1. Local variables of a method are never initialised automatically. It would be a compile time error if we try to access them without initialisation.
2. int a;
System.out.print(a);
This is compile time error as a is not initialised.
3. int a=4, b;
if(a==4)
b=5;
System.out.print(a);
This is compile time error because variable expression a==4 will be checked at run time. So b will be uninitialised at compile time.
4. int a=4, b;
if(a==4)b=5;
System.out.print(a);
This is right as due to final a==4 is a constant expression and will be checked at compile time.
5. String str = "25";
int no;
no = str; //Compile Time error
This will be a compile time error. To remove error we should use
no= Integer.parseInt(str);
parse is a static method in Integer class which returns the integer conversion of the string.
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