To get a trial key
fill out the form below
Team License (a basic version)
Enterprise License (an extended version)
* By clicking this button you agree to our Privacy Policy statement

Request our prices
New License
License Renewal
--Select currency--
USD
EUR
GBP
RUB
* By clicking this button you agree to our Privacy Policy statement

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

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

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

Message submitted.

Your message has been sent. We will email you at


If you haven't received our response, please do the following:
check your Spam/Junk folder and click the "Not Spam" button for our message.
This way, you won't miss messages from our team in the future.

>
>
>
V699. It is possible that 'foo = bar ==…
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
OWASP errors (C#)
Additional information
Contents

V699. It is possible that 'foo = bar == baz ? .... : ....' should be used here instead of 'foo = bar = baz ? .... : ....'. Consider inspecting the expression.

Sep 08 2014

The analyzer has detected an expression of the 'foo = bar = baz ? xyz : zzy' pattern. It is very likely to be an error: the programmer actually meant it to be 'foo = bar == baz ? xyz : zzy' but made a mistake causing the code to do assignment instead of comparison.

For example, take a look at the following incorrect code fragment:

int newID = currentID = focusedID ? focusedID : defaultID;

The programmer made a mistake writing an assignment operator instead of comparison operator. The fixed code should look like this:

int newID = currentID == focusedID ? focusedID : defaultID;

Note that the code below won't trigger the warning because the expression before the ternary operator is obviously of the bool type, which makes the analyzer assume it was written so on purpose.

result = tmpResult = someVariable == someOtherVariable? 1 : 0;

This fragment is quite clear. It is equivalent to the following lengthier one:

if (someVariable == someOtherVariable)
  tmpResult = 1;
else
  tmpResult = 0;
result = tmpResult;

This diagnostic is classified as:

This website uses cookies and other technology to provide you a more personalized experience. By continuing the view of our web-pages you accept the terms of using these files. If you don't want your personal data to be processed, please, leave this site.
Learn More →
Accept