No. Dude, the super class is always called.
SCJP 6 | SCWCD 5 | Javaranch SCJP FAQ | SCWCD Links
SCJP 6 | SCWCD 5 | Javaranch SCJP FAQ | SCWCD Links
Anikt Garg wrote:No its not. If your constructor calls another constructor of the same class using this then the compiler doesn't insert a call to super class constructor.
“The difference between 'involvement' and 'commitment' is like an eggs-and-ham breakfast: the chicken was 'involved' - the pig was 'committed'.”
“The difference between 'involvement' and 'commitment' is like an eggs-and-ham breakfast: the chicken was 'involved' - the pig was 'committed'.”
O. Ziggy wrote:the constructor will not generate a super() statement if the first line of the constructor is a call to this()
Rohit Ramachandran wrote:No. Dude, the super class is always called. Irrespective of what the first statement of the constructor is.
SCJP 6 | SCWCD 5 | Javaranch SCJP FAQ | SCWCD Links
“The difference between 'involvement' and 'commitment' is like an eggs-and-ham breakfast: the chicken was 'involved' - the pig was 'committed'.”
To conclude if I say no matter what you do, it is guaranteed that at least one version of the base class constructor would be invoked
SCJP 6 | SCWCD 5 | Javaranch SCJP FAQ | SCWCD Links