CODEWITHSUNDEEP
c#winformsbit-manipulationbitwise-operators
Kjartan
Kjartan

Reputation: 19151

Understanding bitwise comparisons between non-boolean parameters

I'm working with some old code, trying to improve it, and I came across the following, which I am having trouble understanding:

controlToUpdate.Font = 
    new System.Drawing.Font(someFont,
                            someFontSize,
                            controlToUpdate.Font.Style ^ 
                                (controlToUpdate.Font.Style & FontStyle.Bold));

Specifically, I am confused as to what the last parameter does. As I understand it, the following should do a bitwise comparison, and return the result: 

controlToUpdate.Font.Style ^ (controlToUpdate.Font.Style & FontStyle.Bold)

..but what does that mean in this situation? What are the possible results, that may be passed as the third parameter to new Font(...), and how can I rewrite this more clearly, while keeping with the intent of the original programmer?

Sidenote: Is this a normal way to do things when working with Windows Forms? I'm a little new in that area - is the intent here obvious to coders more experienced in this field?

Upvotes: 1

Views: 204

Answers (3)

Tigran
Tigran

Reputation: 62265

If this is normal, or not, depends on what was the reason of doing this, but basically this means: 

controlToUpdate.Font.Style ^ (controlToUpdate.Font.Style & FontStyle.Bold)

(controlToUpdate.Font.Style & FontStyle.Bold): bitwise AND, so it's enough having a 0, and it will return 0 (can think about this like a multiplication) 

1 0 = 0
0 1 = 0 
1 1 = 1 
0 0 = 0

So (controlToUpdate.Font.Style & FontStyle.Bold) will return true, only if controlToUpdate.Font.Style is Bold too

after we have

controlToUpdate.Font.Style ^: a bitwise XOR operator, where the same value gives 0

1 1  = 0 
0 0  = 0
1 0  = 1 
0 1 =  1

So, considering previous output (say it's a Bold) the result will be false or 0,so Regular font style.

In practise, this is a way to enforse Regular type font, independently of the real style set on the control.

Upvotes: 2

Marc Gravell
Marc Gravell

Reputation: 1064114

controlToUpdate.Font.Style & FontStyle.Bold

performs an "and" to return FontStyle.Bold if the style (controlToUpdate.Font.Style) includes bold, and 0 if the style does not include bold: basically, it gets just the "bold" bit. 

controlToUpdate.Font.Style ^ (controlToUpdate.Font.Style & FontStyle.Bold)

performs an "xor"; if the bold bit was set, it removes it; if the bold bit was not set, then it does nothing (since "xor" with 0 is a no-op).

So basically, that complicated code just removes the "bold" bit (if it is set). A simpler implementation would have been:

controlToUpdate.Font.Style & ~FontStyle.Bold

How that works: here, the ~FontStyle.Bold inverts all the bits; FontStyle.Bold is 1:

000....000001

so ~FontStyle.Bold is:

111...1111110

we then "and" that with our current style, which means it keeps all of the old style except the bold-bit.

Upvotes: 3

Oded
Oded

Reputation: 499352

The FontStyle enumeration is a Flags enumeration, making them into bitmaps.

Using bitwise operators, allows you to find out which flags are "on", "off" and of course, change them.

This is very common - for example, to find out if the style is bold or italic, you would use:

FontStyle.Bold | FontStyle.Italic

Upvotes: 2

Related Questions