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V602. The '<' operator should probab…
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Analyzer diagnostics
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V602. The '<' operator should probably be replaced with '<<'. Consider inspecting this expression.

Jan 27 2012

The analyzer has detected a potential error that may be caused by a misprint. It is highly probable that the '<<' operator must be used instead of '<' in an expression.

Consider the following code sample.

void Foo(unsigned nXNegYNegZNeg, unsigned nXNegYNegZPos,
         unsigned nXNegYPosZNeg, unsigned nXNegYPosZPos)
{
  unsigned m_nIVSampleDirBitmask =
    (1 << nXNegYNegZNeg) | (1 <  nXNegYNegZPos) |
    (1 << nXNegYPosZNeg) | (1 << nXNegYPosZPos);
  ...
}

The code contains an error, since it is the '<' operator that is written by accident in the expression. This is the correct code:

unsigned m_nIVSampleDirBitmask =
  (1 << nXNegYNegZNeg) | (1 << nXNegYNegZPos) |
  (1 << nXNegYPosZNeg) | (1 << nXNegYPosZPos);

Note.

The analyzer considers comparisons ('<', '>') odd if their result is used in binary operations such as '&', '|' or '^'. The diagnostic is more complex but we hope you understand the point in general. On finding such expressions the analyzer emits the V602 warning.

If the analyzer produces a false positive error, you may suppress it using the "//-V602" comment. But in most cases you'd better rewrite this code. It's not a good practice to handle expressions of the 'bool' type using binary operators: it makes the code unevident and less readable.

This diagnostic is classified as:

You can look at examples of errors detected by the V602 diagnostic.