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V809. Verifying that a pointer value is…
Analyzer Diagnostics
General Analysis (C++)
General Analysis (C#)
General Analysis (Java)
Diagnosis of micro-optimizations (C++)
Diagnosis of 64-bit errors (Viva64, C++)
MISRA errors
AUTOSAR errors
Additional information
Contents

V809. Verifying that a pointer value is not NULL is not required. The 'if (ptr != NULL)' check can be removed.

Dec. 6, 2012

The analyzer has detected a code fragment that can be simplified. The 'free()' function and 'delete' operator handle the null pointer correctly. So we can remove the pointer check.

Here's an example:

if (pointer != 0)
  delete pointer;

The check is excess in this case, as the 'delete' operator processes the null pointer correctly. This is how to fix the code:

delete pointer;

We cannot call this fix a true optimization, of course. But it allows us to delete an unnecessary string to make the code shorter and clearer.

There's only one case when the pointer check does have sense: when the 'free()' function or 'delete' operator are called VERY many times, and the pointer, at the same time, ALMOST ALWAYS equals zero. If user code contains the check, system functions won't be called. It will even reduce the run time a bit.

But in practice, a null pointer almost always indicates some error. If the program works normally, pointers won't equal zero in 99.99% of cases. That's why the check can be removed.

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