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Is unevaluated division by 0 undefined behavior?

Posted by: admin November 29, 2017 Leave a comment

Questions:

I’m having a disagreement with some co-workers over the following code:

int foo ( int a, int b )
{
    return b > 0 ? a / b : a;
}

Does this code exhibit undefined behavior?

EDIT: The disagreement started from what appears to be a bug in an overly-eager optimizing compiler, where the b > 0 check was optimized out.

Answers:

No.


Quotes from N4140:

§5.16 [expr.cond]/1

Conditional expressions group right-to-left. The first expression is
contextually converted to bool. It is evaluated and if it is true, the
result of the conditional expression is the value of the second
expression, otherwise that of the third expression. Only one of the
second and third expressions is evaluated.

Further:

§5 [expr]/4

If during the evaluation of an expression, the result is not
mathematically defined or not in the range of representable values for
its type, the behavior is undefined.

This clearly does not happen here. The same paragraph mentions division by zero explicitly in a note, and, although it is non-normative, it’s making it even more clear that its pertinent to this situation:

[ Note: most existing implementations of C++ ignore integer overflows.
Treatment of division by zero, forming a remainder using a zero
divisor, and all floating point exceptions vary among machines, and is
usually adjustable by a library function. —end note ]

There’s also circumstantial evidence reinforcing the above point: the conditional operator is used to conditionally make behavior undefined.

§8.5 [dcl.init]/12.3

int f(bool b) {
  unsigned char c;
  unsigned char d = c; // OK, d has an indeterminate value
  int e = d; // undefined behavior
  return b ? d : 0; // undefined behavior if b is true
}

In the above example, using d to initialize int (or anything other than unsigned char) is undefined. Yet it is clearly stated that the UB occurs only if the UB branch is evaluated.


Going out of language-lawyer perspective: if this could be UB, then any division could be treated as UB, since the divisor could potentially be 0. This is not the spirit of the rule.

Questions:
Answers:

There is no way of dividing with zero in the example code. When the processor executes a / b, it has already checked that b > 0, therefore b is non-zero.

It should be noted that if a == INT_MIN and b == -1, then a/b is undefined behaviour too. But this is prevented anyway because the condition evaluates to false in that case.

Although I am not really sure you meant return b != 0 ? a / b : a; and not return b > 0 ? a / b : a; If b is less than zero, the division is still valid, unless it is the condition described above.

Questions:
Answers:

Does this code exhibit undefined behavior?

No. It doesn’t. The expression

return b > 0 ? a / b : a;  

is equivalent to

if(b > 0)
    return a/b;     // this will be executed only when b is greater than 0
else
    return a;  

Division only performed when b is greater than 0.

Questions:
Answers:

If this were UB then so would

if(a != null && *a == 42)
{
 .....
}

And the sequencing of ifs , ands and ors is clearly designed to specifically allow this type of construct. I cant imagine your colleagues would argue with that