working with binaries

[NeMewSys]

I'm learning how to work with binaries in C.

I tested this code:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
  int x=3;
  x = x &#187; 1;
  printf("lol: &#37;d",x);
  system("PAUSE");	
  return 0;
}

But u get the next error in the line "x = x » 1;"

stray '\187' in program

I was expecting that, if x = 3 (11) that x»1 would make x = 1 (01).
What am i doing wrong here?

Edit:
If i have x = ~x, then x=3 (11) will be -4 and not 0 (00), why?

[matsp]

You should use two > signs, not the single character that looks like two at once.

--
Mats

[Elysia]

Edit:
If i have x = ~x, then x=3 (11) will be -4 and not 0 (00), why?

Because it inverts all bits.
3 = 0000 0011
So ~3 is 1111 1100

[QuantumPete]

You're using French-style closing speech marks. You need two greater-than symbols.

QuantumPete

[NeMewSys]

Because it inverts all bits.
3 = 0000 0011
So ~3 is 1111 1100

Wait, i got confused now about a thing, in your example you show the int x with 8 bits

0000 0000 = 8bits

Does int has 8 bits or 16 bits?
wich bit is the "negative flag" of the int?

Ps. I loled when you told me it was >> not », xD funny errors.

Thanks!

[Elysia]

Obviously, int can be 16 bits or 32 bits depending on the system, but the same theory applies there. All 0s become 1s and all 1s become 0s.
And the sign bit is the highest bit (bit 16 or 32).

16 bits (3): 0000 0000 0000 0011
16 bits (~3): 1111 1111 1111 1100
32 bits (3): 0000 0000 0000 0000 0000 0000 0000 0011
32 bits (~3): 1111 1111 1111 1111 1111 1111 1111 1100

[NeMewSys]

One last thing, in all the examples where i see that the programmer compares the last bit of X with 1, they do it this way:


if ( x & 01) ...
in php i used to do this way if (x & 1)...
are they the same?

Thanks!

[laserlight]

Yes, they are the same, except that 01 is in octal.

[Elysia]

1 = decimal (base 10)
01 = octal (base 8)
0x1 = hexdecimal (base 16)

When working with binary, we usually use hexdecimal because of its close relationship with binary (base 2).

[NeMewSys]

Every number that begins with a zero is octal by default?

[tabstop]

By definition (C's definition), anyway. If you try to type "08" in a program, bad things will happen.

[Elysia]

Yes, it's quite silly, really, but that's the way it was done.

[NeMewSys]

By definition (C's definition), anyway. If you try to type "08" in a program, bad things will happen.

that was my fear.

[Elysia]

Well, 08 is illegal because 8 cannot be represented in base 8, so it would just give a compile error.

[brewbuck]

Yes, it's quite silly, really, but that's the way it was done.

Why is it silly?