V2569. MISRA. The 'operator &&', 'operator ||', 'operator ,' and the unary 'operator &' should not be overloaded.

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++.

The built-in operators '&&', '||', '&' (address-of), and ',' have a specific evaluation order and semantics. When overloaded, they can no longer maintain their specific behavior, and the programmer may not know about that.

1) When overloaded, logical operators no longer support lazy evaluation. When using built-in operators, the second operand is not evaluated if the first operand of '&&' is 'false' or if the first operand of '||' is 'true'. Overloading these operators makes such optimization impossible:

class Tribool
  Tribool(bool b) : .... { .... }
  friend Tribool operator&&(Tribool lhs, Tribool rhs) { .... }
  friend Tribool operator||(Tribool lhs, Tribool rhs) { .... }

// Do some heavy weight stuff
bool HeavyWeightFunction();

void foo()
  Tribool flag = ....;
  if (flag || HeavyWeightFunction()) // evaluate all operands
                                     // no short-circuit evaluation
    // Do some stuff

The compiler will not be able to optimize this code and will have to execute the "heavy-weight" function, which could have been avoided if the built-in operator had been used.

2) Overloading the unary operator '&' (address-of) can also lead to non-obvious issues. Consider the following example:

// Example.h
class Example
        Example* operator&()      ;
  const Example* operator&() const;

// Foo.cc

#include "Example.h"

void foo(Example &x)
  &x; // call overloaded "operator&"

// Bar.cc
class Foobar;

void bar(Example &x)
  &x; // may call built-in or overloaded "operator&"!

The behavior observed in the second case is considered unspecified according to the C++ standard ($, which means that applying the 'address-of' operator to the 'x' object may result in randomly calling the built-in operator or its overloaded version.

3) The built-in operator "comma" evaluates the left operand and ignores the resulting value; it then evaluates the right operand and returns its value. The built-in comma operator also guarantees that any side effects of the left operand will have taken place before it starts evaluating the right operand.

There is no such guarantee in the case of the overloaded version (before C++17), so the code below may output 'foobar' or 'barfoo':

#include <iostream>

template <typename T1, typename T2>
T2& operator,(const T1 &lhs, T2 &rhs)
  return rhs;

int main()
  std::cout << "foo", std::cout << "bar";
  return 0;

This diagnostic is classified as:

  • MISRA-CPP-5.2.11
  • MISRA-CPP-5.3.3

Bugs Found

Checked Projects
Collected Errors
14 526
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