Hi,
Disregard my previous message, I got your idea =) When we are shifting
right then left, then all bits except 7th will be 0 and andmask may be
any, while it contains 0x80. Yes everything is ok here, thanks
---
With best regards, Konstantin
On Tue, Jul 23, 2013 at 5:05 PM, Konstantin Vladimiro
Hi,
But this is awfully wrong. In the general case (value >> 2) & 0xff !=
(value >> 2) & 0x80
Take value to be 0x3ff for example. Then 0xff != 0x80 itself. This
leads to wrong result.
---
With best regards, Konstantin
On Tue, Jul 23, 2013 at 4:57 PM, David Given wrote:
> Konstantin Vladimirov
Konstantin Vladimirov wrote:
[...]
> x = (y & ~(1 << 7)) | (((value >> 9) & 1) << 7);
[...]
> x = y & 4294967167 | (value >> 9) << 7 & 255; <- WAT?
((value >> 9) & 1) << 7
== ((value >> 9) << 7) & (1 << 7)
== ((value >> 9) << 7) & 0x80
== ((value >> 9) << 7) & 0xff
...I think.
That l
Hi,
Simple code:
void somehowuse(unsigned value);
void
foo( unsigned value,
unsigned bytemask,
unsigned psw,
unsigned y)
{
unsigned x;
psw = (psw & ~(1 << 10)) | (((value >> 9) & 1) << 10);
x = (y & ~(1 << 7)) | (((value >> 9) & 1) << 7);
somehowuse(x);
somehowuse(psw);
}
Compile to assemble
