short circuit evaluation

[Lassie]

bool finished = is_done();

if (finished && try_again()) { }

Is my understanding of short circuit evaluation correct that if finished is
false that try_again() will never be executed?
Or does it get evaluated sometimes?

[Jean-Marc Bourguet]

"Lassie" <zh*@woaini.com writes:

if (finished && try_again()) { }

Is my understanding of short circuit evaluation correct that if finished
is false that try_again() will never be executed?

Yes.

--
Jean-Marc

[vippstar]

On Sep 16, 6:55 pm, "Lassie" <[e-mail] mwrote:

bool finished = is_done();

if (finished && try_again()) { }

Is my understanding of short circuit evaluation correct that if finished is
false that try_again() will never be executed?
Or does it get evaluated sometimes?

No, it does not evaluate if finished is 0.

Thus, this is correct:

return 0 && 1 / 0;

The expression (0 && 1 / 0) is evaluated like this:

(a && b)

First, a is evaluated. If a is false, then the whole expression is
replaced by 0. (false)
If a is true, then b is evaluated, and the expression is replaced by
true or false, depending on what b was evaluated.

This is also correct

return 1 || 1 / 0;

[Richard]

vi******@gmail. com writes:

On Sep 16, 6:55 pm, "Lassie" <[e-mail] mwrote:

>bool finished = is_done();

if (finished && try_again()) { }

Is my understanding of short circuit evaluation correct that if finished is
false that try_again() will never be executed?
Or does it get evaluated sometimes?

No, it does not evaluate if finished is 0.

Thus, this is correct:

return 0 && 1 / 0;

The expression (0 && 1 / 0) is evaluated like this:

(a && b)

First, a is evaluated. If a is false, then the whole expression is
replaced by 0. (false)
If a is true, then b is evaluated, and the expression is replaced by
true or false, depending on what b was evaluated.

This is also correct

return 1 || 1 / 0;

Wow what a long winded way of explaining something.

Short answer : as soon as one "false" is matched from left to right the
&& statement "exits" and returns false.

[sh.vipin]

On Sep 16, 8:55*pm, "Lassie" <[e-mail] mwrote:

bool finished = is_done();

if (finished && try_again()) { }

Is my understanding of short circuit evaluation correct that if finished is
false that try_again() will never be executed?
Or does it get evaluated sometimes?

no it will not be evaluated ever,
but are you trying it in "C".
In "C" there is no "bool" built in data type .
If you type casted "bool" as some other built in data type then please
post as much relevant code as possible in queries in future

--
vIpIn

[Kenny McCormack]

In article <ga**********@r egistered.motza rella.org>,
Richard <rg****@gmail.c omwrote in regards to our friend vippy:
....

>Wow what a long winded way of explaining something.

That is the way - if you are a reg wannabee.

>Short answer : as soon as one "false" is matched from left to right the
&& statement "exits" and returns false.

Another way:
&& means "evaluate while true[*]; return last evaluated expression"
|| means "evaluate until true[*]; return last evaluated expression"
[*] Or end of list reached.

[vippstar]

On Sep 16, 7:23 pm, sh.vi...@gmail. com wrote:

On Sep 16, 8:55 pm, "Lassie" <[e-mail] mwrote:

bool finished = is_done();

if (finished && try_again()) { }

Is my understanding of short circuit evaluation correct that if finished is
false that try_again() will never be executed?
Or does it get evaluated sometimes?

no it will not be evaluated ever,
but are you trying it in "C".
In "C" there is no "bool" built in data type .
If you type casted "bool" as some other built in data type then please
post as much relevant code as possible in queries in future

There's a bool macro in <stdbool.hin C99.

[jameskuyper]

sh.vi...@gmail. com wrote:

On Sep 16, 8:55�pm, "Lassie" <[e-mail] mwrote:

bool finished = is_done();

if (finished && try_again()) { }

Is my understanding of short circuit evaluation correct that if finished is
false that try_again() will never be executed?
Or does it get evaluated sometimes?

no it will not be evaluated ever,
but are you trying it in "C".
In "C" there is no "bool" built in data type .

No, but it is a standard typedef. This is just a code fragment, which
doesn't even provide declarations for is_done() and try_again(). I
would assume that the missing code also contains a line which says

#include <stdbool.h>

If you type casted "bool" as some other built in data type then please
post as much relevant code as possible in queries in future

Agreed: explaining a problem by providing a complete program that
demonstrates it always makes for more productive responses.

[vippstar]

On Sep 16, 7:38 pm, jameskuy...@ver izon.net wrote:

sh.vi...@gmail. com wrote:

On Sep 16, 8:55 pm, "Lassie" <[e-mail] mwrote:

bool finished = is_done();

if (finished && try_again()) { }

Is my understanding of short circuit evaluation correct that if finished is
false that try_again() will never be executed?
Or does it get evaluated sometimes?

no it will not be evaluated ever,
but are you trying it in "C".
In "C" there is no "bool" built in data type .

No, but it is a standard typedef. This is just a code fragment, which
doesn't even provide declarations for is_done() and try_again(). I
would assume that the missing code also contains a line which says

#include <stdbool.h>

It is actually a macro, it's explicity mentioned in the standard to be
so.
See 7.16 p 2

[Keith Thompson]

ja*********@ver izon.net writes:

sh.vi...@gmail. com wrote:

>On Sep 16, 8:55�pm, "Lassie" <[e-mail] mwrote:

bool finished = is_done();

if (finished && try_again()) { }

Is my understanding of short circuit evaluation correct that if
finished is false that try_again() will never be executed?
Or does it get evaluated sometimes?

no it will not be evaluated ever,
but are you trying it in "C".
In "C" there is no "bool" built in data type .

No, but it is a standard typedef. This is just a code fragment, which
doesn't even provide declarations for is_done() and try_again(). I
would assume that the missing code also contains a line which says

#include <stdbool.h>

Either that, or it was compiled with a C++ compiler.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
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