PQOTD

[BobJanova]

It will run either ProcessFlyouts, or ProcessCarrierAnimations, or neither, and the variable will tell you if it successfully ran either. I would imagine you either want && or |.

[OriginalGriff]

I assume you meant it to do this:

bool more = ProcessFlyouts();
more = ProcessCarrierAnimations() || more;

:laugh: Would do the same in C and C++ as well.

Those who fail to learn history are doomed to repeat it. --- George Santayana (December 16, 1863 – September 26, 1952) Those who fail to clear history are doomed to explain it. --- OriginalGriff (February 24, 1959 – ∞)

[Kornfeld Eliyahu Peter]

I assume that you want to run the second method if first returns true, if that the case you have to switch the order

bool more = ProcessFlyouts();
more = ProcessCarrierAnimations() || more;

I'm not questioning your powers of observation; I'm merely remarking upon the paradox of asking a masked man who he is. (V)

[Marc Clifton]

Exactly. :)

[Marc Clifton]

deflinek wrote:

However changing || to | will make it called regardless of more's value.

Yup! Marc

[Marc Clifton]

Yup. | Marc

[Marc Clifton]

OriginalGriff wrote:

more = ProcessCarrierAnimations() || more;

Quite so. The bitwise | operator solves the problem too. Marc

[Marc Clifton]

Kornfeld Eliyahu Peter wrote:

you have to switch the order

Yes indeed. Or use the bitwise operator. Marc

[Rage]

MISRA[^] Rule 33 & 34:

The right hand operand of a || shall not contain side effects
The operands of a logical || shall be primary expressions

The example you gave is the very reason for the above two rules.

~RaGE();

I think words like 'destiny' are a way of trying to find order where none exists. - Christian Graus Entropy isn't what it used to.

[OriginalGriff]

Yes - but I don't like bitwise operators with bools...

Those who fail to learn history are doomed to repeat it. --- George Santayana (December 16, 1863 – September 26, 1952) Those who fail to clear history are doomed to explain it. --- OriginalGriff (February 24, 1959 – ∞)

[Rage]

Yep, but this is a workaround, not a best practice though.

~RaGE();

I think words like 'destiny' are a way of trying to find order where none exists. - Christian Graus Entropy isn't what it used to.

[Kornfeld Eliyahu Peter]

I do not like that change from C/C++...I mean the | and || changes... X|

I'm not questioning your powers of observation; I'm merely remarking upon the paradox of asking a masked man who he is. (V)

[BobJanova]

There's actually some nice concise uses of

var result = DoSomething() && DoSomethingThatDependsOnIt();
var result = DoSomething() || HandleFailureCase();

It's even better in languages where you can use non-boolean types with the boolean operators.

[Keith Barrow]

Though Nish's and OG's examples will work, I think it is symantically better to do this:

bool more = ProcessFlyouts();
bool foo = ProcessCarrierAnimations();
more = more || foo;

Less terse code (normally bad) - but explicit in calling ProcessCarrierAnimations.

PB 369,783 wrote:

I just find him very unlikeable, and I think the way he looks like a prettier version of his Mum is very disturbing.[^]

[PIEBALDconsult]

more |= ProcessCarrierAnimations(); ?

You'll never get very far if all you do is follow instructions.

[Bassam Abdul Baki]

Marc Clifton wrote:

You should be able to glean what the programmer intends to happen from the code. :)

Riiiiiiiiiiight!!! :)

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[Rage]

Let's say I have never seen these two lines of code, OK ?

~RaGE();

I think words like 'destiny' are a way of trying to find order where none exists. - Christian Graus Entropy isn't what it used to.

[Simon_Whale]

Have you been to QA lately? :laugh:

Every day, thousands of innocent plants are killed by vegetarians. Help end the violence EAT BACON

[Andy Brummer]

var more = more || initalizeMore(); is a common practice in javascript.

Curvature of the Mind now with 3D

[Forogar]

This isn't a change from C++. It works the same way for both operators. Doesn't it? Now I'm going to have to read up on it, dammit!

- I would love to change the world, but they won’t give me the source code.