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Rule 4-5-1 prohibits 'b1 |= b2;' ? - Printable Version

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+---- Thread: Rule 4-5-1 prohibits 'b1 |= b2;' ? (/showthread.php?tid=1277)

Rule 4-5-1 prohibits 'b1 |= b2;' ? - grunwald - 28-09-2016

Neither `|` nor `|=` is listed as allowed operator.
Does this mean the following code a violation of Rule 4-5-1?

Code:
bool b1 = false;

void f(int x)
{
    b1 |= (x == 0);
}


Re: Rule 4-5-1 prohibits 'b1 |= b2;' ? - dg1980 - 29-09-2016

grunwald Wrote:Neither `|` nor `|=` is listed as allowed operator.
Does this mean the following code a violation of Rule 4-5-1?

Code:
bool b1 = false;

void f(int x)
{
    b1 |= (x == 0);
}
x |= y is just short for x = x | y and bitwise operators are forbidden by the rationale in this rule.
Code:
FlexeLint for C/C++ (Unix) Vers. 9.00L, Copyright Gimpel Software 1985-2014
--- Module: diy.cpp (C++)
     1  //lint -indirect(au-misra-cpp-alt.lnt) -e514 -e1786 -e9141 -e970 -e952
     2  bool b1 = false;
     3  
     4  void f(int x)
     5  {
                          _
     6      b1 |= (x == 0);
diy.cpp  6  Note 9111:  boolean expression used with non-permitted operator ''|'' [MISRA C++ Rule 4-5-1]
diy.cpp  6  Note 9111:  boolean expression used with non-permitted operator ''|'' [MISRA C++ Rule 4-5-1]


Re: Rule 4-5-1 prohibits 'b1 |= b2;' ? - mishak - 29-09-2016

Remember that any non-zero value will be considered as 'true' within a logical context, which can lead to surprises:

Code:
a = 0x01u;   // This is logically 'true'
b = 0x10u;   // This is logically 'true'

if ( a &  b )   // Result of bit-wise 'and' is 'false'
if ( a && b )   // Result of logical 'and' is true
Rule 4-5-1 helps to enforce a consistent use of boolean types, helping to allowing defects to be detected.

Re: Rule 4-5-1 prohibits 'b1 |= b2;' ? - misra cpp - 28-10-2016

Yes, b1 |= b2 is prohibited

As pointed out by dg1980, |= is a bitwise operator and the rationale of the rule prohibits the use of bitwise operators