AndAlso

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M

Matt

Hi, I am new to vb.net. In using the If condition, can we
use either of And or AndAlso. Whats the difference ? In VB
I have always used And.
For eg.
if num > 10 AndAlso num < 20

Thanks.
Matt
 
Matt said:
Hi, I am new to vb.net. In using the If condition, can we
use either of And or AndAlso. Whats the difference ?
Click to expand...

The "AndAlso" operator uses "short-circuit" evaluation - if the
First condition is False, the second isn't even evaluated. Useful
for things like this:

dataSet = FunctionThatReturnsADataSet_OrNothing()

' Did I get a DataSet with just one row in it?
If Not ( dataSet Is Nothing ) _
AndAlso dataSet.Tables( 0 ).Rows = 1 _
Then

"And" always works out /both/ expressions.

HTH,
Phill W.
 
*And* evaluates both Expression1 and Expression2. Whereas, *AndAlso*,
provides a means to perform short-circuiting the evaluation process is some
cases.

False AndAlso True will only evaluate the 'False' and knows it does not
need to evaluate the 'True' because it already has its answer. This can help
performance if used in a loop.


Regards - OHM
 
In addition to the comments mentioned, there is also an "orelse" operator
which does lazy evaluation using the or operator


e.g.

If True orelse MyReallySlowFunction() Then


MyReallySlowFunction() will never get called because "orelse" has evaluated
True and so does not need to evaluate MyReallySlowFunction()


HTH,

Trev.
 
In the case of "if x and y": this is true if both x and y are true.

In the case of "if x andalso y": this is true only if x is true
regardless of the value of y, and if x is true and y is true.

Basically ANDALSO is the same as regular AND except that the first value
you are testing MUST be true to evaluate as true.

Is that confusing enough?
 
In the case of "if x and y": this is true if both x and y are true.

In the case of "if x andalso y": this is true only if x is true
regardless of the value of y, and if x is true and y is true.
Click to expand...

Umm, no. Read the rest of the thread for details, but if x is true
and y is false, ' x andalso y' will evaluate to false.
Basically ANDALSO is the same as regular AND except that the first value
you are testing MUST be true to evaluate as true.
Click to expand...

Again, no. (and this definition is different than your first
definition, but both are wrong).
Is that confusing enough?
Click to expand...

Yes, which is why it's not done that way.

More specifically, 'x and y' and 'x andalso y' will always evaluate to
the same value for boolean values, either true or false.

Postscript: Actually, I suppose that's only true if evaluation order is
guaranteed with the 'And' operator, and I can't find anything that says
it is. So a minor caveat.
 
Just a useless trivia note:

The real terms for these operations are short-circuiting logical disjunction
(orelse), and short-circuiting logical conjunction (andalso), which most
people refer to as "short-circuited", or "lazy." In .NET it's used for
logical operators, but the concept is broad and could be applied to any
chain of logic.

It basically means, stop processing as soon as possible. Like playing
basketball best 3 out of 5, you don't need to actually play all five games
unless both teams win twice.

~
Jeremy
 
David said:
Umm, no. Read the rest of the thread for details, but if x is true
and y is false, ' x andalso y' will evaluate to false.
Click to expand...

What you just said is what I said, or tried to say. Perhaps you didn't
understand what I wrote, or I didn't word it correctly. So we are in
agreement here.
Again, no. (and this definition is different than your first
definition, but both are wrong).
Click to expand...

According to the table given in the help file, what I said is correct.
The first value you are testing must be true in order for it to evaluate
to true regarless of the second value. And only when both values are
true does it evaluate to true. See the table below.

If expression1 is / And expression2 is / Value of result is
True True True
True False False
False (not evaluated) False
 
What you just said is what I said, or tried to say. Perhaps you
didn't understand what I wrote, or I didn't word it correctly. So we
are in agreement here.
Click to expand...

Perhaps so, but why don't we rephrase your original statement?

In the case of "if x and y": this is true if both x and y are true.

In the case of "if x andalso y": this is true if both x and y are true.


Period. End of discussion as far as the value of the operation goes.

But then what does this phrase fragment mean?
According to the table given in the help file, what I said is correct.
Click to expand...

No, the table's not the point here. Both 'And' and 'AndAlso' share
the same table. The problem above is the word "except", since in
both cases the first value MUST be true for the phrase to evaluate
as true. In other words, "the first value you are testing MUST
be true to evaluate as true" is true of both 'And' and 'AndAlso',
so the word "except" makes no sense here.

Now if you really understand this and there's just some grammatical
confusion, then no big deal. But you keep trying to point out
the And/AndAlso difference by saying something about the value
of 'X And Y' vs. 'X AndAlso Y', and that's just not the difference
between the two operators.
The first value you are testing must be true in order for it to
evaluate to true regarless of the second value.
Click to expand...

Likewise the second value must be true in order for it to evaluate to
true regardless of the first value.

Those two combined is simply is a very longwinded way of simply
saying...
only when both values are true does it evaluate to true.
Click to expand...

And the lines above are true of the 'And' operator as well, so that
hardly points out the difference between them.
 
It seem to me that the best explanation for AndAlso is this:


If x andalso y then...
whatever
endif


is equivalent to


If x then
if y then
whatever
endif
endif

On the other hand, And can give different results than does AndAlso or the
multiple if scenario. When using the same operator as a bitwise logical
"And" you have to be cautious.

I ran the following code

\\\\\\\\\\\\\\\\\\\\
Dim flag As Integer

flag = &H10 + &H8

If (flag And &H10) And (flag And &H8) Then

MsgBox("true branch of if using 'And'")

Else

MsgBox("false branch of if using 'And'")

End If

If (flag And &H10) AndAlso (flag And &H8) Then

MsgBox("true branch of if using 'AndAlso'")

Else

MsgBox("false branch of if using 'AndAlso'")

End If

If (flag And &H10) Then

If (flag And &H8) Then

MsgBox("true branch of if using 'Nested If'")

Else

MsgBox("false branch of if using 'Nested If'")

End If

End If

////////////

The results are False, True and True (in order).

Regards,
Ot
 
You're right. Now go have a cup of hot chocolate.
Click to expand...

Is a cuppa Joe OK?

I know it seems a bit nit-picky to belabor points like this, but the
problem is that many more people read Usenet than post to it. I know
I've been steered wrong before when I read a post and thought to myself
"Well, if that were untrue somebody would have responded, so that must
be how things are".
 
On the other hand, And can give different results than does AndAlso or the
multiple if scenario. When using the same operator as a bitwise logical
"And" you have to be cautious.

I ran the following code

\\\\\\\\\\\\\\\\\\\\
Dim flag As Integer

flag = &H10 + &H8

If (flag And &H10) And (flag And &H8) Then
Click to expand...

Luckily, this won't compile with Option Strict On
 
Hi David,

What good that you can set it off is it not, than you can show good the
essentials of things.

But did you think there was something wrong with the sample?
I thought it was a good explanation?

Cor
 
Yes, indeed. If you use Cast you can get what you want.

If CBool(flag And &H10) And CBool(flag And &H8) Then

MsgBox("true branch of if using 'And'")

Else

MsgBox("false branch of if using 'And'")

End If

This version yields "true", regardless of the setting of the Strict Option.

David said:
Luckily, this won't compile with Option Strict On
Click to expand...


Yes, indeed. If you use Cast you can get what you want. As I said "be
cautious."

If CBool(flag And &H10) And CBool(flag And &H8) Then

MsgBox("true branch of if using 'And'")

Else

MsgBox("false branch of if using 'And'")

End If

This version yields "true", regardless of the setting of the Strict Option.
 
I should really learn the difference between ctl-C and ctl-X. I intended
to cut/paste my reply to the bottom. Oh well.
 
What good that you can set it off is it not, than you can show good the
essentials of things.

But did you think there was something wrong with the sample?
I thought it was a good explanation?
Click to expand...

So did I, I was just adding some info. It's not just a good
explanation, but also a good example of one of the problems with the
overloaded 'And' operator.
 
Yes, indeed. If you use Cast you can get what you want.

If CBool(flag And &H10) And CBool(flag And &H8) Then

MsgBox("true branch of if using 'And'")
Else
MsgBox("false branch of if using 'And'")
End If

This version yields "true", regardless of the setting of the Strict Option.
Click to expand...

Yep, although I usually recommend to people that they stop using the
'And' operator for boolean expressions and use AndAlso and OrElse
exclusively. I realize there's a readability argument for 'And', but I
find AndAlso just as readable and it's less error-prone.
 
Provided that you are evaluating Boolean values, One scenario
where AND may be a better option is when you
have the evaluation of a function.

FunctionA(X) And FunctionB(X)

It may be required that FunctionB is called, in which case, if
you wanted to use AndAlso, then you would have had to have
called the function first and assigned its Value to another variable
then use that as the right hand side of AndAlso.

AndAlso, makes for better readability as it is an evaluation of
Boolean values, not a bitwise AND.

Additionally, if using AndAlso, use the condition most likely
to return False, on the left hand side, this improves performance.

Regards - OHM
 
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