 ## 3 replies

Last post Oct 07, 2019 05:47 AM by Yongqing Yu

• ### How to convert this VB code to C#?

Oct 04, 2019 06:17 PM|Peter Cong|LINK

I have the following VB codes:

Dim u as Integer

Dim a as Integer = &H2  //need help here

.........

u = some integer number

Dim D as Boolean = True

D = CBool(a And u)   //need help here

If Not D then

return true

else

return false

end if

,,

can anybody help me to convert these codes to C#?

Thanks

• ### Re: How to convert this VB code to C#?

Looks like homework...

`Dim a as Integer = &H2  //need help here`

You could simply run the code through the Visual Studio debugger.  Anyway, the &H means Hex and 2 in Hex is same as 2 decimal.

`int a = 2;`

Or in Hex notation

`int a = 0x2;`

This is binary logic.

`D = CBool(a And u)   //need help here`

The idea is to determine if the variable u has bit 2 set.

The C# code is below.

```int a = 0x02;
int u = 0x0A;

bool result = (a & u) > 0;
Console.WriteLine(result);```

The result is true because bit two is set in u.

• ### Re: How to convert this VB code to C#?

Oct 04, 2019 10:17 PM|Peter Cong|LINK

#### mgebhard

Looks like homework...

`Dim a as Integer = &H2  //need help here`

You could simply run the code through the Visual Studio debugger.  Anyway, the &H means Hex and 2 in Hex is same as 2 decimal.

`int a = 2;`

Or in Hex notation

`int a = 0x2;`

This is binary logic.

`D = CBool(a And u)   //need help here`

The idea is to determine if the variable u has bit 2 set.

The C# code is below.

```int a = 0x02;
int u = 0x0A;

bool result = (a & u) > 0;
Console.WriteLine(result);```

The result is true because bit two is set in u.

Thanks a lot for your quick help, I will try it next Monday and let you then,

Have a great weekend,

• ### Re: How to convert this VB code to C#?

Oct 07, 2019 05:47 AM|Yongqing Yu|LINK

Hi Peter,

According to your VB code, you can convert these code to c# as follows:

```int u;
int a = 0x2;
//...
u = 12;//some integer number
bool D = true;
D = Convert.ToBoolean(a & u);
if(!D)
{
return true;
}
else
{
return false;
}```

Best Regards,

YongQing.

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