The analyzer has detected a potential error in a construct consisting of conditional statements.
Consider the following example:
if (a == 1) Foo1(); else if (a == 2) Foo2(); else if (a == 1) Foo3();
In this code, the 'Foo3()' method will never get control. We are most likely dealing with a logical error here and the correct version of this code should look as follows:
if (a == 1) Foo1(); else if (a == 2) Foo2(); else if (a == 3) Foo3();
This diagnostic is classified as:
You can look at examples of errors detected by the V6003 diagnostic.