we cannot create object of any class in switch case directly. Why?

Question

#include <iostream>
using namespace std;

class Student{
string name;
};

int main() {
int n=1;
switch(n){
case 1:
Student s1;
break;
default:
break;
}
return 0;
}

Answer 1

Notice that the switch statement body is enclosed in braces { }. These define the beginning and end of a basic block of code. Any variables (or objects) defined with a basic block exist only within that block. It is a principle known as scope. When execution goes outside the block, the variables defined within the block go out of scope and are discarded. They become inaccessible.

Answer 2

you can use pointers

#include <iostream>
using namespace std;

class Student{
string name{"John"};
public:
Student()=default;
Student(const string& n):name(n){
cout<<name<<endl;
}
void sayname(){
cout<<"I am "<<name<<endl;
}
};

int main() {
int n=1;
Student* s1;
switch(n){
case 1:
s1 = new Student();
break;
case 2:
s1 = new Student("Jane");
break;
default:
break;
}
s1-> sayname();
}

Answer 3

add braces:

switch(n) {
case 1:
{
Student s1;
break;
}
default:
break;
}

Answer 4

john ds
s1 still only exists within the block scope of the braces.

Answer 5

Bob_Li my bad, i thought he had a compiler error and as long he not use it anywhere else outside the block i assumed he asking why cannot compile it

Answer 6

Please show your code.

Answer 7

using unique_ptr

https://sololearn.com/compiler-playground/c3YvXqrdi4WS/?ref=app