Pour obtenir une clé
d'essai remplissez le formulaire ci-dessous
Demandez des tariffs
Nouvelle licence
Renouvellement de licence
--Sélectionnez la devise--
USD
EUR
RUB
* En cliquant sur ce bouton, vous acceptez notre politique de confidentialité

Free PVS-Studio license for Microsoft MVP specialists
To get the licence for your open-source project, please fill out this form
** En cliquant sur ce bouton, vous acceptez notre politique de confidentialité.

I am interested to try it on the platforms:
** En cliquant sur ce bouton, vous acceptez notre politique de confidentialité.

Votre message a été envoyé.

Nous vous répondrons à


Si vous n'avez toujours pas reçu de réponse, vérifiez votre dossier
Spam/Junk et cliquez sur le bouton "Not Spam".
De cette façon, vous ne manquerez la réponse de notre équipe.

>
>
>
V2566. MISRA. Constant expression evalu…
Analyzer diagnostics
General Analysis (C++)
General Analysis (C#)
General Analysis (Java)
Diagnosis of 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
Contents

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

23 Déc 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 true. 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
Unicorn with delicious cookie
Nous utilisons des cookies pour améliorer votre expérience de navigation. En savoir plus
Accepter