;
travel_time day1= {5,45}; //5 hours 45 minutes
travel_time day2 = {4,55}; //4 housr 55 minutes
travel_time trip= sum(&day1,&day2);
cout<< "Two-day total:";
show_time(&trip);
travel_time day3= {4,32};
cout<< "Three-day total:";
travel_time trip1=sum(&trip,&day3);
show_time(&(trip1));return 0;
}
travel_time sum(travel_time* t1,travel_time*t2)
{
travel_time total;
total.mins= (t1->mins + t2->mins) %Mins_per_hr;
total.hours= (t1->hours + t2->hours) + (t1->mins + t2->mins) /Mins_per_hr;returntotal;
}void show_time(travel_time*t)
{using namespacestd;
cout<< t->hours << "hours,"
<< t->mins << "minutes\n";
}
结果和方案一相同。
出现的错误：
show_time(&(sum(&day1,&day2)));这样写时，出现taking address of temporary fpermissive错误。
原因是：
Your middle result which is a temporary variable since it will disappear
If you assign the result of sum(&day1,&day2) to a variable then it will no longer be a temporary and you can then take the address of it.
因此，不能对未赋值的临时变量取地址
原文：http://www.cnblogs.com/shuqingstudy/p/7226413.html