This is the code:
int main()
{
char c = 0x00;
//c |= 0x0A;
c |= 0xA0;
while(c != 0x00)
{
cout << (c & 1) << endl;
c = (c >> 1);
}
}
Why does this code work when I do or with 0X0A and not with 0xA0, as a number 0xA0 is to big to fit a signed char but why am I not allowed to set the bits for 0xA0?
When I print the loop it never breaks and it only print ones? How come?
It is because of the sign extension of when performing a right shift.
0xA0 <=> 10100000 binary (and is a negative number because the MSB is set to 1)
(0xA0 >> 1) <=> 11010000
replace char c by unsigned char c
or mask the MSB after the shift.
int main()
{
char c = 0x00;
//c |= 0x0A;
c |= 0xA0;
while(c != 0x00)
{
cout << (c & 1) << endl;
c = (c >> 1) & 0x7f;
}
}
Note that char is usually a signed (range -128..127) but some c/C++ compiler define char as unsigned (AIX xlC compiler is known case).
(0xA0 >> 1) <=> 11010000 isn't necessarily right - it's implementation defined.
Your implementation must add a 1 in the most significant bit when right-shifting a negative number (the operator's behaviour is implementation defined, so usually depends on which CPU instructions are available).
0xA0 = 1010 0000 binary
The first time you sample c & 1 - the least significant bit - it's actually a 0. Then...
c = (c >> 1)
...shifts this to 1101 0000, then 1110 1000, 1111 0111, 1111 1011, 1111 1101, 1111 1110, 1111 1111 - which is then stable and never changes further. Hence, you keep printing 1s and never satisfy the termination condition of c == 0.
When I print the loop it never breaks and it only print ones? How come?
Well, it will have printed a few 0s at the very start, but your screen's probably scrolled past them in a blink and thereafter you only see the 1s.
c & 1).