C# windows base Class related

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Hi All,
I have some confusion on the following code

try
{

}
catch (Exception ex)
            {
                con.Close();
                MessageBox.Show(ex.Message);
            }

If I replace the '.' with a comma

MessageBox.Show(ex.Message);

With

MessageBox.Show(ex,Message);

It shows different thing, can any one please define this?

Thanks & regards
Indrajit dasgupta

[edit]Code blocks added, general tidy up- OriginalGriff[/edit]

Posted 5-May-11 0:14am

IndrajitDasgupat

Updated 5-May-11 0:27am

OriginalGriff

v2

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Abhinav S · Answer 1 · 2011-05-05T00:29:00

Solution 1

When you put a comma in the method definition, you get a list of parameters you can enter into that method.
This is part of intellisense in order to help the user choose the appropriate overloaded methods.

If you put a ".", you get a list of members that you can use for that object.

Posted 5-May-11 0:29am

Abhinav S

Comments

Hemant__Sharma 5-May-11 6:47am

Right. My 5

Abhinav S 5-May-11 7:03am

Thanks.

RAJI @Codeproject · Answer 2 · 2011-05-05T00:31:00

public static DialogResult Show(
	string text,
	string caption,
	MessageBoxButtons buttons,
	MessageBoxIcon icon,
	MessageBoxDefaultButton defaultButton,
	MessageBoxOptions options,
	string helpFilePath
)

these are MessageBox parameters.

when you call

MessageBox.Show(ex.Message);

Only string text is Displayed.

when you call

MessageBox.Show(ex,Message);

string text,
string caption are displayed

OriginalGriff · Answer 3 · 2011-05-05T00:31:00

Solution 5

MessageBox has a number of overloads for the Show method:
By replacing the '.' with ',' you are trying to use an overload with two parameters.
The compiler will probably complain that it cannot cast an Exception to a IWin32Window

Posted 5-May-11 0:31am

OriginalGriff

musefan · Answer 4 · 2011-05-05T00:30:00

So the comma(,) is used to specify separate parameters, so with a dot you are calling

MessageBox.Show(String)

and with comma you are calling

MessageBox.Show(String, String) //or similar

the second of course will give you an error message of Message not defined (I assume) and that Show does not have an overload that takes 'Exception' as first paramater

Ashishmau · Answer 5 · 2011-05-05T00:31:00

Solution 3

with

C++

MessageBox.Show(ex.Message);

u r using Message property of exception

with

C++

MessageBox.Show(ex,Message);

u r passing two parameters.
so both r different thing

Posted 5-May-11 0:31am

Ashishmau

Nagy Vilmos · Answer 6 · 2011-05-05T00:33:00

Solution 6

In the first instance you are calling MessageBox.Show(String), the second shouldn't build.

Posted 5-May-11 0:33am

Nagy Vilmos