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Webinar: Parsing C++ - 10.10

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V3181. The result of '&' operator is '0…
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V3181. The result of '&' operator is '0' because one of the operands is '0'.

Sep 07 2022

The analyzer has detected that there is a bitwise 'AND' (&) operation with an operand equal to 0. A code fragment may contain incorrect operator or operand.

Example:

public enum TypeAttr
{
  NotPublic = 0x0,
  Public = 0x1,
  NestedPublic = 0x2,
  NestedPrivate = 0x3
}

public static bool IsNotPublic(TypeAttr type)
{
  return (type & TypeAttr.NotPublic) == TypeAttr.NotPublic;
}

The 'IsNotPublic' method checks if the argument of the 'TypeAttr' type has the 'NotPublic' flag.

Such a method of checking doesn't have any practical sense since the 'TypeAttr.NotPublic' flag has a zero value, which means using this flag as an operand of the '&' operator always results in a zero value. Thus, in the presented implementation, we always get the true condition.

The correct implementation of the check may look as follows:

public static bool IsNotPublic(TypeAttr type)
{
  return type == TypeAttr.NotPublic;
}

The analyzer will also issue a warning for the use of the operand that is '0' with the '&=' operator. This code also looks questionable, because if one of the operands is '0', the result of the expression is also '0'.

This diagnostic is classified as:

You can look at examples of errors detected by the V3181 diagnostic.