STRONG.TYPE.JOIN.CONST

Comparison of strong type with constant

The STRONG.TYPE family of checkers detects situations in which programmer-enforced strong typing (type-defined abstract types) is broken or ignored, allowing the underlying ANSI type semantics to dominate.

The STRONG.TYPE.JOIN.CONST checker looks for an instance in which a strongly typed value is compared with a constant using a binary operator. In this rule, constants can be considered: integral constants, quoted strings, or expressions of the form &v, in which v is a static or automatic variable.

Vulnerability and risk

A compiler following the ANSI standard won't report a warning for this sort of issue, as it checks only the underlying types, not the surface, or programmer-defined, types. As a result, it's possible that a logic error can occur.

Vulnerable code example

Copy 
 typedef int Weight;

 int main() {
   Weight w;
   if (w == 321) ;
   return 0;
 }

Klocwork flags line 5, indicating that the strongly typed value w is compared with a constant using binary operator ==.

Fixed code example

Copy 
 typedef int Weight;

 int main() {
   Weight w;
   if (w == (Weight) 321) ; 
   return 0;
 }

In the fixed code, the comparison is made between two strongly typed Weight values.

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