How can i exist nosted "switch" without existing from the program (C++)?

Question

hi guys , i need help please
i wrote a c++ program for my exam test ,
in this program i used a nosted "switch "
how can i exist from the interior "switch" and go to the exterior "switch" without exiting from the program?

Simple example :

while(j!=2)
{
cin>>j ;
switch(j)
{
case 1 :

while(d!=2)
{
cin>>d ;
switch(d)
{
case 1 :
cout<<"hi"<<endl;
case 2 :
cout<<"hello"<<endl ;
}
}
case 2 :
cout<<"money " ;

}

}

For example i chose the first case of the interior "switch" After the result appears, I want to return to the exterior "switch "

Answer 1

Thirt13n my problem is not the declaration , i mean for example i chose the case 1 of the exterior switch , then the case 1 of the interior switch , he return hi
The i want to return to the exterior switch and chose case 2 he return money

Answer 2

hey Ramzi Jebalia
if your while loop satisfy condition then it goes to infinity !
I suggest do not use while loop !

well you do something like this ....
sol: use goto label inner switch's case:1 , label it before external switch's case 2, then use return statement after cout;

switch(d)
{ case 1:
cout<<blabla;
goto some_label;
case 2:
cout<< ;}
}
some_label : case 2 :
cout<<"money";
return 0;
}

Answer 3

Ramzi Jebalia
depends on j and d , but you have to declare variable before using them !

Answer 4

Thirt13n thank youuu , i will try it🙏

Answer 5

That's what "break" statement is for.

while(j!=2)
{
cin>>j ;
switch(j)
{
case 1 :

while(d!=2)
{
cin>>d ;
switch(d)
{
case 1 :
cout<<"hi"<<endl; break;
case 2 :
cout<<"hello"<<endl ; break;
}
}
case 2 :
cout<<"money " ;
}
}

and as far as "goto" statement is considered, I would recommend avoid using it as much as you can to avoid transforming your program to an unmanageable mess.