How this code run.

Question

char f1 (char c)
{
return c=='z' ? 'a' : c+1;
}
c=f1(c);
return c;
}
int main ()
{
char x='x';
cout<<f2(x);
cout<<f2(x);
cout<<f2(x);
}

Answer 1

This doesn't run.

// This is an incorrect call to the function and is also called outside of main, which you can't do. c was also not declared as a char yet. If you move this inside of the function f1, then f1 becomes recursive. But this would never be executed if it comes after the return inside of f1.
c=f1(c);

//This return is not within a function and has no purpose. Because c was not declared yet, this will also be an error
return c;

//This closing bracket is not closing anything. It is an extra bracket and will cause issues
}

//In main a function called f2 is called, which doesn't exist
cout << f2(x)

To fix this, you can have the following:

char f1 (char c)
{
cout << c << endl;
return c == 'z' ? 'a' : c + 1;
}

int main()
{
char x = 'x';
x = f1(x);//outputs 'x', x = 'y'
x = f1(x);//outputs 'y', x = 'z'
x = f1(x);//outputs 'z', x = 'a'
cout << x << endl;//outputs 'a'
return 0;
}

https://code.sololearn.com/cPukOKDTM2X1