How to assign values to a UINT8 variable?

[boki]

Dear All,
If I have a UINT8 variable, how can I assign it as my expected value?

UINT8 a;
a=(00001001);

?
Thank you very much.

Best regards,
Boki.

[John Harrison]

"boki" <bo******@ms21.hinet.net> wrote in message
news:4c**************************@posting.google.c om...

Dear All,
If I have a UINT8 variable, how can I assign it as my expected value?

UINT8 a;
a=(00001001);

?

Use hexadecimal

a = 0x09;

C++ does not support binary.

john

[Tom Widmer]

On 10 Sep 2004 04:38:18 -0700, bo******@ms21.hinet.net (boki) wrote:

Dear All,
If I have a UINT8 variable, how can I assign it as my expected value?

UINT8 a;
The standard C++ type is unsigned char. It isn't necessarily 8 bits,
but it is on common platforms such as Windows and Linux.
a=(00001001);

?

C++ does not have binary literals. Instead the best option is to use
hex literals:

//hex
a = 0x9;
//or octal
a = 011;
//or decimal
a = 9;

Tom

[boki]

John and Tom, thank you two very much.

Best regards,
Boki.
Tom Widmer <to********@hotmail.com> wrote in message news:<5b********************************@4ax.com>. ..

On 10 Sep 2004 04:38:18 -0700, bo******@ms21.hinet.net (boki) wrote:

Dear All,
If I have a UINT8 variable, how can I assign it as my expected value?

UINT8 a;

The standard C++ type is unsigned char. It isn't necessarily 8 bits,
but it is on common platforms such as Windows and Linux.

a=(00001001);

?

C++ does not have binary literals. Instead the best option is to use
hex literals:

//hex
a = 0x9;
//or octal
a = 011;
//or decimal
a = 9;

Tom