Conditional execution based on short-circuit logical operation

1
votes

As the evaluation of logical operators && and || are defined as "short circuit", I am assuming the following two pieces of code are equivalent:

p = c || do_something();

and

if (c) {
   p = true;
}
else {
   p = do_something();
}

given p and c are bool, and do_something() is a function returning bool and possibly having side effects. According to the C standard, can one rely on the assumption the snippets are equivalent? In particular, having the first snippet, is it promised that if c is true, the function won't be executed, and no side effects of it will take place?

2
cshort-circuiting
Check out stackoverflow.com/questions/3958864/… – tonysdg
That's pretty much what short circuiting is all about. And it is guaranteed by the standards ONLY when the operation is between bool. NOT when the operator is overloaded. – spalac24
@user3267581 What do you mean by overloaded operator? Are you having C++ in mind? – Eugene Sh.
Yes, sorry, didn't notice the question was C specific. – spalac24
I believe in C++ it should be possible for an overloaded || be implemented as short circuited as well.. But it's a different question – Eugene Sh.

2 Answers

2
votes

After some search I will answer my question myself referencing the standard: The C99 standard, section 6.5.14 Logical OR operator is stating:

Unlike the bitwise | operator, the || operator guarantees left-to-right evaluation; there is a sequence point after the evaluation of the first operand. If the first operand compares unequal to 0, the second operand is not evaluated.

And a similar section about &&. So the answer is yes, the code can be safely considered equivalent.

1
votes

Yes, you are correct in your thinking. c || do_something() will short-circuit if c is true, and so will never call do_something().

However, if c is false, then do_something() will be called and its result will be the new value of p.