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V6019. Unreachable code detected. It is possible that an error is present.
The analyzer detected code that will never be executed. It may signal the presence of a logic error.
This diagnostic is designed to find blocks of code that will never get control.
Consider the following example:
void printList(List<Integer> list) {
if (list == null) {
System.exit(-999);
System.err.println("Error!!! Output empty!!! list == null");
}
list.forEach(System.out::println);
}The ''prinln (...)' function will never print the error message, as the 'System.exit(...)' function does not return control. The exact way of fixing this error depends on the logic intended by the programmer. The function could be meant to return control, or maybe the expressions are executed in the wrong order and the code was actually meant to look like this:
void printList(List<Integer> list) {
if (list == null) {
System.err.println("Error!!! Output empty!!! list == null");
System.exit(-999);
}
list.forEach(System.out::println);
}Here is another example:
void someTransform(int[] arr, int n, boolean isErr, int num, int den)
{
if (den == 0 || isErr)
{
return;
}
...
for (int i = 0; i < n; ++i)
{
if (!isErr || arr[i] <= 0)
continue;
arr[i] += 2 * num/den;
}
...
}In this fragment the code 'arr[i] += 2 * num/den;' was not executed. Checks of a variable 'isErr' in the beginning of the method and loop are contradictory and changes of this variable between the checks aren't. Therefore, the operator 'continue' will be executed for each step of the loop. Most likely, this happened because of refactoring.
Example of suspicious code:
void someTransform(int[] arr, int n, boolean isErr, int num, int den)
{
if (den == 0 || isErr)
{
return;
}
...
for (int i = 0; i < n; ++i)
{
if (arr[i] <= 0)
continue;
arr[i] += 2 * num/den;
}
...
}This diagnostic is classified as:
You can look at examples of errors detected by the V6019 diagnostic. |
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