C++ 如何使主函数可以访问函数中的变量?
下面我有一些简单的代码,我无法正确运行。本质上,我有一个自定义函数C++ 如何使主函数可以访问函数中的变量?,c++,function,boost,compiler-errors,C++,Function,Boost,Compiler Errors,下面我有一些简单的代码,我无法正确运行。本质上,我有一个自定义函数Create(),它根据用户的输入创建变量(点、线、圆)。然后我在main函数中调用这个函数,并尝试调用我在Create()中创建的变量。这显然行不通。如何解决这个问题 using boost::variant; //Using declaration for readability purposes typedef variant<Point, Line, Circle> ShapeType; //typedef f
Create()
,它根据用户的输入创建变量(点、线、圆)。然后我在main函数中调用这个函数,并尝试调用我在Create()
中创建的变量。这显然行不通。如何解决这个问题
using boost::variant; //Using declaration for readability purposes
typedef variant<Point, Line, Circle> ShapeType; //typedef for ShapeType
ShapeType Create()
{
int shapenumber;
cout<<"Variant Shape Creator - enter '1' for Point, '2' for Line, or '3' for Circle: ";
cin>>shapenumber;
if (shapenumber == 1)
{
ShapeType mytype = Point();
return mytype;
}
else if (shapenumber == 2)
{
ShapeType mytype = Line();
return mytype;
}
else if (shapenumber == 3)
{
ShapeType mytype = Circle();
return mytype;
}
else
{
throw -1;
}
}
int main()
{
try
{
cout<<Create()<<endl;
Line lnA;
lnA = boost::get<Line>(mytype); //Error: identified 'mytype' is undefined
}
catch (int)
{
cout<<"Error! Does Not Compute!!!"<<endl;
}
catch (boost::bad_get& err)
{
cout<<"Error: "<<err.what()<<endl;
}
}
使用boost::variant//为可读性目的使用声明
typedef变体形状类型//ShapeType的typedef
ShapeType创建()
{
整数形状枚举器;
Coutshapper;
if(shapeEnumber==1)
{
ShapeType mytype=Point();
返回mytype;
}
else if(ShapeEnumber==2)
{
ShapeType mytype=Line();
返回mytype;
}
else if(ShapeEnumber==3)
{
ShapeType mytype=Circle();
返回mytype;
}
其他的
{
投掷-1;
}
}
int main()
{
尝试
{
cout您需要存储返回值:
ShapeType retShapeType = Create() ;
std::cout<<retShapeType<<std::endl;
....
lnA = boost::get<Line>( retShapeType );
ShapeType retShapeType=Create();
标准::cout