switch-case statement without break

Question

According to this book I am reading：

Q What happens if I omit a break in a switch-case statement?

A The break statement enables program execution to exit the switch construct. Without it, execution continues evaluating the following case statements.

Suppose if I have codes looking like

switch (option}{     case 1:     do A;     case 2:     do B;     default:     do C;     break; } 

Does this mean if I choose case 1, the A and C are done. If I choose case 2, B and C are done. If i choose neither, then only C is done.

if so, what happens if we omit the break after do C.

I assume these are bad programming practice, but I am curious what would happen to get a deeper understanding how it all works. Thanks

Answer 1

You execute everything starting from the selected case up until you see a break or the switch statement ends. So it might be that only C is executed, or B and then C, or A and B and C, but never A and C

Answer 2

• If you don't include break in any of case then all the case below will be executed and until it sees break.

• And if you don't include break in default then it will cause no effect as there are not any case below this 'Default' case.

• And not using break generally considered as a bad practice but some time it may also come handy because of its fall-through nature.For example:

  case optionA:

  //optionA needs to do its own thing, and also B's thing. //Fall-through to optionB afterwards. //Its behaviour is a superset of B's. 

  case optionB:

  // optionB needs to do its own thing // Its behaviour is a subset of A's. break; 

  case optionC:

  // optionC is quite independent so it does its own thing. break;