C++ 'AND' evaluation - standard guaranteed?

Question

Possible Duplicate:
Safety concerns about short circuit evaluation

What does the standard say about evaluating && expressions - does it guarantee that evaluation of parameters will stop at the first false?

E.g.:

Foo* p;
//....
if ( p && p->f() )
{
    //do something
}

is the f() guaranteed not to be called if p == NULL?

Also, is the order of evaluation guaranteed to be the order of appearence in the clause?

Might the optimizer change something like:

int x;
Foo* p;
//...
if ( p->doSomethingReallyExpensive() && x == 3 )
{
    //....
}

to a form where it evaluates x==3 first? Or will it always execute the really expensive function first?

I know that on most compilers (probably all) evaluation stops after the first false is encountered, but what does the standard say about it?

score 3 · Accepted Answer · edited May 23 '17 at 12:27

What does the standard say about evaluating && expressions - does it guarantee that evaluation of parameters will stop at the first false?

Yes. That is called short-circuiting.

Also, is the order of evaluation guaranteed to be the order of appearence in the clause?

Yes. From left to right. The operand before which the expression short-circuited doesn't get evaluated.

int a = 0;
int b = 10;
if ( a != 0 && (b=100)) {}

cout << b << endl; //prints 10, not 100

In fact, the above two points are the keypoint in my solution here:

Find maximum of three number in C without using conditional statement and ternary operator

score 3 · Answer 2 · answered Oct 01 '11 at 17:35

In the ANSI C standard 3.3.13:

Unlike the bitwise binary & 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 equal
to 0, the second operand is not evaluated.

There is an equivalent statement in the C++ standard

score 1 · Answer 3 · answered Oct 01 '11 at 17:33

1

&& (and ||) establish sequence points. So the expression on the left-hand side will get evaluated before the right-hand side. Also, yes, if the left-hand side is false/true (for &&/||), the right-hand side is not evaluated.

answered Oct 01 '11 at 17:33

C. K. Young

219,335
46
382
435

score 0 · Answer 4 · answered Oct 01 '11 at 17:42

What does the standard say about evaluating && expressions - does it guarantee that evaluation of parameters will stop at the first false?

Also, is the order of evaluation guaranteed to be the order of appearence in the clause?

5.14/1. Unlike &, && guarantees left-to-right evaluation: the second operand is not evaluated if the ﬁrst operand is false.

This only works for the standard && operator, user defined overloads of operator && don't have this guarantee, they behave like regular function call semantics.

Might the optimizer change something like: if ( p->doSomethingReallyExpensive() && x == 3 ) to a form where it evaluates x==3 first?

An optimizer may decide to evaluate x == 3 first since it is an expression with no side-effects associated if x is not modified by p->doSomethingReallyExpensive(), or even evaluate it after p->doSomethingReallyExpensive() already returned false. However, the visible behavior is guaranteed to be the previously specified: Left to right evaluation and short-circuit. That means that while x == 3 may be evaluated first and return false the implementation still has to evaluate p->doSomethingReallyExpensive().

C++ 'AND' evaluation - standard guaranteed?

4 Answers4