I have this switch system and I'm using eclemma to test the branch coverage. We are required to have at least 80% in branch coverage for everything so I'm trying to test as much as possible. However, eclemma tells me this switch system is not fully tested in terms of branch coverage.
pos = p.getCurrentPosition().substring(0, 1);
switch (pos) {
case "G":
goalkeepers++;
break;
case "D":
defense++;
break;
case "M":
midfield++;
break;
case "F":
offense++;
break;
case "S":
substitutes++;
break;
case "R":
reserves++;
break;
}
I used straightforward JUnit tests to go trough each of these cases. Still eclemma marks this as yellow and says "7 of 19 branches missed". I would say there are only 7 ways to go through this switch system (the 6 individual cases+all undefined).
I tried searching for similar questions on stack overflow. Some of them had as solutions to use if/else for full coverage. I'm not sure if this is the only way possible to get this coverage.
Can anybody explain where all these 19 branches come from and how I could test these remaining 7 to get a 100% branch coverage on this switch case?
The Java compiler translates the switch-case code either to a
tableswitch
or to alookupswitch
. Thetableswitch
is used when there are only a few gaps are between the different cases. Otherwise, thelookupswitch
is used.In your case a
tableswitch
is used because the hash codes of your cases are closely spaced (unlike in the code referenced by owaism):The numbers to the left of the colon are the ordered hash codes and the filled gaps between them, the numbers to the right are the jump destinations. (In Java, the hash code of a character is its ASCII value.)
68
is the hash code of "D" (the lowest one), and83
is the hash code of "S" (the highest one).69
is the value of one of the gaps between the real cases and will jump to the default case.However, I assume that EclEmma excludes these branches from the coverage computation of a
tableswitch
(it would lower the coverage even more because of the gaps). So we have 0 (counted) branches yet.Next, an equals comparison of the string value is performed at each jump destination (except at the one of the default case). As your switch-case consists of 6 cases, we have 6 six jump destinations with an equals comparison.
The byte code of the comparison for case "G" is below:
EclEmma counts two branches: either the input string and the case string are equals or they are not. Thus, we have 6 * 2 branches for the comparisons. (The default case does not branch.)
Next, if the two strings are equal the index of the case will be stored (byte code lines
105-106
for the case "G"). Then a jump to the secondtableswitch
will be executed. Otherwise, the jump will be executed directly.This switch operates on the previously stored case index and jumps to the code in the case (case "G" has index
0
, the default case has-1
). EclEmma counts 7 branches (6 cases plus the default case).Consequently, we have 0 counted branches in the first
tableswitch
, 12 branches in theequals
comparisons and further 7 branches in the secondtableswitch
. All in all, this results in 19 branches.Your tests do not cover any of the 6 not-equals branches. In order to cover these, you would need to find a string for each case which is not equal to the case condition but has the same hash code. It is possible, but definitively not sensible...
Probably, the branch counting of EclEmma will be adjusted in the future.
Moreover, I guess you don't have a test case which does not match with any of the cases (thus the (implicit) default case is not covered.)
Check out the following Link: http://sourceforge.net/p/eclemma/discussion/614869/thread/80e770df/
Below is snippet from the above link:
This for an example having a switch with 3 cases: