programming concepts with IF ELSE

Question

1.00/5 (1 vote)

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I have a problem like below

CASE 1:

C#

bool flag1= false;
bool flag2= true;

If(flag1 && flag2)
{
}

else
{

}

CASE 2:

C#

bool flag1= false;
bool flag2= true;

If(flag1 || flag2)
{
}

else
{

}

in both case how control will follow from if else block.

Posted 22-Oct-13 7:11am

Member 10350030

Updated 22-Oct-13 23:17pm

v4

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Comments

phil.o 22-Oct-13 14:43pm

Is it me, or both cases are the same? One should read (flag2 && flag1) for case 2 obviously.

More exactly, in case 1, since flag1 is false, flag2 won't ever be evaluated.
On the contrary, in case 2, flag2 will be evaluated to true, then flag1 will be evaluated to know if both are true.

BillWoodruff 23-Oct-13 0:06am

I wonder if this question may have been edited and the Conditional And Operator && substituted for the Binary And Operator & in one of the two cases ? The solutions below suggest to me that's the case.

@Member 10350030 What exactly is the "problem" here ? Did the answers you received satisfy you ?

Member 10350030 23-Oct-13 5:16am

In Case2 there is some modification

bool flag1= false;
bool flag2= true;

If(flag1 || flag2) {//or}
{
}

else
{

}

there is logical OR.

5 solutions

Solution 5

&& Operator default behavior is to make condition necessary to met and not only one both can met will execute. property of true or false or false and true so that if not met then else condition execute.

Posted 22-Oct-13 10:52am

Muhammad Aqib Shehzad

Solution 1

&& means and - sot this isn't a problem, you must map your request the make a condition (in the current case only when bot flags are true you will get the if, any other combination runs the else branch)...

Posted 22-Oct-13 7:20am

Kornfeld Eliyahu Peter

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lukeer · Accepted Answer · 2013-10-22T23:56:00

Solution 6

Case1 executes the else block, because one of the flags is not set to true. AND requires all flags to equal true.

Case2 executes the if block, because one of the flags is set to true. OR is satisfied with one of the flags equalling true.

Posted 22-Oct-13 23:56pm

lukeer

CPallini · Accepted Answer · 2013-10-22T10:02:00

You code won't even compile (you used falg1 instead of flag1 in if expression :-) ).
In both cases, executing the else block is the correct behavior, since one of the and operands is false. However there is a slight difference ( as correctly noted by phil.o):

In CASE 1, flag2 is not even evaluated.
In CASE 2, both flag1 and flag2 are evaluated

This is a nice property, called short-circuit, of the && operator[^].

Sergey Alexandrovich Kryukov · Accepted Answer · 2013-10-22T10:34:00

Solution 4

An addition to Solution 3: why "flag2 is not even evaluated" can be important?

C#

static bool Check1() { System.Console.WriteLine("checking condition 1"); return false; }
static bool Check2() { System.Console.WriteLine("checking condition 2"); return true; }

if (Check1() && Check2()) { } // will print only "checking condition 1"

// but 

if (Check1() & Check2()) { } // will print only "checking condition 1" and then "checking condition 2"

So, using '&&', instead of '&', "optimizes out" and redundant checks where it is possible. The difference is manifested with there is a side effect in a Boolean operand.

In practice, I would advise:

avoid using complex Boolean operation with any Boolean functions which may cause side effects;
In case the above is taken into account, prefer '&&' and '||' over '&' or '|'.

—SA

Posted 22-Oct-13 10:34am

Sergey Alexandrovich Kryukov

Comments

Kornfeld Eliyahu Peter 23-Oct-13 1:31am

When you think C# as the first-choice you make the same mistake C# designers made when removing the bit-wise handling didn't remove & and | but changed it's meaning...
In fact you complicated the case for a man just started his way (using boolean variables there are no meaning of the optimization you bring here up).

Sergey Alexandrovich Kryukov 23-Oct-13 1:48am

This is a good point, about not the first-choice technology, but I don't do this mistake. This is first thing which concerned me. For example, meaning of '&&' and '&' is different. So what? Every language's syntax and semantics should be reviewed and not mixed up with those in some other language.

Now, complicate what? Would you suggest to hide some information?
Remember, this is not the solution, this is an illustration and advice. At this moment, I see no sense in your criticism based on "complication", otherwise I would greatly appreciate it...

Thank you.
—SA

CPallini 23-Oct-13 3:16am

Good. My laziness is fixed now. :-)
5.

Sergey Alexandrovich Kryukov 23-Oct-13 9:22am

Thank you, Carlo. :-)
—SA

BillWoodruff 23-Oct-13 3:41am

I wrote to Anders H. and told him that having two distinct use-cases for & and | was just a terrible choice, and I suggested they implement BitAnd, and BitOr, or Bit& and Bit| instead, but I haven't heard back from him. No idea why :)

Sergey Alexandrovich Kryukov 23-Oct-13 9:25am

Do you mean & for Boolean and bitwise &?
I prefer how it is. This is not a problem at all. How about many different meaning of +=: concatenation, all integer types, all floating-point, and adding a delegate to delegate instance's invocation list.
—SA