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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 rule is classified as:
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You can look at examples of errors detected by the V6003 diagnostic. |
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