Hi, i'm using a richtextbox named r1, and i want to write in Bold and
italic, i tryed this for bold and it work
Dim f As New Font("Verdana", 30, FontStyle.Bold)
r1.Font = f
r1.Text = "Hello"
but if i try this for bold and italic it doesn't work
Dim f As New Font("Verdana", 30, FontStyle.Bold And FontStyle.Itali c)
r1.Font = f
r1.Text = "Hello"
so how can i write in bold and italic?
Thx :) 4 13376
paraidy napisal(a):
Hi, i'm using a richtextbox named r1, and i want to write in Bold and
italic, i tryed this for bold and it work
Dim f As New Font("Verdana", 30, FontStyle.Bold)
r1.Font = f
r1.Text = "Hello"
but if i try this for bold and italic it doesn't work
Dim f As New Font("Verdana", 30, FontStyle.Bold And FontStyle.Itali c)
r1.Font = f
r1.Text = "Hello"
so how can i write in bold and italic?
Thx :)
Hi,
Try this code
Dim NewFont As New Font(RichTextBo x1.Font, FontStyle.Bold +
FontStyle.Itali c)
RichTextBox1.Fo nt = NewFont
Hope this help.
Regards,
sweet_dreams
sweet_dreams ha scritto:
Hi,
Try this code
Dim NewFont As New Font(RichTextBo x1.Font, FontStyle.Bold +
FontStyle.Itali c)
RichTextBox1.Fo nt = NewFont
Yes it worked, thx a lot :)
"paraidy" <sa*******@tisc ali.itwrote in message
news:11******** *************@b 28g2000cwb.goog legroups.com...
>
sweet_dreams ha scritto:
>Hi,
Try this code
Dim NewFont As New Font(RichTextBo x1.Font, FontStyle.Bold + FontStyle.Ital ic) RichTextBox1.F ont = NewFont
Yes it worked, thx a lot :)
The + works, but what you are doing is a bitwise operation against an
enumeration value. The FontStyle has the FlagsAttribute associated with it
so we know we can treat them as binary flags. (look up the FlagsAttribute
for more information on the following information)
What it should be is:
Dim NewFont As New Font(RichTextBo x1.Font, FontStyle.Bold Or
FontStyle.Itali c)
RichTextBox1.Fo nt = NewFont
The reason your FontStyle.Bold And FontStyle.Itali c didn't work is:
FontStyle.Bold = 1 = 0001 ' in binary
FontStyle.Itali c = 2 = 0010 ' in binary
So...the result of your AND statement = 0001 And 0010 = 0000 (all off).
What you want is an OR = 0001 Or 0010 = 0011 (last two bits which represents
the two flags, Bold and Italic, turned on). It just so happens that using
regular addition on those values will give you the same result :)
HTH you understand it better ... for future reference,
Mythran
Mythran wrote:
"paraidy" <sa*******@tisc ali.itwrote in message
news:11******** *************@b 28g2000cwb.goog legroups.com...
sweet_dreams ha scritto:
Hi,
Try this code
Dim NewFont As New Font(RichTextBo x1.Font, FontStyle.Bold +
FontStyle.Itali c)
RichTextBox1.Fo nt = NewFont
Yes it worked, thx a lot :)
The + works, but what you are doing is a bitwise operation against an
enumeration value. The FontStyle has the FlagsAttribute associated with it
so we know we can treat them as binary flags. (look up the FlagsAttribute
for more information on the following information)
What it should be is:
Dim NewFont As New Font(RichTextBo x1.Font, FontStyle.Bold Or
FontStyle.Itali c)
RichTextBox1.Fo nt = NewFont
The reason your FontStyle.Bold And FontStyle.Itali c didn't work is:
FontStyle.Bold = 1 = 0001 ' in binary
FontStyle.Itali c = 2 = 0010 ' in binary
So...the result of your AND statement = 0001 And 0010 = 0000 (all off).
What you want is an OR = 0001 Or 0010 = 0011 (last two bits which represents
the two flags, Bold and Italic, turned on). It just so happens that using
regular addition on those values will give you the same result :)
HTH you understand it better ... for future reference,
Mythran
Understood now, thx a lot :) This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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