the meaning of " >> "

Hello .
I dont know if it is the board that I can send this question or not , sorry if I am wrong .
I have a source code , but I am not sure that if it is in assembly or some another programming language .

do u know what is the meaning of " >> " or " << " ?

I know in ' C ' it is shift but in these codes ....?

here are some parts of the code :
{IPchecksum & IPTypeOfService & IPLength & IPTypeOfService are variable name}

Code:

---

.set IPchecksum = IPchecksum + (IPVersionLength<<8) + IPTypeOfService
.set IPchecksum = IPchecksum + (IPTypeOfService<<8) + IPLength

---

Code:

---

.MACRO UDPSendByte                ;send to UDPport byte (given as parameter1)
        .if ((@0 >> 0)&1)
          out    UDPPort,RegForUDP1
          out    UDPPort,RegForUDP0
        .else
          out    UDPPort,RegForUDP0
          out    UDPPort,RegForUDP1
        .endif

---

Code:

---

in      temp0, UDPPort                  ;load port state
        cbr    temp0, (1<<TxPlusPin)|(1<<TxMinusPin)          ;clear TX+ and TX-
        mov        temp1,temp0                        ;store the default state for recovering
        out    UDPPort,temp1                        ;back to port as default state - no voltage between TX pins

        sbr    temp0, (1<<TxMinusPin)          ;set TX- (TX+ was cleared)
        mov    RegForUDP0, temp0              ;and store as log0 state

---

and also what is the meaning of '@' , is it like pointers in 'c' ??
are they assembly codes ?

thanks

In C++, they are insertion (<<) and extraction (>>) operators. insertion operators are commonly used to put things onto a stream, and extaction operators to take them off of the stream. for example:

Code:

---

std::cout<<"Hello World";

---

puts "Hello World" onto the standard output stream, and

Code:

---

std::cin>>integer;

---

takes something (presumably an int) off the stream.

as for what langauge that code is, I can't help you there. I don't believe it's C++ judging from the comments though.

That looks like assembler to me... but it has interesting bits of a higer level langauge in it - .if .else .endif etc. As far as i can tell that is propiorty code going thats meant to be run on a dsp on a level 3 network switch or someting similar.

But in your case the << and >> will be bit shifters.

Code:

---

.set IPchecksum = IPchecksum + (IPVersionLength<<8) + IPTypeOfService
.set IPchecksum = IPchecksum + (IPTypeOfService<<8) + IPLength

---

This bit mearly turns two 8 bit words into a single 16 bit word, by moving IPVersionLength bits 8 bits up (same as multiplying by 256, but faster).

Code:

---

.MACRO UDPSendByte                ;send to UDPport byte (given as parameter1)
        .if ((@0 >> 0)&1)
          out    UDPPort,RegForUDP1
          out    UDPPort,RegForUDP0
        .else
          out    UDPPort,RegForUDP0
          out    UDPPort,RegForUDP1
        .endif

---

I'm a bit flumouxed by this bit. I think that the @0 is a pin number on the chip, but why bit shift it by 0 places? it does nothing. But it seems to signfy most/least signficant byte first.

Code:

---

in      temp0, UDPPort                  ;load port state
        cbr    temp0, (1<<TxPlusPin)|(1<<TxMinusPin)          ;clear TX+ and TX-
        mov        temp1,temp0                        ;store the default state for recovering
        out    UDPPort,temp1                        ;back to port as default state - no voltage between TX pins

        sbr    temp0, (1<<TxMinusPin)          ;set TX- (TX+ was cleared)
        mov    RegForUDP0, temp0              ;and store as log0 state

---

Again this isn't how I would do it. Presumably both TxMinusPin and TxPlusPin are #defined somewhere, so why not just use their straight values instead of confusing matters with bit shifting?

The << and >> also work as bit shifters in C and C++.
A<<1 == A * 2
and
A>>1 == A /2 (rounded down).
You can also get some weird changing of the sign.

Try this proggie out, with an input of 65535.

Code:

---

#include <stdio.h>
#include <iostream>

int main()
{
  int d, c;
  cout << "Please enter a number between -2,147,000,000 and 2,147,000,000" << endl;
  cin >> c;
  cout << "Heres the number bit shifted up(l col), down(m col) and each bit (LSB first, r col)" << endl;

  for(d=c; c|d;c<<=1, d>>=1)
    printf("\n%13d%13d%2d", c, d, d&1);

  return 0;
}

---

tnx alot for helping .
I found some tutorial on internet and used it for the meaning of the operators .
it is AVR microcontroler assembler .
u r right : >> and << are shift operators and
@ is used for parameters of macro .
now I know the operator operations but dont know the logic behind some sentences
for example why the programmer has written these ?

Code:

---

.if ((@0 >> 0)&1)

---

or in some another part for declaring 'IPLength' he used this (UDPSendByte is type of it ):

Code:

---

UDPSendByte  (IPLength >> 8)
UDPSendByte  (IPLength >> 0)

---

can any one tell me why in 2 lines ? and with 2 shifts ?

Quote:

---

Originally Posted by arian

now I know the operator operations but dont know the logic behind some sentences
for example why the programmer has written these ?

Code:

---

.if ((@0 >> 0)&1)

---

or in some another part for declaring 'IPLength' he used this (UDPSendByte is type of it ):

Code:

---

UDPSendByte  (IPLength >> 8)
UDPSendByte  (IPLength >> 0)

---

can any one tell me why in 2 lines ? and with 2 shifts ?

---

Why he put extraneous bits in I have no idea. But the last two lines of code first send the IPLenth in two parts - first the high byte and then the low byte (presumably its a 16bit number).

you said : " first the high byte and then the low byte " .
sorry but I dont undrestand , can you plz explain more ?

Say we have a 16-bit integer like so:
1010111101101001

And let's call it i. Then,
i >> 8 is 0000000010101111
and
i >> 0 is 1010111101101001

What UDPSendByte is probably doing is casting it to a byte, truncating the first 8 bits, which means that those two lines are sending the following bytes respectively,
10101111
01101001

Does that make sense?

yes tnx

You're welcome.