Our website uses cookies to enhance your browsing experience.
Accept
to the top
close form

Fill out the form in 2 simple steps below:

Your contact information:

Step 1
Congratulations! This is your promo code!

Desired license type:

Step 2
Team license
Enterprise license
** By clicking this button you agree to our Privacy Policy statement
close form
Request our prices
New License
License Renewal
--Select currency--
USD
EUR
* By clicking this button you agree to our Privacy Policy statement

close form
Free PVS‑Studio license for Microsoft MVP specialists
* By clicking this button you agree to our Privacy Policy statement

close form
To get the licence for your open-source project, please fill out this form
* By clicking this button you agree to our Privacy Policy statement

close form
I am interested to try it on the platforms:
* By clicking this button you agree to our Privacy Policy statement

close form
check circle
Message submitted.

Your message has been sent. We will email you at


If you do not see the email in your inbox, please check if it is filtered to one of the following folders:

  • Promotion
  • Updates
  • Spam

Webinar: C++ semantics - 06.11

>
>
>
V2566. MISRA. Constant expression evalu…
menu mobile close menu
Analyzer diagnostics
General Analysis (C++)
General Analysis (C#)
General Analysis (Java)
Micro-Optimizations (C++)
Diagnosis of 64-bit errors (Viva64, C++)
Customer specific requests (C++)
MISRA errors
AUTOSAR errors
OWASP errors (C#)
Problems related to code analyzer
Additional information
toggle menu Contents

V2566. MISRA. Constant expression evaluation should not result in an unsigned integer wrap-around.

Dec 23 2019

This diagnostic rule is based on the software development guidelines developed by MISRA (Motor Industry Software Reliability Association).

This diagnostic rule applies only to code written in C. As specified by the C standard, an overflow of values of unsigned types results in a wrap-around. Using this mechanism in evaluation of expressions at runtime is a well-known practice (unlike signed types, where an overflow leads to undefined behavior).

However, an unsigned integer wrap-around in expressions evaluated at compile time may be misleading.

Example of non-compliant code:

#include <stdint.h>
#define C1 (UINT_MAX) 
#define C2 (UINT_MIN) 
....
void foo(unsigned x)
{
  switch(x)
  {
    case C1 + 1U: ....; break;
    case C2 - 1U: ....; break;
  }
}

According to this rule, an unsigned integer wrap-around that occurs when evaluating a constant expression of unsigned type, it will not be treated as an error if the expression will never be evaluated:

#include <stdint.h>
#define C UINT_MAX
....
unsigned foo(unsigned x)
{
  if(x < 0 && (C + 1U) == 0x42) ....;  
  return x + C; 
}

The '(C + 1U)' expression resulting in an overflow will not be executed since the 'x < 0' condition is always false. Therefore, the second operand of the logical expression will not be evaluated.

This diagnostic is classified as:

  • MISRA-C-12.4
  • MISRA-CPP-5.19.1