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Hi

I would like to know if the standard specifies the evaluation order
of several conditions, like in the following example:

if (is.eof() || (c == '\n'))

Does is.eof() is evaluated first or is the other way around ?

thanks
Bogdan

Op 24-Nov-11 22:22, Bogdan schreef:
> Hi
>
> I would like to know if the standard specifies the evaluation order
> of several conditions, like in the following example:
>
> if (is.eof() || (c == '\n'))
>
> Does is.eof() is evaluated first or is the other way around ?

(is.eof()) is evaluated first, and (c == '\n') will not be evaluated at
all if (is.eof())==true.

On Nov 24, 9:25*pm, Dombo <do...@disposable.invalid> wrote:
> Op 24-Nov-11 22:22, Bogdan schreef:
>
> > Hi
>
> > * *I would like to know if the standard specifies the evaluation order
> > of several conditions, like in the following example:
>
> > if (is.eof() || (c == '\n'))
>
> > Does is.eof() is evaluated first or is the other way around ?
>
> (is.eof()) is evaluated first, and (c == '\n') will not be evaluated at
> all if (is.eof())==true.

Just to clarify, the operators "||", "&&" and "," are special in this
respect - they guarantee that the first operand is evaluated first,
and that the second is (for || and &&) only evaluated if required. For
other operators such as "+", the operands could be evaluated in either
order.

On 11/24/2011 2:11 PM, Paul N wrote:
> On Nov 24, 9:25 pm, Dombo<do...@disposable.invalid> wrote:
>> Op 24-Nov-11 22:22, Bogdan schreef:
>>
>>> Hi
>>
>>> I would like to know if the standard specifies the evaluation order
>>> of several conditions, like in the following example:
>>
>>> if (is.eof() || (c == '\n'))
>>
>>> Does is.eof() is evaluated first or is the other way around ?
>>
>> (is.eof()) is evaluated first, and (c == '\n') will not be evaluated at
>> all if (is.eof())==true.
>
> Just to clarify, the operators "||", "&&" and "," are special in this
> respect - they guarantee that the first operand is evaluated first,
> and that the second is (for || and&&) only evaluated if required. For
> other operators such as "+", the operands could be evaluated in either
> order.

Unless the operator has been overridden for a UDT, in which case it's
equivalent to calling:

operator&&(left, right)

In which case evaluation order is undefined. In the OP's case, it's
bool, so it is in fact, left to right.

Hi,
that's a special form of lazy evaluation, called short-circuit evaluation. Have a look here http://en.wikipedia.org/wiki/Short-circuit_evaluation.

Greetings from Rottenburg,
Rainer