Re: 2nd code for - boolean error ? - C# Discussion Boards

2-Jul-05 0:56

Hi all,

here is the real code I am using for with improved code and conditions but I am running also into the bracket.

but the last contition is false BAKER is not Meyers and so the I should never run into the bracket (MessageBox ..)

string a = "MR";
string b = "MRS";
string c = "BAKER";
string d = "Meyers"; 

if
  (
	((string.Equals(a.TrimEnd().TrimStart().ToString().ToUpper(),"MR")) && 
	(string.Equals(b.TrimEnd().TrimStart().ToString().ToUpper(),"MRS"))) ||
	((string.Equals(a.TrimEnd().TrimStart().ToString().ToUpper(),"MRS")) && 
	(string.Equals(b.TrimEnd().TrimStart().ToString().ToUpper(),"MR"))) &&
	(string.Equals(c.TrimEnd().TrimStart().ToString().ToUpper(),
	d.TrimEnd().TrimStart().ToString().ToUpper()))  
  )
{
	MessageBox.Show("ERROR");
}

where is the fault ?!?!?

please help

Thanks Frank

leppie2-Jul-05 6:35

leppie

2-Jul-05 6:35

((string.Equals(a.TrimEnd().TrimStart().ToString().ToUpper(),"MR")) == a.Trim().ToUpper() == "MR"

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Robert Rohde2-Jul-05 6:50

Robert Rohde

2-Jul-05 6:50

Let me break your statement into three parts:

bool a = (string.Equals(a.TrimEnd().TrimStart().ToString().ToUpper(),"MR")) 
&& (string.Equals(b.TrimEnd().TrimStart().ToString().ToUpper(),"MRS"));
bool b = (string.Equals(a.TrimEnd().TrimStart().ToString().ToUpper(),"MRS")) 
&& (string.Equals(b.TrimEnd().TrimStart().ToString().ToUpper(),"MR"));
bool c = string.Equals(c.TrimEnd().TrimStart().ToString().ToUpper(),
d.TrimEnd().TrimStart().ToString().ToUpper());

Combined it would look like this:

if (a || b && c) {
MessageBox.Show("ERROR");
}

What the compiler makes out of this statement is this:

if (a || (b && c)) {
MessageBox.Show("ERROR");
}

Thats because the logical AND has a higher priority than the logical OR. Its like this simple numeric example:
3 + 2 * 4 = 3 + (2 * 4)

In conclusion the statement will be true if a is true no matter what the other subparts are.

fracalifa2-Jul-05 7:01

2-Jul-05 7:01

Thank you for your answer,

but I think I did:

if ( ( a || b) && c ) {
MessagBox.Show("ERROR")
}

Please look at the (many) brackets - therefor I placed the brackets around the or-expression to force the compiler to check the or-term fist and this result with the &&-expression !

And that's what's going wrong

Frank

Guffa2-Jul-05 7:22

Guffa

2-Jul-05 7:22

Yes, you thought that you did, but you failed. Wink | ;)

Instead of ((a || b) && c) you created ((a) || (b) && c).

string a = "MR";<br />
string b = "MRS";<br />
string c = "BAKER";<br />
string d = "Meyers"; <br />
<br />
a = a.Trim().ToUpper();<br />
b = b.Trim().ToUpper();<br />
c = c.Trim().ToUpper();<br />
d = d.Trim().ToUpper();<br />
<br />
if (((a == "MR" && b == "MRS") || (a == "MRS" && b == "MR")) && c == d) {<br />
	MessageBox.Show("ERROR");<br />
}

---
b { font-weight: normal; }

DavidNohejl2-Jul-05 7:31

DavidNohejl

2-Jul-05 7:31

no you did not. It will result in

 if
  (
	(1) ||	( 1 ) && (0)  
  )
{
	MessageBox.Show("ERROR");
}

Must be this:

if
				(
				(((string.Equals(a.TrimEnd().TrimStart().ToString().ToUpper(),"MR")) && 
				(string.Equals(b.TrimEnd().TrimStart().ToString().ToUpper(),"MRS"))) ||
				((string.Equals(a.TrimEnd().TrimStart().ToString().ToUpper(),"MRS")) && 
				(string.Equals(b.TrimEnd().TrimStart().ToString().ToUpper(),"MR")))) &&
				(string.Equals(c.TrimEnd().TrimStart().ToString().ToUpper(),
				d.TrimEnd().TrimStart().ToString().ToUpper()))  
				)
			{
				MessageBox.Show("ERROR");
			}

will result in

if
  (
	((1) ||	( 1 )) && (0)  
  )
{
	MessageBox.Show("ERROR");
}

and will not display message box.

But still, it's a mess.

David

DavidNohejl2-Jul-05 6:57

DavidNohejl

2-Jul-05 6:57

May I suggest you little simplification?

a.TrimEnd().TrimStart().ToString().ToUpper() -> a.Trim().ToUpper()

David

Re: OT

fracalifa2-Jul-05 7:04

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2-Jul-05 7:04

Thank's but that doesn't help in my problem !

regards

Frank

Re: OT

DavidNohejl2-Jul-05 7:17

DavidNohejl

2-Jul-05 7:17

fracalifa wrote:
that doesn't help in my problem !

Pardon, but it does. Low readability (daily-wtf calibre!) of your code is what caused problem here.

ok, it is prevence but not solution, that's why I say it is OT (off topic).

David

Re: OT

Colin Angus Mackay3-Jul-05 1:36

Colin Angus Mackay

3-Jul-05 1:36

Improving the readability of code makes a big difference in being able to find and correct errors quickly. It also means that if another person has to maintain the code they can do so with less effort.

error in bool condition ?

fracalifa1-Jul-05 23:45

1-Jul-05 23:45

Hi all,

coming from "C" I'm a little bit confused on the result of the following boolean term.
I think this can never become TRUE but it does - why ?

if( ((("A"=="A") && ("B"=="B")) || (("A"=="B") && ("B"=="A"))) && (true==false) )
{
MessageBox.Show("Error"); // this code is not reachable ?????
}

A and B is really a comparison of strings

reduced to my understanding is that the results shoud be

if( (((TRUE)&&(TRUE)) ||((FALSE && FALSE))) && (FALSE) ) // should ever be FALSE
........\---------/........\----------/.........|
........... TRUE......||.......FALSE.......&& FALSE
.............\------------------/...............|
.....................TRUE..................&& FALSE = FALSE //!!

the second problem is that the code in the brackets is not reachable ???? Confused | :confused:

the count of backets is correct !
that's leave my understanding of C#

where is my error in thinking ?

Thanks for your help

Frank

Dario Solera1-Jul-05 23:52

Dario Solera

1-Jul-05 23:52

In c# the == operator has been redefined for strings, so when you compare two strings written like "abc" the compiler is able to compare them. Using variables, you have to use the string.Equals(...) or the string.CompareTo(...) methods to have reliable results. In VB.NET you can use the = operator also for string variables.

Robert Rohde2-Jul-05 0:04

Robert Rohde

2-Jul-05 0:04

Have you actually tried this code? The fact that the compiler tells you thats not reachable definitely means that it will never get executed. Ive tried the code and the messagebox is not showing.

fracalifa2-Jul-05 0:29

2-Jul-05 0:29

That's one of the two problems,
I reduced two brackets - that it runs

but I am trying to implement the first suggestion and than .... ?

User 66582-Jul-05 1:03

User 6658

2-Jul-05 1:03

The first term (("A"=="A") && ("B"=="B")) is true, thus the whole OR-expression || will become true.
Because the first term is always true, there is no need for the compiler to evaluate the second term.

regards

modified 12-Sep-18 21:01pm.

fracalifa2-Jul-05 1:10

2-Jul-05 1:10

... yes but ..

I put the OR-expression into brackets followed by an AND-expression so the compiler has to check for
( 1 || 0 ) && ( 0 ) so there is a need to do this - because the last term is not true !

regards

User 66582-Jul-05 6:21

User 6658

2-Jul-05 6:21

Oh sorry, I misread the brackets.

Well, because the last term is always false, the compiler detects that the whole clause never can become true. The or-clause becomes always true, bug then the and-clause will become false anyway.

modified 12-Sep-18 21:01pm.

fracalifa2-Jul-05 6:45

2-Jul-05 6:45

Hi Greeeg,

please look to following code I have implemented:
Please try this code:

string a = "MR";
string b = "MRS";
string c = "BAKER";
string d = "Meyers";

if
(
((string.Equals(a.TrimEnd().TrimStart().ToString().ToUpper(),"MR")) &&
(string.Equals(b.TrimEnd().TrimStart().ToString().ToUpper(),"MRS"))) ||
((string.Equals(a.TrimEnd().TrimStart().ToString().ToUpper(),"MRS")) &&
(string.Equals(b.TrimEnd().TrimStart().ToString().ToUpper(),"MR"))) &&
(string.Equals(c.TrimEnd().TrimStart().ToString().ToUpper(),
d.TrimEnd().TrimStart().ToString().ToUpper()))
)
{
MessageBox.Show("ERROR");
}

The result of the first 4 lines of the if-expression becomes TRUE -- o.k.
but the comparison of c.Trim and d.Trim is FALSE so the whole expression (due to the &&) has to become false but the code is running into the MessageBox bracket

why

thank you for your help
Frank

User 66582-Jul-05 7:52

User 6658

2-Jul-05 7:52

this is strange indeed, looks like the && operator is evaluated BEFORE the || operator - I thought they are handled equally.
If you want the || clause to be evaluated before, better put brackets around the first two expressions.
This is a test I made:

			string a = "MR";
			string b = "MRS";
			string c = "BAKER";
			string d = "Meyers";

			bool b1,b2,b3,b4,b5;

			b1 = string.Equals(a.TrimEnd().TrimStart().ToString().ToUpper(),"MR");
			b2 = string.Equals(b.TrimEnd().TrimStart().ToString().ToUpper(),"MRS");
			b3 = string.Equals(a.TrimEnd().TrimStart().ToString().ToUpper(),"MRS");
			b4 = string.Equals(b.TrimEnd().TrimStart().ToString().ToUpper(),"MR");
			b5 = string.Equals(c.TrimEnd().TrimStart().ToString().ToUpper(), d.TrimEnd().TrimStart().ToString().ToUpper());

			bool b6, b7;

			b6 = (((b1 && b2) || (b3 && b4)) && b5); // FALSE
			b7 = ((b1 && b2) || ((b3 && b4) && b5)); // TRUE

//			if((b1 && b2) || (b3 && b4) && b5)
			if(b6)
			{
				MessageBox.Show("ERROR");
			}

regards

/edit:
This supports my theory:
Operator precedence and associativity

modified 12-Sep-18 21:01pm.