STRONG.TYPE.JOIN.ZERO

Comparison of strong type with zero

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.ZERO checker looks for instances of comparison between a strongly typed value and zero using a binary operator.

In this rule, zero is defined as any zero constant that has not been cast to a strong type. For example, the checker considers the following to be zero:

  • 0L
  • (int)0

The checker considers the following examples not to be zero:

  • (SPEED)0
  • (SPEED*)0

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 float Speed;

 int main() {
   Speed s;
   if (s == 0) ; 
   return 0;
 }

Klocwork flags line 5, indicating that a strongly typed value, s, is compared with zero.

Fixed code example

Copy 
 typedef float Speed;

 int main() {
   Speed s;
   if (s == (Speed) 0) ; 
   return 0;
 }

In the fixed code, the comparison is made clearly with two strongly typed values.

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